Automata, Languages and Machines. Volume B. (Pure and Applied Mathematics)

Y wpx.

-p

X , then

Proof. Let s E Sx. Since X

a

PROPOSITION 7.2.

Proof. Let Y

If X

<tp X.


Y and Y < X , then y is an equivalence.

Since

are surjective partial functions and since Qx and Q y are finite, it follows that Qx and Q y have the same cardinality and p and y are bijections. Since py: Qx -+Qx is a permutation of Q x , there exists an exponent n > 1 such that (w)"is the identity. I t follows that p-l = y ~ ( q y ) ~ - * . Since the composition of coverings is again a covering, we obtain Y -+lX as required I COROLLARY 7.3. X

-

Y i f and only i f X

< Y and Y < X

The corollary shows that the notion of equivalence of ts's is the one appropriate to consider in conjunction with the preorder relation In the sequel, our main concern will be with equivalence classes of ts's; an individual ts will be, most of the time, just a representative of its equivalence class. The set of all these equivalence classes will be denoted by TS. This set is (partially) ordered by the relation In TS we have

<.

<.

1. Transformation Semigroups

18

the subsets TM, CTS, CTM, and TG defined respectively by the tm’s, cts’s, ctm’s, and tg’s. More precisely, CTS consists of those equivalence classes of ts’s containing at least one cts. T h e same remark applies to the sets CTM and TG. For TM this remark is irrelevant in view of Exercise 7.1. That the notions of equivalence and isomorphism are really distinct is shown in Exercise 7.2.

-Y

EXERCISE 7.1.

Show that

if X

EXERCISE 7.2,

Let X

(Q, S ) be a ts and let X ’ = (Q, S’) with

=

and X is a tm, then Y also is a tm.

S’ = {s‘ I s‘ E PF(Q), s’ c s for some s E S } Show that X

-

X . Show that X

M

X’ holds iff S

EXERCISE 7.3. Show that X mp Y

12X

= S‘.

Y and both 9 and 9-l

wP,

are proper. 8. Join, Sum, and Direct Product

Let Q , , Q2 be finite sets and let

be their disjoint union. T h e semigroups PF(Q,) and PF(Q,) will both be regarded as subsemigroups of PF(Q), PF(Q,) consisting of all partial functions s: Q -+ Q such that Q1s c Q1, Q2s = 0 . Similarly for PF(Q,). Now let X,= (Q,, S,), X 2 = (Q2, S,) be ts’s; Then S, and S, may be regarded as subsemigroups of PF(Q).We define the join of X I and X, to be

with Q = Q1 u Q, ind S the subsemigroup of PF(Q) generated by Sl U S 2 . Explicitly we have S=

S, Sl s 2

u S, u 8

if if if

S, # 0 # S, S, = 0

S, = 0

8. Join, Sum, and Direct Product

19

The following relations are clear

Xi c X,vX,

for i = 1 , 2

x,vx,=x,vx, ( X , v X,) v x, = x, v ( X , v X 3 ) xvo=x X,

< Y,

and X ,

< Y,

imply

X,vX,

< Y,v

Y,

xl*v x,. c ( X , v X,). X , V X , c x,vx, Given s1 E PF(Q,), s, setting

E

PF(Q,), we define (s,, s,)

E

PF(Ql v Q2)by

This permits us to regard PF(Q,)xPF(Q,) as a subsemigroup of

PF(Qi u 9 2 ) . Given ts’s X , = (Q,, S , ) and X , X , and X , to be

Xi

=

(Q,, S , ) we define the sum of

+ Xz = (Q,S )

with

Q = Qi v Q z S,xS,

if S, # 0 # S, if S, = 0 if S, = 0

T h e sum has the following formal properties

I. Transformation Semigrou ps

20

If X , and X , are tm's, then so is X , + X,. If X , and X , are complete, then so is X , + X,. If X , and X , are tg's, then so is X , X,.

+

+ x,

+

There seems to be no clear connection between R, and X , X , . The underlying set for both X , v X , and X , X , is the disjoint union Q1u Q2. We now define the direct product X l x X , with Q l x Q 2 as underlying set. We first consider the case when both X , and X,. are complete. I n this case we set

+

X i x X 2 = (QixQz,S i x S 2 )

(8.1)

In the case when X , or X , or both are not complete we define

The complement of Q1x Q2 in $3; consequently we have

x QZcis a right

ideal in X , x X , and

XlXX2 = ( Q l X Q 2 , W where S' is a quotient semigroup of S, x S,. Each defines the partial function s' defined by (41 9

42)s' =

(sl,

s,) E S , x S ,

(41s19 q z s 2 )

with the result interpreted as 0 if either qlsl = 0 or q2sz = 0. See Exercise 8.1 for a fuller description of S'. The direct product has the following formal properties

XlXX,wx2xx,

x,x ( X ,x X,) = ( X Ix X,) x x, XXl'=X x x o =0 X,

< Y,

and X ,

< Yz

imply

( X ,x X,). c X,'

X l x X , < Y,x Y ,

x x;

x1xx2

XlXX, c ( X ,x x,). < X1"X x , c

8. Join, Sum, and Direct Product

21

If X , and X , are tm's, then so is X I x X,. If X , and X , are complete, then so is X , x X , . If X , and X , are tg's, then so is X , x X , .

XX(Y,V Y , ) = ( X , x Y , ) v ( X x Y , ) XX(Y1

+ YZ) = ( X X YI) + ( X X Y2)

From these rules others may be derived. For instance

X,<X,xX,

if

<

1'<X2.

Indeed, X , = X , x 1' X , x X,. and x are associative and I n view of the fact that the operators v , commutative, we shall use the following notation

v

+,

xi=x,v ... v x ,

lsisn

c

xi=x,+ . . . + x n

lsicn

n x,

lsicn

=x,x

. .. x x ,

We shall also use the notation X ( k )for the k-fold direct product X X . . . X X with X " ) = X and X'O)= 1'. We also note that if X I and X , are represented by (Ql,Zl) and (929 Cd, then

X,v X, X , X,

is represented by

+

(Q1U Q2,ZlU 2 2 )

is represented by

(Q, u Q , , ZlXZ2)

X,x X,

is represented by

(Q,x Q , , Zlx C,)

Of course the unions Q1 u Q 2 , Z, u Z, are assumed to be disjoint. Our notation X X Y for the direct product of ts's is not altogether consistent with the convention identifying a semigroup S with the ts (S', s). Indeed if S and T are semigroups, then we have the inclusion ((SX T)., S X 7')c ( S ' X T',SX 7')= (S, S)X (T', T )

T h e inclusion cannot be replaced by an equality except when both S and T are monoids or if S = 1 or if T = 1. Thus the product S X T when S and T are viewed as semigroups is in general different from the same product viewed as a product of ts's.

I. Transformation Semigroups

22 EXERCISE 8.1.

t2

s2,

Let Xi= (Qi, Si)for i = 1, 2. DeJine for sl, t , E S , ,

s, (s19 s2)

-

( 2 , 9 tz)

provided (41 9

4 2 ) ( s 1 9 s2)

=

(41 ?

QZ)(tl,tz)

for all ( q l , q2) E Q, x Q 2 . Show that if s, # 8 and s, # 8, then (s, , s2) (tl , t,) implies s, = t, , s2 = t,. Thus, in particular, i f neither S, nor S , contains 8, then S,X S, acts faithfully on Q1x Q,.

-

EXERCISE 8.2.

Let n: Q, x Q, -+ Q, be the projection (q, , qZ)z= qz.

Prove

XIXX, 4

x 2

n-'

State necessary and sufficient conditions for x 2

<-&xx,

to hold. EXERCISE 8.3.

Show that

x+ Show that

if

1'

EXERCISE 8.4.

Y < ( X + l ' ) X ( Y + 1')

< X , then x+

Y <Xx(Y

+ 1')

Show that

<2 ( k ) Q' < 2"k) p< Q

Z.(k)

provided card Q 5 2k. 9. Some Simple Inequalities

We recall that Sxdenotes the monoid obtained from the semigroup S by adjoining a unit element, regardless whether S has one or not. PROPOSITION 9.1.

For any semigroup S (SI,

S)

< (S, S )x c

23

9. Some Simple Inequalities Proof. We denote by 1 the unit element of

element of

SI.

S' and by e the unit

Define rp: S X 2 - S '

if i = O , s E S if i = 1, s = 1 otherwise

s

e

0

For each transformation t of (SI, S) consider the transformation (t, u ) where (T is the unique transformation of C. Then for all s E S' (s, i)p,t c st = (st, 0)rp

so that (t, u ) covers t

Let

=

(s, i)(t, a)p,

I

X = (8,S ) be a ts. For any

qE

Q, the ideal

will be called the principal ideal generated by q. It is the smallest right ideal containing q. PROPOSITION 9.2. Let X = ( Q , S ) and let P be a subset of Q such that

PuPS=Q Then

c X I (P)

(9.1)

x<

(9.2)

X < p ' x ( S u lQ,S)

PEP

Let P = ( p , , . . . ,p , } . Since we shall have to deal with the disjoint union R = ( p l ) u . . . u (p,) of sets which are in general nondisjoint subsets of Q, we shall denote the elements of R as pairs (i, q ) with 1 5 i 5 n, q E (pi).Define the surjective function Proof.

9: R + Q

(i, q)q = q

for

1 5i I n, q

E

(pi)

For each s E S, let si be the transformation of X I ( p i ) defined by S. Set t = (sl, . . . ,sn). Then t is transformation of the ts ZX I (pi).

1. Transformation Semigroups

24

Further,

so that t covers s. This proves (9.1). Next consider the surjective function 9: P x ( S

u 1)-Q

(P,1)v = P (P,s)v = Ps For each s E S we then have

(P,1)PS = P S ( p , djvs = P S I S

(P,1 ) ( 1 9 (p,

s)v = (P, = P S s)v = ( p , s's)v = P S I S

Thus the transformation ( 1 , s) of P' x (S u 1 , , S ) covers s Since ( S u l,, S ) c

(SI,S), Propositions 9.1

I

and 9.2 combine to give

COROLLARY 9.3.

X
For any ts X

=

(Q,S )

X
Let X = (9,S ) and let P be a subset of Q such

that

PS=Q Then

x
v:

Then each

J E

PxS'+Q

S is covered by (1, s)

I

9. Some Simple Inequalities COROLLARY 9.6.

25

Let X

=

(Q, S ) be a ts. If Q S = Q, then

X
I

COROLLARY 9.7. If X = (Q, S ) is a tm, then

X
1

The difference between Proposition 9.5 and statement (9.2) of Proposition 9.2 is very subtle. In the latter (S u l,, S ) appears, while ProposiS). The difference shows up only when tion 9.5 uses S, i.e. the ts (S, S is a monoid with a unit element different from 1,. PROPOSITION 9.8.

Let X = (Q, S ) be a complete ts. Then

s < X'k' with k = card Q. We recall that X ( k )is the k-fold direct product X X . . . x X. Proof. QX

Let Q = { q l , . . . ,q k } and Qk be the k-fold Cartesian product Define the function

. . . x Q.

i: S + Q k si = (qlS,

li = ( q l ,

* * * 9

qks)

. . . , qk)

if

s' # S

Clearly i is injective. For each s E S , t = (s, . . . ,s) is a transformation of X ' k )and xit = mi. Thus i : S -+ X ( k is ) an injection and consequently

s < X'k' 1 A ts X

=

(Q, S ) is called monogenic if Q = p S for some p

PROPOSITION 9.9 If X = (Q, S ) is a monogenic ts, then

x<s If, further, X is complete, then

x < s <X'k' with k = card Q.

E

Q.

1. Transformation Semigroups

26

The first part follows from Proposition 9.5, while the second one follows from Proposition 9.8 I We recall that 77,denotes the monoid with elements u, 1, with 1 as unit element and with uu = u. Note that viewed as a tm, U , is isomorphic with C’. Also observe that U , M 1 I where 1 is the single element monoid. EXERCISE 9.1.

Show that if S is a monoid, then there exists an injective

morphism SI +s

x u,

EXERCISE 9.2. Let S be a non-empty semigroup and let So be the semigroup obtained from S by adjoining a zero regardless whether S has a zero or not. Thus So= S u 0 (disjoint union) and SO = 0 = OS for all s E So. Show that there exists then a surjective morphism

s x u, + so EXERCISE 9.3.

inequality S’

X

Construct an example of a semigroup S for which the

< S X U , fails.

EXERCISE 9.4. Show that i f X Yl or X Y,.

<

<

< Y , + Y , and tjC X is monogenic, then < n’x Y and if X

EXERCISE 9.5.

Show that i f X

EXERCISE 9.6.

Show that the conclusion of Proposition 9.8 fails for

x < Y.

x = (1, e).

is monogenic, then

10. The Wreath Product

be tds. Our objective is to define a new product X o Y called the wreath product. It is “larger” than the direct product X X Y , and contrary to the direct product, it is no longer symmetric in X and Y . As in the case of the direct product we begin the description with the case when X and Y are complete.

10. The Wreath Product

27

We consider the set W = SPxT

(10.1)

where S p is the set of all functions j : P

X

(10.2)

0

Y

=

---f

S. T h e wreath product is then

(Qx P, W )

with the action of W on Q x P defined by the formula

(4,P ) ( f , t ) = ( d P f 1, P t )

(10.3)

Clearly if (4,P)( f,t ) = (q, p ) ( f ', t') for all (9,p ) E Q x P then (f,t ) = ( f ', t'). Thus W may be regarded as a subset of F ( Q x P), the monoid of all functions Q X P -+ Q X P. We show that W is closed under composition. For this we consider another element (g,u ) E W. Acting with (g, u ) on the right-hand side of (10.3) we obtain

(!7(Pf>((Pt)g)!Ptu) Thus setting

Ph = (Pf)((PM

(10.4) we obtain h

E

S p and

Thus W is closed under composition. T h e formulae above show that we may define the composition intrinsically in W by setting (10.5)

with h given by (10.4). The associativity of the composition in W may be verified by a simple computation. As in the case of the direct product the passage from the case when X and Y are assumed to be complete to the general case is achieved by completion. Thus we define

x

0

Y

= (X" 0

Y")I Q X P

We observe that the complement of Q x P in Q"xP" is a right ideal. Consequently X o Y = (QxP, V )

28

1. Transformation Semigroups

where V is a quotient semigroup of the semigroup

W = S P c xT = S x c x p c In particular, SXOY

<SXcoYc

If (f,t) E W and ( q , p ) E Q X P, then formula (10.3) holds provided we interpret the result as 0 if either p t = 0 in X or q ( p f ) = 0 in Y. Consequently X o Y is represented by (QxP, S p c x T ) with action given by (10.3). T h e wreath product has the following formal properties

xxY

Y X 0 ( Y 0 2) = ( X 0 Y ) 0 2 c x o

Xol'=X

1'oY= Y

xoo=o oox=o X<X'

Y

and

< Y' Y)'

c

X'

XOY

c

x

(X

0

X

imply 0

0

o

Y <X'

o

Y'

Y'

B

(XOY)"<X"OYC If X and Y are tm's, then so is X o Y. If X and Y are complete, then so is X o Y

X (XI

0

(Y1v Y 2 )c ( X

+ Xz)

0

0

Y,) v ( X

y c (Xl 0 Y )

0

+ (X,

Yz) O

Y)

( X l O X 2 ) x (Yl O Y2) c ( X l X Y , ) O ( X 2 x Y2P T h e associative law is best established by introducing a triple wreath product X 0 Y 0 2. More generally, given ts's Xi= (Qi , Si) for 1 5 i 5 n, the n-fold wreath product

x=x*o ... o x ,

10. The Wreath Product

29

(the reversed numbering simplifies the notation) is defined as follows. T h e underlying set is

Q = Q,,x . . . xQ1 We denote by W the set of all n-tuples

f =(fn,

.

a

.

>f*)

where fi:

. . xQ:

Q4-1~.

-+

Si

is a function. If i = 1, then Q:-l x . . . x QIc is interpreted as 1 and f, is simply an element of S1.Each f E W defines a partial function

f:Q+Q by setting for q = (qn, . . . , q l )

X is then the ts represented by (Q, W ) . The rule relating

< and

o

requires some more detail.

PROPOSITION 10.1. If

X<X’,

Y
9

then

x

0

Y

< X’

0

Y’

rl

where

Proof. Given a transformation (f,t ) of X o Y , choose t‘ E T‘ which covers t and choose f ‘: P” + S‘ so that for each p‘ E P , the transforma-

tion p ‘ f ‘ E S‘ covers (p‘y)f whenever p‘y # 0. Then

( 4 ’ , P ’ ) d f ?t ) = (Q’%P’Y)(f, t ) =

((dV)(P’YIf,P ’ Y 4

= (4YPi’)Yh P‘t‘Y) = (4’(Plf 7

9

P‘t’h

= ( 4 ’ , P ‘ ) ( f ’9 t’)V

I

30

x

1. Transformation Semigroups PROPOSITION 10.2. If X and Y are transformation groups then so is 0

Y.

Proof. Let (f,t ) be a transformation of Define g : P S by setting

XoY

with

f : P -+ S.

---f

Pg

=

W - ' l f I-'

Then

k,P N f , t)(g,t-l) = ( d P f 1, pt)(g, t-l) =

M P f )(Pf )-l, ptt-')

= (g,P)

Thus (g, t-l) is the inverse of ( f , t )

In Section 7 we denoted by TS the set of all equivalence classes of ts's. TS had then the structure of a (partially) ordered set. A glance at the list of formal rules shows that using the wreath product X o Y as a multiplication, TS is a monoid with the class of 1' as unit element. Further, the order and the multiplication are compatible (ie. x < y implies xz < y z and zx xy), so that TS becomes an ordered monoid. Within these ordered monoids we have the ordered submonoids CTS, TM, CTM, and TG.

<

EXERCISE 10.1. With the notations established at the beginning of this section consider ( f , t ) E W and dejine

If,tl= { P I P E P , P t = % Pf=O)

and p t = pt'

and

pf = p f '

for all

P @ If, t I

Deduce that if Y is complete and 8 g S , then W acts faithfully on Q x P. EXERCISE 10.2. Show that

10. The Wreath Product

31

where the exponent ( k )indicates the k-fold direct product while k indicates the k-fold wreath product. EXERCISE 10.3. Let X and Y be ts’s, let

V be a subsemigroup of Sxoy

and let W be the image of V under the natural projection n: Sxor -+ S, given by (f,s)n = s. Show that

EXERCISE 10.4. Let X Xv Y o Z where X is a cts while Y and Z are ts’s. Show that there is a complete ts Z’ c Z such that X Y o Z‘ where y’ is the restriction of y .

<,

EXERCISE 10.5. Let

?t:

Q X xQ y -+ Q y be the projection. Prove that XOY-qY n-’

State necessary and suficient conditions for Y<XOY n

to hold. EXERCISE 10.6. Consider the set N of all integers k

2 0 with the natural

order and with multiplication of integers as multiplication. Thus N is an ordered monoid. Show that the function k + k is a morphism N + TS of ordered monoids. Formulate analogous statements for CTS, TM, CTM, and TG. EXERCISE 10.7. Let Q be a right 2-module, P a right r-module, and let X and Y be the transformation monoids represented by (Q, 2) and ( P , r).Consider the finite alphabet D which is the disjoint union of P x Z and r. Convert Q x P into a right Q-module

Show that ( Q x P, Q) represents X o Y . Why is the consideration of transformation monoids essential for this argument ?

1. Transform.ation Semigroups

32

EXERCISE 10.8. Let X and Y be ts's represented by (Q,S ) and (P, T ) where S and T are semigroups, Q is an S-module and P is a T-module.

Show that X

o

Y is represented by ( Q x P, W ) with W = S px T.

EXERCISE 10.9.

Show that for any ts's X and Y

Sxeoye

= sxoye

EXERCISE 10.10. Show that the smallest elements in the ordered monoids TS, TM, CTS,CTM, TG are respectively 0, o', 1, l', 1'.

Show

EXERCISE 10.11.

x

0

0 = 0 0,x= 0

XOO = O o X = EXERCISE 10.12.

0' 0

if if

S,#0 sx=0

Estabfish

FoFmFvF+F+

F

References

The new departure of this chapter is the use of partial functions in the definition of ts's. Except for that, the results of this chapter should be regarded as folklore. The seemingly slight generalization achieved by the use of partial functions will prove to be very significant.

CHAPTER

11 Decomposition Theorems

T h e main result of this chapter is the Krohn-Rhodes Decomposition Theorem for transformation semigroups. Some results dealing with transformation groups which motivate the Krohn-Rhodes Theorem are developed. A stronger version of the Krohn-Rhodes Theorem, which we call the Holonomy Decomposition Theorem, is presented in the latter half of the chapter. 1. Decompositions

An inequality (1.1)

x<x,

0

. . . ox,,

where X , XI, . . . , X, are ts’s will be called a decomposition of X . For such a decomposition to be of interest, it is clearly necessary that the XI, . . . , X, be in some way “simpler” or “smaller” than X . Sometimes one strives to arrive at decompositions (1.1) in which X , , . . . , X,, themselves are no longer usefully decomposable. The notion of decomposition as defined above is at variance with usual notions in algebra in which one expects an isomorphism instead of an inequality. I n our context, requiring an isomorphism in (1.1) is unreasonable; any ts X for which card Qx is a prime would be indecomposable. In Chapter VI we shall relate decomposition (1.1) with decompositions of sequential functions and sequential machines. This will reinforce the evidence that (1.1) is the correct type of decomposition to consider. 33

II. Decomposition Theorems

34 2. Decomposition of Groups

Let G be a finite group and H a subgroup of G. We denote by GIH the set of all right cosets Hg, with g E G. Clearly GIJI is a right G-module under the action (Hg)g' = H(gg'). T o make this action faithful we must divide G by the subgroup consisting of all elements g' such that Hgg'= Hg for all g E G . This condition is equivalent with gg' E: Hg or g' E g-lHg. Thus the subgroup is

which is the largest subgroup of H which is invariant in G . We thus obtain the tg

(GIH, GIHa) If H is an invariant subgroup, then H Q = H a n d the tg above becomes the quotient group GIN. PROPOSITION 2.1. Let G be aFnitegroup with subgroups K c H c

G.

Then (GIK, GIKQ)< ( H I K , H I K H )O (GIH, GIHG) Proof.

Let H g , , . . . , Hgn be an enumeration of the cosets of GIH.

Define q ~ :H / K x G / H -+ G / K

by setting

( K h , Hgi)q = Khgi This is a bijection. Given g E G and 1 5 i 5 n we have

for a unique 1 5 j 5 n. Thus

for a unique hi

E

H. Setting (Hgi)f = hi we obtain a function

2. Decomposition of Groups

COROLLARY 2.2.

35

If H is a subgroup of a Jinite group G, then G


o

( G / H ,G / H c )

This follows from Proposition 2.1 by taking K = 1 COROLLARY 2.3.

I

If H is an invariant subgroup of a Jinite group G,

then G
(2.1)

I

Let G be a Jinite group and let

G = G , ~ G , - , ~. . . ~ G l ~ G o = l

be subgroups such that GiPlis an invariant subgroup of Gi for 1 5 i 5 n. Then G (Gl/G,) 0 . . . 0 (GnIGn-1) I

<

In particular, the Jordan-Holder Theorem tells us that the sequence

(2.1) may be chosen so that all the quotients Hi = Gi/Gi-, ,

1 5i 5n

are simple (non-trivial groups). We thus obtain THEOREM 2.5.

For each Jinite group G we have the decomposition G
. . . OH,

where the Hi are simple groups and Hi

< G for

1 5 i 5 n. [

We shall call the above decomposition of G a Jordan-Hiilder decomposition.

It. Decomposition Theorems

36

EXERCISE 2.1. Let X = (Q, G ) be a tg. Show that the following conditions are equivalent

(i) There exists go E Q such that qoG = Q. (ii) qG = Q for all q E Q. (iii) For any qo, q, E Q there exists g E G such that qog = q l . (iv) There exists a subgroup H of G such that

X w (GIH, G / H G ) Property (i) expresses the fact that X is monogenic. Property (iii) asserts that X is transitive. EXERCISE 2.2. Show that G is a simple group, then HG = 1 for any proper subgroup H . Thus ( G / H ,G ) is a transitive tg. EXERCISE 2.3. Let q : G + G, be a surjective morphism of groups. Let H be a subgroup of G and let Hl = Hq. Show that Hfl= H G q and that q induces a covering

EXERCISE 2.4. Consider (QP

G)

< (f', G )

where both sides are assumed to be transitive tg's. Show that the companion relation q' is an automorphism of G. Further show that q is a function. Conclude that card Q divides card P. 3. Some Useful Decompositions PROPOSITION 3.1. Let X = (Q, S ) be a tm and let G be the maximal subgroup of S. Then S - G is a two-sided ideal in S and

X

< ( Q , S - G)'

0

G

37

3. Some Useful Decompositions

Clearly 9 is surjective since 1, E G. T o show that 9 is the required covering we must show that each s E S is covered by some pair (f,h ) with

f : G+(S-G)Ul,

~

E

G

We note that on one hand we have

while on the other hand

Thus s is covered by (f,h ) if€

gs

(3.1)

= (gf )(gh)

If s E G, we choose h = s, and gf then setting h = 1, (3.1) becomes

gv-'

for all g E G =

If X

=

E

G. If

s E S - G,

= gf

Since S - G is a two-sided ideal, gsg-l above defines f I COROLLARY 3.2.

1 for all g

E

S - G and thus the formula

(Q, G) is a tg, then

K<@oG PROPOSITION 3.3.

If X = (Q, S ) is a

ts

a n d p E Q is a state such that

then with Y

X I (Q - p ) .

1

Proof. We recall that

Thus C

=

(2, 0) with

0

C is the ts given by the diagram

=

6.

II. Decomposition Theorems

38

Since Q - p is a right ideal in X , the ts Y is represented by (Q - p , S). Define (P: ( Q - P ) X 2 - Q

For any s E S define f : 2

+S p

Of = s,

by setting h,

If = p s

= 0, then If may be defined arbitrarily. Then for any q E Q - p

If p s

Thus (f,

0)

covers s

1

As an application consider a semigroup S that is not a monoid. Then we considered two ts's. One was

s = ( S u 1, S ) and the other one was ( S , S ' ) represented by ( S , S ) . It is clear that ( S , S ' ) is a restriction of the ts ( S u 1, S ) to the subset S of S u 1 and that (S u 1)s c S for all s E S. Thus we are in the situation of Proposition 3.3 with Q = S u 1, p = 1. This yields PROPOSITION 3.4.

Let S be a non-empty semigroup that is not a monoid,

then

s<(s,s')oc 1 PROPOSITION 3.5. Let X = ( Q , S ) be a ts, L a left ideal in S, and T a subsemigroup of S such that L u T = S. Then

X with Y

=

( T u l,, T ) .

< (Q, L)'

0

4. The Krohn-Rhodes Decomposition

39

Proof. Define

y : Qx(Tu l ) + Q

by setting (9, t)fP = qt

(4, 1)v = Q

Thus ( f , x) covers s iff ts c ( t f ) ( t x )

for all

t E T

u1

Consequently we may set tf=ts,

x = i

if

S E L

tf=1,

x=s

if

SET

If s E L n T , either choice works

I

COROLLARY 3.6. Let X = ( Q , S ) be a tm, L a left ideal in S , and T

a submonoid of S such that L

u T = S.

Then

x<(Q,L)'oT

I

4. The Krohn-Rhodes Decomposition THEOREM 4.1. (Krohn-Rhodes Decomposition). Each transforrnation semigroup X = ( Q , S ) admits a decomposition

x<x,

0

...

ox,

where for each index 1 5 i 5 n either

OY

Xiis a simple group and X i < S.

II. Decomposition Theorems

40

It will be convenient to derive Theorem 4.1 from THEOREM 4.2. Each transformation semigroup X admits a decomposition

X<X,

0

= (Q, S ) with S

#0

... o x n

where for each index 1 5 i 5 n, Xi = ( Q 6 ,GJ is a transformation group with Gi S.

<

T o derive Theorem 4.1 from Theorem 4.2 observe that by Corollary

3.2

Xi < Qi' Gi 0

If card Qi 5 2k, then, by Exercise I, 10.2, Qi'is covered by the k-fold wreath product of T . T o the group G, we apply the Jordan-Holder decomposition of Theorem 2.5. If S, = 0, then X = Q F kfor some k 2 1.

<

Proof of Theorem 4.2. Given a ts X = ( 8 , S ) we consider the pair of integers X q = (card S, card Q )

Such pairs are ordered lexicographically with priority for the first coordinate. Thus (m, n ) < (m', n') if m < m' or m = m 1 and n < n'. Let

A,,

.

.

a

,

Ak

be the set of all those images Q s with s E S which are maximal (i.e. such that Qs is not a proper subset of Q t for any t E S). The argument now branches into three cases.

Case 1. k = 1 and A, = Q. Since Qs = Q for some s E S, S contains at least one permutation of Q. Thus X is a tm. Let G be the maximal subgroup of S (i.e. the group of all permutations in S). If G = S, then X is a required decomposition of X. Thus we may assume S - G # 0. By Proposition 3.1

<x

X
Y = (Q,S

- G)

4. The Krohn-Rhodes Decomposition

41

Since card(S - G) < card S we have Y q < X q and thus by recursion we may assume that

P, . . . Pn

Y<

0

0

is a decomposition satisfying the conditions of Theorem 4.2 for Y. Since Fi are tm's, we also have

P, . . . Pn

Y'<

0

0

and consequently

X
... oFnoi!7

X.

Case 2. k = 1 and A , f Q. Let p E Q - A,. Since Q s c A, for all s E S, it follows that QS c A, and thus QS c Q - p . If Q - p = 0,

<

then Q = p and thus X 1'. This is then the required decomposition of X . If Q - p # 0, then Proposition 3.3 yields

X
Y = X I (Q-P) Since Q - p is a right ideal for X , we have

(Q - P , S ' )

Y=

where S' consists of all the restrictions s' of the transformations of X. I t follows that S' is a quotient semigroup of S and thus

S'

<S

card s' 5 card S

and

This implies

y31 <Xrl and thus by recursion we may assume that

Y<

PI . . . Pn 0

0

is a decomposition satisfying the conditions of Theorem 4.2 for 9. Since P, o . . . o Y, is closed it follows that

P < P, ... 0 F,, 0

II. Decomposition Theorems

42

Thus

x < 8,0 . . . Pn0T 0

satisfies the requirements of Theorem 4.2 for X , in view of the fact that S'

< s.

Case 3.

K > 1. Define Li

= {S

I s E S, QS c A i }

for 1 5 i 4 k . Then Li is a left ideal in S and the maximality of the sets Ai implies

s = L , u ... ULk Define

T = L,

L = L,,

U

... ULk

Then T is a subsemigroup of S and L u T

=

S . Proposition 3.5 yields

X < Z o P with

Z=(Q,L), If A,

=

Y=(Tul,,T)

L , . Thus card L < card S and

Qs, then s

217 If A, = Qt, then t Consequently

$

<x17

Li for i > 1. Thus t y17

@

T and card T < card S.

<x17

By recursion we may assume that 2 and Y have decompositions

z
0

...

ozn,

Y < P,

0

. . . P, 0

as prescribed by Theorem 4.2. Since also

Z
0

...

P < 8, 0 . . . Fm

ozn,

x
0

ozno

is a required decomposition of X

I

P, o . . .

0

Y, -

6. Height, Pavings, and Holonomy

43

5. Comments on the Proof

T h e proof of the Krohn-Rhodes Theorem just concluded is rather brief, partly because the three Propositions 3.1, 3.3, and 3.5 used in the proof were stated and proved earlier. T h e proof also satisfies the requirement of being algorithmic (i.e. effective). This means that for any particular given ts X the steps of the proof provide an algorithm which, in a finite number of steps predictable in advance, will lead to a decomposition of X satisfying the requirements of Theorem 4.1 or 4.2. However, the brevity of the proof has been achieved at the expense of the economy of the corresponding algorithm. Indeed, the algorithm is very poor and even for relatively uncomplicated ts’s leads to unnecessarily long and repetitious decompositions. In Section 6-8, we shall establish a more sophisticated result called the Holonomy Decomposition Theorem. This will lead to a much more efficient way of obtaining the decomposition required by Theorem 4.2. Because of this, the treatment of examples is postponed to Section 9. 6. Height, Pavings, and Holonomy

Let X

=

(Q, S) be a ts. Given subsets a, b of Q, we shall write b s a

if either b c a or b c as for some s E S. Equivalently, b 2 a iff b c as for some s E S u 1,. If c 5 b and b 5 a , then c 5 a. Indeed, if b c US and c c bt, then c c a(st). We thus obtain a preorder in the set of all subsets of Q. From this follows an equivalence relation defined by

a-b

iff

a s b and b s a

T h e following assertions are clear:

(6.1) b 2 a implies card b 5 card a. (6.2) b a implies card b = card a.

-

-

PROPOSITION 6.1. Let a b and let b c as for some s E S. Then b = as and there exists S E S such that a = b f , and

q = qsf

for all q E a

q = qis

for all q E b

II. Decomposition Theorems

44

-

Proof. Since card b = card a, the inclusion b c as implies b = as. Condition b a implies that a c bt for some t E: S u 1,. By the argument above, we have a = bt and thus a = ast. Since st defines a permutation of a, some power (st)”, n > 1, is the identity on a. Define S = t(st)”-l. Then qsS = q for all q E a. This implies qsSs = qs and therefore q‘is = q’ for all q’ E as = b. Further b i = asS = a COROLLARY 6.2.

If a

such that

-

b, then there exist elements u, zi E S

b = QU,

u 1,

u = bfi

uzi is the identity on a

fiu is the identity on b

1

We shall be interested in the family A consisting of the following subsets of Q:

(6.3) all sets of the form Qs with (6.4) the set Q

sE

S

(6.5) the empty set 0 (6.6) all the singletons of Q. In this set A we have the preorder and the equivalence as defined above (for all subsets of 9). A height function for .the ts X is function h defined on the set A with integer values and satisfying the following conditions :

(6.7)

Oh = -1.

(6.8)

qh = 0 for all singletons q in Q.

-

(6.9) a b implies ah = bh. (6.10) b < a implies bh < a h . (6.11) If 0 5 i 5 Qh, then ah = i for some a

E

A.

The notation b < a signifies that b 5 a but that a I ’ b does not hold. Condition (6.10) asserts that two elements of A of the same height are either equivalent or incomparable. Among all the height functions for X there is a smallest one which may be defined as follows: ah = i if A contains a chain

a = ~ ~ > a ~ ... - ~> a>, > a , # O

6. Height, Pavings, and Holonomy

45

and no longer such chain exists. Necessarily then a, is a singleton. The height Xh of the ts X is the integer Qh, where h is this smallest height function. Given a E A such that card a > 1 we denote by B, the set of all b E A such that b c a, b f a and which are maximal with respect to this property. Thus if b c c c a and c E A , then either b = c or c = a. Since the singletons of Q are in A, it follows that

The set B, is called the paving of a ; the sets b E B, are called the bricks of the paving of a. Next assume that as = a for some s E S. Let A, be the set of all elements of A contained in a. Since s defines a permutation of a, it also defines an injective function A, -+A,. Since A, is finite, this function is bijective. Thus s defines a permutation of the set B,. There results a ts

where G, consists of all permutations of B, defined by the elements s E S such that as = a. It may very well be that there are no such elements s; in this case G, = 0. If, however, G, # 0, then G, is a group and Ha is a tg. In particular, this always holds if X is a tm since the identity transformation defines an element of G,. In general, when we have a ts Y = ( P , T ) such that either Y is a tg or P # 0 and T = 0, then we shall say that Y is a generalized tg. I t is easy to see that a ts is a generalized tg iff Y'is a tg. The generalized tg Ha is called the holonomy ts of a, while G, is called the (generalized) holonomy group of a. We have

since G, is a quotient of the subsemigroup (s

IsE

s, as = a }

In particular, if a = Q, then GQ is a quotient group of the group of invertible elements of S if X is a tm, otherwise GQ= 0. It would be an error to regard Ha as contained in the ts X , which has a as the underlying set rather than the paving B,.

P R O P O S I T I O N 6.3. If u

-

b, then Ha M Hb.

II. Decomposition Theorems

46

Proof. We may assume a # b. By Corollary 6.2 we then have elements u, ii E S such that b = au, a = bzi and uii is the identity on a ziu is the identity on

For each s E S such that as

=a

b

we then have

b(iisu) = am = au

=b

Similarly, if bt = b, then a(utE) = a. It follows that u maps Ba bijectively onto Bb and yields the required isomorphism Ha w Hb I 7. The Holonomy Decomposition Theorem

Let X = (Q, S ) be a ts and let h be a height function for X with Qh = n > 0. Given 0 < i 5 n let

be a set of representatives of the equivalence classes of elements of A of height exactly i. We define

Hiv= Hal v . . . v Hak Hi+ == Hal . . . Hak Hix = Hi1 x . . . x Hik

+

+

By Proposition 6.3, these ts's are independent of the choice of the representatives a,, . . . , a k . THEOREM 7.1.

(Holonomy Decomposition).

(7.1)

x
(7.2)

X < H T o ... o H 7

(7.3)

X < H T o . . . OH?

Observe that the rule X v Y (7.2) follows from (7.1).

<X + Y

implies Hiv

< Hi+and

thus

7. The Holonomy Decomposition Theorem

47

The height function h for X is arbitrary. Normally the minimal height function would be used, since this would minimize n. However, we also could use a height function h such that for each 0 < i 5 n there is exactly one equivalence class of height i. This yields COROLLARY 7.2. Let a , , . . . , ar be representatives of the equivalence classes of elements of A of cardinality >1, arranged in such a way that ai < aj implies i < j . Then

Observe that by necessity a, = Q. Since each Ha,is a (generalized) tg whose action group divides S, the corollary implies Theorem 4.2 and thus also the Krohn-Rhodes Decomposition Theorem. The proof of Theorem 7.1 makes strong use of relational coverings that were introduced in 1,4 but not used until now. A relational covering

XQY will be said to be of rank i (with respect to a given height function h for X ) if for each p E Q y we have

pp, E A

and

pp,h

zi

The key statement is PROPOSITION 7.3. If X d,+ Y is a relational covering of rank i with 1I :i 5 n, then there exist relational coverings

XQH,VoY w

of rank i - 1.

The proof is given in the next section. Taking the proposition for granted, we now prove Theorem 7.1. We begin with the unique relational covering

XQ1' (P

II. Decomposition Theorems

48

given by OF = Q. Clearly q~ has rank n. Applying Proposition 7.3 repeatedly n times we obtain relational coverings

x a p o ...

OH,'

I

XQH?

o...

OH>

7

of rank 0. However a relational covering of rank 0 is a covering. Thus Q may be replaced by and Theorem 7.1 follows.

<

8. Proof of Proposition 7.3

Let

be a relational covering of rank i (I 5 i 5 n) with respect to a fixed height function h in X . We write X = (8,S ) , Y = (P, T) and define

PI = { P I P

E

Po = { P I P

E P,

p , Pvh = i>

Pvh < i>

From (6.11) we know that A contains elements of height exactly i. Let a , , . . . , ak (K > 0) be representatives of the equivalence classes of such elements of A. Thus for each p E P, we have

Pv

-

aj

for a unique index 1 < j 5 k. Using Corollary 6.2 we select elements u p ,tip E S v 1, such that u p p=p v ,

p p i p = aj

qupiip= q

for all q

qiipup= q

for all q ~ p g ,

E

aj

We propose to establish the relational covering

8. Proof of Proposition 7.3

49

of rank i - 1 with a suitable relation y : (Bal u

...

u B,Jx P - Q

Since the union B = B,, u . . . u B,, is treated as disjoint, each element of B will have the form ( j , b ) with 1 < j 5 k, b E B,. The relation y is now defined as follows: (j,b,p)y=

r

ifPCPo

if p ~ P ~ , p p - a j otherwise

bu,

Clearly y is surjective, ( j ,b, p ) y E A, and ( j , b, p ) y h < i. Further

(i,W w = PP?

(8.2) t

Next consider s E S. Since p is a relational covering, there exists T such that ps c t p . We shall show that

E

(8.3 1

YS

= ( f , t)Y

where f is a suitable function defined on P with values in the action semigroup W of H,". The definition of pf for p E P breaks into three cases.

Case 1. p t

E

Po. Then

(i,b, p ) ( f , t

) =~ ((i,W f1, P ~ ) Y= ptcp

Since

( j , b, P ) W S

= PV = p t v

(8.3) holds for any choice of pf.

-

Case 2. p t E PI and pys # ptp. Since ptyh = i we have p t p al for some 1 5 15 k. Since p p s is a proper subset of p t y , it follows that ppstipt is a proper subset of ptq.?tiPt= a l . Consequently, there exists a N brick c E B,, such that ppszipt c c. We define pf to be ( I , c ) , i.e. the constant transformation of with value (I, c ) . Then, for ( j ,b ) E B

( j , b, PI($

t ) =~ ( ( j ,4 ( p f ) ,P ~ ) Y= (tc, pt)w = CUpt

and

( j , b, P)YS Thus (8.3) holds.

= p v s = pps~p,tUpt = CUpt

II. Decomposition Theorems

50

Case 3 . p t E P , and ppls = ptpl. This implies ptpl Sppl. Since ptqh = i and pplh 5 i, the conditions imposed on a height function imply pplh = i and ppl -ptpl. Let ppl -ptpl- a j . Then aiupsiipl= pplsiipl = ptpliipt = a j so that upsiiptis a transformation of the holonomy tg Hu, and therefore that we shall denote by pf. With this defines a transformation of definition we have

If 1 f j , then ( I , b, p ) y = 0. Thus (8.3) holds. This concludes the proof of (8.1). We also show (8.1‘) with 7 : (B,,x

. . .X

B ~ J X P - ~ Q

defined as follows

where b = ( b l , . . . ,bk). Clearly 7 is a surjective relation, ( b , p ) q E A , and ( b , p ) y h < i. Further

Given s E S, we choose t E T so that

pls c

tpl. We assert that

where g is a suitable function defined on P with values in the action monoid of T h e construction of g is a slight modification of the construction o f f above and is omitted I

F.

EXERCISE 8.1.

Describe

~S’S

of height 51.

9. Examples

51

EXERCISE 8.2. Prove

+ sup(Xh, Y h )

(Xv Y)h5 1

(X+ Y)h<Xh+

Yh+ 1

( X xY)h5 Xh + Yh When does equality hold? EXERCISE 8.3.

Compare the holonomy ts’s and the height of X with

those of Xc. 9. Examples EXAMPLE 9.1. Let F, be the monoid of all functions f: n n ( n 2 1). We consider the tm (n, F,). T h e class A consist of all the subsets of n, and any two subsets of the same cardinality are equivalent. Thus for each 1 < K 5 n we may choose k as a representative set of height K - 1. The set k is paved by all of its subsets of cardinality k - 1 and the holonomy ts of k is the tg --f

Hk-l =

(k,Sk)

where S, is the group of all permutations of k. Consequently the holonomy decomposition of (n, F,) is

(n,Fn)

< (2,S,)

0 *

9

*

0

(n, Sn)

It will be shown that within a suitable context this decomposition is the best possible one (see XII,6). T h e argument and the decomposition remain unchanged if we replace F, by the larger monoid PF, of all partial functions n -+ n. An interesting phenomenon takes place if we exclude all the permutations S , from F,,, i.e. consider the ts (n, F, - S,). T h e holonomy ts’s Hkp1 for k < n remain unchanged. However Hn-l instead of (n, S,) becomes (n, 0) = n. Thus we obtain (n,F ,

-

S,)

< (2,S,) . . . o

o

(n- 1, Sn-l)0 ii

More generally given any 1 5 p < n, consider the. subsemigroup Fn,p of all functions f: n + n such that card(nf ) 5 p . Then n is paved by all of its subsets of cardinality exactly p , so that the top holonomy ts is

52

II. Decomposition Theorems

(r, 0) where

Y =

(i)is

the binomial coefficient. Thus

(n,I z , , p )

< (2, S,)

0

f

’

-

0

(p, S,) 0 i:

Again nothing changes if we consider partial functions rather than functions. EXAMPLE 9.2. We consider the ts C ( p , , )given by the diagram

The integer p 2 1 is called the period, while Y 2 0 is called the stem. The ts C ( p , ritself ) is called the cyclic ts of type ( p , I ) . If Y = 0, then C t p , ois ) a tm and is the cyclic group 2, of order p . If Y > 0, then C ( p , r ) is a ts (but not a tm) and = Z ( p , r )is the cyclic monoid with stem r and period p . The action semigroup S ( p , rof ) C ( p , ris) the cyclic semigroup with generator a and relation ar = a r + p . With the exception of the is the ts defined by S t p , r fIn . the exceptional case (p, Y ) = (1, l), C(p,r) case (9, r ) = (1, l ) , C(l,l)is the ts C whose action semigroup is the semigroup 1 containing a single element. Thus C,,,,, and C,,,,, have isomorphic semigroups. The family A for C ( p , rconsists ) of the empty set, all singletons, and sets

ai= { a j I i s j < r + p }

for 0 5 i 5 Y . Note that ao is interpreted as the state 1. Each set ai for 0 5 i < r is paved by the two sets {d, ai+l}and the holonomy ts is Had= 2 with no transformations. The set a, = {a+,. . . , ar+p--l}is paved by its singletons and Ha, = 2,. Thus the Holonomy Decomposition Theorem yields C(p,r)

< zp zr 0

This decomposition is far from being the best possible one. We shall now show by direct arguments how a “better” decomposition can be achieved. First we note that there exists an injective morphism S(p,r)

+

ZpX

S(1.r)

9. Examples

53

of monoids. The morphism maps the generator 0 into (t,7) where T is the generator of Z p and 7 is the generator of S(l,r). This implies C(p.r)

<

z p

x C(1,r)

We now consider the ts C,,.,) given by the diagram

If Y > 0, then Proposition 3.3 can be applied with the state 1 playing the role of the state p in the proposition. Removing the state 1 we obtain ,, the states renamed. Thus Proposition 3.3 yields the the ts C ( , ~ , - with inequality ~ ( l , r ) e(l.r-1) 0 C

<

c = 2 we obtain C(l,r)< 2.-1 c

Iterating this result and noting that

0

with

Zo

interpreted as 1'. Thus finally

See Exercise III,2.2 for another decomposition of C(l,r). EXAMPLE 9.3. Given an integer n > 1 convert the set n into a semigroup K , by setting = i for 0 < i, j < n. Thus K , is the reversal of the semigroup 6.T o obtain the corresponding ts we adjoin a unit element e to K , obtaining the ts (n u e, K,). The family A consists of the sets n u e, n,the empty set, and singletons. The set n u e is paved by (n,e ) and Hnve M 2. The set n is paved by its singletons and H , = n'. We thus obtain the decomposition

K,&< ii' o Z EXAMPLE 9.4.

Consider ts F given by the diagram

10' The family A consists of the set 2, the two singletons, and the empty set. The holonomy ts H I is 2 and thus we obtain

F
II. Decom position Theorems

54

If we replace F by its completion FC, the diagram becomes

and the family A consists of the sets (0, 1, O}, (0, a}, (1, n},all the singletons, and the empty set. T h e set (0,1, O}is paved by the two sets (0, O } and (1, O}and its holonomy ts is 2. T h e two sets (0, O}and (1, O}are equivalent and their holonomy ts’s are 2. Thus we obtain FC
EXAMPLE 9.5. I t is interesting to see what the algorithm gives when applied to the already decomposed ts

X=ZoZ We have

X

=

(Q, S ) ,

Q

=

(00, 01, 10, l l }

and S consists of eight elements, four of which are constants. Each pair (f,s) in S can be written as a triple (Of, If, s) and we obtain the following table of the non-constant transformations of S I 0 0 01 10 11

i3,?,6

00 10 00 10 10 00

1,8,0 10 00 8,i, i 01 11 i,8, i 11 01

01

11

11

01

The family A consists of the set Q, the two doubletons (00, lo} and (01, ll}, all the singletons, and the empty set. The set Q is paved by the two doubletons and H , has no transformations. Thus H , = 2. The two doubletons are non-equivalent and their holonomy ts’s are 2. Thus we obtain

x<moz=40z

9. Examples

55

If instead we consider

X=Z02. we obtain four additional non-constant transformations as follows : 00 01 10 11

a, i, 1

00 11 01 00 01 10 11

i,a,i 10

a,a,

1

i,i,1

00 10 00 10

11 01 01 11

This produces two more doubletons (00, 11) and {lo, 01} and Q is paved by the four doubletons. Thus H , = 4. The four doubletons now become equivalent and their holonomy ts is 2'. We thus obtain

x<2.04 We let the reader work out the cases X = 2 o 2 and X = 2. o 2.. In the last case Exercise I,10.7 should be used to tabulate the generators of

s,.

EXAMPLE 9.6. Let Q = 6 and let X = (Q,.S) be the ts generated by the transformations u, z: Q c Q given by the table 0 1 2 3 4 5 2 0 1 2 0 3 4 3 2 2 2 2

The elements of A are the sets

Q,

a

=

b

Qo = 4,

=

Qoz = 3,

c = QT = { 2 , 3 , 4 )

the empty set, and all the singletons. No doubletons appear in A. Q is paved by {a, c, 5 ) and H , = 3. a is paved by {b, 3 ) and Ha = 2. For the sets b and c we have

bt

= C,

cu = b

56 so that b

II. Decomposition Theorems

-

c. Further bo = b and bra = b yielding the permutations (0, 1, 2) -+(2, 1 , O ) and (0, 1, 2) + (0,2, 1). Thus the entire permutation group S, of 3 is generated. Consequently H b = (3, S,).

We thus obtain x < ( 3 , o o z o 3

This example was discussed by Ginzburg (1968, pp. 137-142), and is given here for purposes of comparison.

X

EXAMPLE 9.7. Consider the ts

= (Q,

S) defined by the diagram

The set A with 0 and singletons omitted and with equivalent elements not repeated has the form

Q

Thus the height is 3, Q is paved by 6 and ($4) and H3 = 2. The set 6 is paved by the doubletons {03), (14), and {25), and H , = 2,. The set { p q } is paved by singletons and Htpql= 2,. Finally (03) is paved by singletons and HtO3) = 2,. Thus X

<

Z

2

O

~

O

2

57

References

EXAMPLE 9.8. For each n 2 1 consider the ts X n = (Q,, S,) given by the diagram

e

1

n

+

with n 1 states e, 1, . . . , n and n transformations al, . . . , a,. I n addition to the set Q,,, singletons, and 0, the family A consists of all the doubletons ei with 1 5 i 5 n. Further

(ei)aj = j e

if j f i

(&)ai = i Thus the doubletons are all equivalent to one another. The height is 2. T h e set Qn is paved by the doubletons ei and the holonomy ts has no transformations. Thus H , = n. The doubleton e l is paved by its singletons. If n 2 3, then we have (el)a, = 2e, (2e)a, = e3, (e3)a1 = le. Thus (el)o,a,a, = le. Consequently He, = 2,. Thus

~,
= 2,

if n > 2

then He, = 2’ and

Finally X , = C . References

T h e material in Section 2 is folklore going back probably to Frobenius. Krohn-Rhodes (1965) established the decomposition theorem using semigroups and state dependent sequential machines. This combination is unfortunate and the resulting proof is very difficult. A much improved version of the proof and also another more algebraic proof appear in Krohn-Rhodes-Tilson (1968). Lallement (1971) published an algebraic modification of the proof applicable to monoids. Meyer-Thompson (1 969)

58

II. Decomposition Theorems

published another, considerably simpler, algebraic proof. Our proof given in Section 4 is a further simplification of Meyer-Thompson. Zeiger (1967) attempted a proof along a different line. His proof contains some inaccuracies but a correct version of Zeiger’s method appears in Ginzburg (1968). It is the analysis of this proof that has led us to the Holonomy Decomposition Theorem. Abraham Ginzburg, “Algebraic Theory of Automata,” Academic Press, New York, 1968. Kenneth Krohn and John L. Rhodes, Algebraic theory of machines, I. Prime decomposition theorem for finite semigroups and machines, T7ans. Amer. Math. Soc. 116 (1965), 450-464. Kenneth Krohn, John L. Rhodes, and Bret R. Tilson, The prime decomposition theorem of the algebraic theory of machines, in “Algebraic Theory of Machines, Languages, and Semigroups” (M. A. Arbib, ed.), Academic Press, New York, 1968, pp. 81-125. Gerard Lallernent, On the prime decomposition theorem for finite monoids, Math. Systems Theory 5 (1971), 8-12. A. R. Meyer and C. Thompson, Remarks on algebraic decomposition of automata, Math. Systems Theory 3 (1969), 110-118.

H. Paul Zeiger, Cascade synthesis of finite state machines, Information and Control 10 (1967), 4 1 9 4 3 3 . H. Paul Zeiger, Yet another proof of the cascade decomposition theorem for finite automata, Math. Systems Theory 1 (1967), 225-228.

CHAPTER

111

Transformation Semigroups (continued)

In Chapter I we introduced only the very basic facts about ts’s, just sufficient to satisfy the needs of Chapter 11. In this chapter we develop additional techniques and establish a number of facts that will be needed in Chapter I V and also later. Thus this chapter will treat a number of somewhat unrelated topics. 1. Classes and Closed Classes

A family X of ts’s will be called a class if (1.1) Y

<X

and X

E

X imply Y

E

X.

A class X is said to be weakly closed if (1.2) X , Y

E

X implies X x Y EX.

A class X is said to be closed if (1.3) X , Y

E

X implies X

o

Y

E

X.

<

Since X X Y X o Y , each closed class also is weakly closed. T h e intersection of any family of classes (or weakly closed classes or closed classes) is again a class (or a weakly closed class or a closed class). Nothing prevents a class from being empty, and in fact 0 is the smallest closed class. This is not to be confused with the class 0 consisting only of the empty ts 0. This also is a closed class and next to 0 it is the smallest one. 59

I l l . Transformation Semigroups (continued)

60

Given any two classes X and Y in TS we define the classes

XxY= {ZIZ<XXY,XEX,YEY) XoY= (ZjZ<XOY,XEX, YEY) The rules

-

(XlX Y1)x (X2X YZ) (Xl

0

Yl) x (X2

0

Y,)

(XIXX2)X (YlX YZ)

= (X,x-X2)

O

( YlX Y2)

imply

(1.4) If X and Y are weakly closed, then so are X x Y and X oY. The associative rules

(XxY)xZ =Xx(YxZ) (xoY)oz=xo(Yoz) are easily verified. Thus the n-tuple products

x, . . . ox,

x,x . . . xx,,

0

may be written without parentheses. The class (1') is a unit element for both products, while the class 0 is a zero. A class X is weakly closed iff X x X c X and is closed iff X o X c X. If Y is any family of ts's, we shall denote by (Y) the least weakly closed class containing Y, and by [Y] the least closed class containing Y. We have (1.5)

(Y)= { X I X < Y , X

(1.6)

[Y]=

. . .x Y , , Y , , . . . , Y , E Y )

( X I X < Y , O ... OY,,Yl,

..., Y , E Y )

We omit the proof of these two simple facts. For each class X and each k 2 1 we denote by X(k)the k-tuple product X x . . . x X and by Xk the k-tuple product X o . . . o X., Consequently

u u Xk

gr) = X'+' =

X'k'

k 21

[XI

=

x+=

ktl

Thus, in particular

[XI

=

(XI+

1

1. Classes and Closed Classes

61

PROPOSITION 1.l. Let X be a weakly closed class in TS containing 2'. I f X = (Q, S ) is a tm and S E X,then X E X.If, further, X is complete and X E X, then s E x.

< Q' x S. Since 2' E X implies X, the first conclusion follows. The second conclusion follows

Proof. Corollary I,9.7 implies X

Q'

E

from the inequality S

< X ( k )given by Proposition

1,924

I

PROPOSITION 1.2. Let X be a weakly closed class in TS containing 2' and C. If X = (Q, S ) is a ts and S E X, then X E X.I f , further, X is complete and X E X, then S E X.

<

Corollary I,9.4 implies X Q ' x Sx C. Since 2' E X implies Q' E X, the first conclusion follows. The second one follows from Proposition I,9.8 I Proof.

PROPOSITION 1.3.

Let X = (8,S ) be a monogenic cts. Then (X)= ( S )

and [ X I = [ S ] . Proof. We recall that X is monogenic provided Q = qS for some q E Q. The desired conclusion follows from the inequalities

X

< S <X ( k ) ,

given by Proposition I,9.9

k

= card

Q

I

PROPOSITION 1.4. Let X be a weakly closed class in TS containing 2'. If X and Y are in X, then X v Y and X Y are in X.

+

< +

+

Since X v Y X Y , it suffices to consider only X Y. We first prove that X 1' E X.If Qx = 0 or Sx = 0, then X 1'< k' with K = 1 card Qx. Since 2' E X,it follows that k' E X and thus X 1' E X. Thus we may assume Qx # 0 and Sx # 0. With this assumption we prove Proof.

+

+

+

+

X+l'<XX2' I

with q: Q x X z - Q x

v1

111. Transformation Semigroups (continued)

62

Indeed, for any s E S,, the transformation (s, 1) of by (s, 1) viewed as a transformation of Xx2'. Having proved that X 1' E X we now have

X + 1' is covered

+

(X+ l')x(Y Identifying q E

Qx with

+ 1') E X

(q, 0) and p E

Qy

with (0, p ) we have

x + Y c (X+ l ' ) x ( Y + ThusXf

YEX

1')

I

Let X be a weakly closed class containing 2'. Then X if and only if X I (q) E X for each q E Qx.

PROPOSITION 1.5.

X

E

We recall that ( q ) denotes the principal ideal q u qS. T h e proposition follows from Proposition 1.4 and Proposition I,9.2 I

Let TG be the family of all transformation groups, and G the family of all groups. Prove EXERCISE 1.I.

(TG) = (G),

[TG]= [GI

EXERCISE 1.2. Show that the conclusions of Propositions 1.1 and 1.5 fail for X = (1'). Show that the conclusion of Proposition 1.2 fails for X = (2.). Show that the conclusion of Proposition 1.4 fails fm X = (E).

2. Sinks in a t s

A state p in a ts X = (9, S ) is called a sink if pS c p or equivalently if p = (p). T h e restriction X I Q - p will be denoted by X - p . A ts may have more than one sink state. If X is not complete, then the state added to X to produce the completion X c is a sink state for X c . Given a ts X = (9,S) we denote by XO the ts (80,S) obtained from X by adjoining a (new) sink to X.Thus Q" = Q u 0 with 0 assumed not to be an element of Q. T h e action of S on QD is defined by q's={;

if q E Q and qs otherwise

E

Q

2. Sinks in a ts

63

Clearly X O is complete, and X u - 0 = X.We note the following connections with the completion operator X c: if X is complete otherwise

x c q XO x

if X is complete otherwise PROPOSITION 2.1.

0 and X x - p' < Y - p.

either p p

=

Let X



-

Y and let p be a sink state for Y . Then = p p is a sink state for X and

p , or else p'

Proof. Assume p p = p'. Let s

E

S, and let t

p's = p p c p i p c p p

=

E

S y cover s. Then

p'

and thus p' is a sink state for X. Define R' = Qx - p p and let R = R'p-l. Then by Exercise I,6.4, X I R' Y I R. Since Y I R c Y - p , it follows that X I R' Y - p. However, X I R' = X if pp = 0 and X I R' = X - p' if p p = p' I

<

<

If X

< Y O and ifX has no sinks, then X < Y I

PROPOSITION 2.3. Let

X be a ts with a sink p and let Y be a complete

COROLLARY 2.2.

ts.

If X -p


X

then

< C'

o

Y

<*

Proof. Let X = (Q, S ) , Y = (R, T ) and let X - p Y. Thus y: R Q - p is a surjective partial function, and for each s E S there exists t E T such that ---f

(2.1)

ys' c t y

where s' is the partial function Q - p (2.2)

rys c rty

-+

Q - p defined by s. Thus

if rys

Define the surjective partial function p: 2 x R - + Q

(0, T)V = P (1,

= 'W

f p

II I, Transformation Semigroups (continued)

64

Given s E S and t E T satisfying (2.1), define

f: R + (1, a}

0

yf =

{1

if rys = p otherwise

We shall show that (f,t ) covers s. We have

The results agree if rys = p . If rys + p , then (2.2) gives the required inclusion. Next we have

Thus again the required inclusion holds. The completeness of Y was used to ensure that rt # 0 I EXAMPLE 2.1.

Consider the ts E given by the diagram

1 5 0 Clearly 0 is a sink and E - 0 = (1,O). Since C' o (1, O) = (2, O), the inequality E c' o ( E - 0) fails. This shows that the completeness of Y in Proposition 2.3 is essential.

<

A class X in TS is said to be completely generated if for each X E there exists a complete Y E X such that X Y.

<

THEOREM 2.4.

X

Let X be a completely generated closed class in TS

containing C'. Then (i) (ii) (iii) (iv)

If X is a ts with a sink state and X E X if and only if Xu E X. X E X if and only if Xc E X. X E X if and only if Sx E X.

if

X

- 0E

X, then X

E

X.

65

2. Sinks in a t s

Proof. (i) follows from Proposition 2.3 and (ii) follows from (i). (iii) follows from (ii). Since C' E X implies 2', C E X, (iv) follows from Proposition 1.2 applied to X c I PROPOSITION 2.5.

p E ~ SThen .

Let X

=

(Q, S ) be a ts andp E Q a state such that

x < x I (PI

O

(X-

whew (P)=PUPS=PS, Proof.

x - ( ~ > = (xQ l- ( P ) >

Define

Given s E S we define

f: (Q - ( P I ) u 0 - s qf= t

if q # 0 and qs E ( p ) , where t is such that pt = qs.

q f = to

if q f 0 and qs E Q - ( p ) , where to E S is such that pt, = p .

qf= s

in all other cases.

Thus the desired equality holds. If q = 0, then

E

S

Ill. Transformation Semigroups (continued)

66

Thus again equality holds. If q = 0 and r This completes the proof I COROLLARY 2.6.

f p , then

(Y,

q ) p = 0.

If R is a minimal non-empty right ideal in a ts X ,

then

X < X l Ro(X-R)" Indeed, if

p

E

R, then p S

=

R and thus p E pS and R

=

(p) I

EXERCISE 2.1. Let X be a ts. Let X ' be the ts obtained from X by removing simultaneously all the sinks of X . The ts X ' may again have sinks, and these are removed yielding XI' etc. A t the end a (possibly empty) ts X o is obtained which has no sinks. Show that if X Y , then X o Yo.

<

EXERCISE 2.2.

<

Consider the cyclic ts C(l,r)introduced in Example

II,9.2. Prove

< c Z(l*',< (C'Y = ZYl*l,= Ul' C(1,?) (C.).-'

O

3. Transitivity Classes

Let X = (Q, S) be a ts. We shall say that X is transitive if for any two distinct states q1 , qa E Q we have q, E qlS. More generally a non-empty subset R of Q is said to be transitive if q, E qlS for any two distinct states q1 ,q, in R. T h e maximal transitive subsets of Q are called the transitivity classes (also called 9-classes) of X . An alternative approach is to define q, 5 qI if q, E (ql) i.e. if either q, = q1 or q, E qlS. This defines a preorder in Q and leads to the equivalence relation q1 q, defined by q, 5 q1 and q1 5 q, (or equivalently (q2) = ( 4 , ) ) . The transitivity classes are then the equivalence classes for this equivalence relation. If R, and R, are transitivity classes, then either R, n R,S = 0 or R, c R,S. In the latter case we write R, 5 R,. I t is easily seen that R, 5 R, and R, 5 R, imply R, = R,. Thus a (partial) order of the transitivity classes of X is obtained.

-

THEOREM 3.1. Let X be a campletely generated closed class in TS containing C'. Then X E X if and only if X I R E X for each transitivity class R in X .

67

4. ldempotents in Semigroups

Assume X is not transitive. Let X = (Q, S) and let R be a minimal non-empty right ideal in X . If r E R, then rS = R since R is minimal. Consequently R is a transitivity class of X and by assumption X I R E X. Let R' be a transitivity class of X - R. Then R' also is a transitivity class of X and ( X - R ) I R' = X I R' E X. Arguing by induction on the cardinality of Q we may thus assume that X - R E X. It now follows from Theorem 2.4 that ( X - R)o E X. By Corollary 2.6 Proof.

X < X I R o ( X - R)o and thus X E X EXERCISE 3.1. Show that the minimal (in the sense of order) transitivity classes of a ts X coincide with the minimal right ideals of X . EXERCISE 3.2. Show that the transitivity classes f o r X o Y and X X Y are R, x R, where R, is a transitivity class of X and R, is a transitivity class of Y . EXERCISE 3.3.

Show that a transitive ts is monogenic. Show that a monogenic tg is transitive. 4. ldempotents in Semigroups

It is rather surprising that thus far in this volume we needed only the most rudimentary concepts concerning semigroups, namely those exposed in 1 , l . We shall now introduce a very modest new quantum of semigroup theory which will suffice until more is introduced in Chapter V. An element e in a semigroup S is called idempotent if e = e2. For each such idempotent e w e have the monoid eSe. I t has e as unit element and is the largest monoid in S with e as unit element. T h e group of invertible elements in eSe is denoted by G,; it is the maximal group in S containing e. If e and e' are distinct idempotents in S , then the monoids eSe and else' are distinct (since they have different.unit elements) but they may overlap. However, the groups G, and G,, are disjoint. Indeed, assume s E G, n G,, and let t and t' be inverses of s in G, and Get respectively. Then e = st = se't = st'st = ere e' so that e

= e',

=

t's

=

t'es = t'sts

contrary to assumption.

= e'e

Ill. Transformation Sernigroups (continued)

68

PROPOSITION 4.1.

Every non-empty ( j n i t e ) semigroup S contains an

idempotent. proof. By Proposition 1,l.l if Ss = S = sS for all s E S, then S is a group and the conclusion is clear. If not, then for some s E S either Ss or sS is a non-empty proper subsemigroup of S. Thus the conclusion follows by induction on card S I

We note that the conclusion of Proposition 4.1 is false for infinite semigroups. Indeed, the semigroup of all strictly positive integers with addition as operation has no idempotents. PROPOSITION 4.2. If S is a semigroup and s E S, then sn is idempotent for some n 2 1.

This follows from Proposition 4.1 applied to the cyclic subsemigroup of S generated by s I Conversely, one could first prove Proposition 4.2 by considering cyclic semigroups, and then deduce Proposition 4.1. PROPOSITION 4.3.

I f S is a monoid with 1 as the onZy idempotent,

then S is a group. Proof. Let s E S. By Proposition 4.2, sn = 1 for some n sn-1

is an inverse for s

> 1. Then

I

PROPOSITION 4.4. Let p: T -+S be a surjectiwe morphism of semigroups. For each idempotent e in S there is an idempotent e' in T such that e = e'p

Since p is surjective, ep-' is a non-empty subsemigroup of T. Thus by Proposition 4.1, ep-' contains an idempotent I Proof.

PROPOSITION 4.5. Let p: T --+ S be a surjective morphism of semigroups. For each monoid (or group) S' in S there exists a monoid (OY group) T in T such that S' = T'p. Proof. Let T' be a subsemigroup of T of lowest possible cardinality such that T'p = S'. Such a subsemigroup exists since S'p-lp = S' and

5. ldempotents in a

69

ts

S'v-' is a subsemigroup of T. Let e be the unit element of S'.By Proposition 4.4, there exists an idempotent f in T' such that fv = e. Since

( f T Y ) v= (fv)(T'v)(fv) = eS'e = S' and since f T ' f c T' it follows from the minimality of T' that f T Y = T'. Thus T' is a monoid with f as unit element. Now assume that S' is a group. Let t E T'. Then

( T ' t )v = ( T ' v ) ( t T )= S ' ( t v ) = S' and since T't c T' it follows from the minimality of T' that T't Similarly, tT' = T'. Thus T' is a group, by Proposition 1,l.l I

=

T'.

COROLLARY 4.6. If S < T where T is a semigroup and S is a group (or a monoid), then there exists a group (or a monoid) T' in T and a surjective morphism v: T' -+ S I

<

PROPOSITION 4.7. If S G where S is a non-empty semigroup and G is a group, then S is a group, and is a quotient group of a subgroup of G.

Proof. There exists a subsemigroup T of G and a surjective morphism T + S. By Proposition I,l.Z, T is a group. It follows readily that S is a quotient group of T I

Let v: S + T be a surjective morphism of monoids which is minimal in the sense that S'g, # T for all proper subsemigroups S' of S. Show that lv-l is a subgroup of S. EXERCISE 4.1.

EXERCISE 4.2. Let S be a semigroup and let e be an idempotent in S such that eSe has lowest possible cardinality. Show that eSe is a group.

5. ldempotents in a t s

Let X = ( Q , S ) be a ts and let e be an idempotent in X (i.e. an idempotent in S). We consider the ts X,, as defined in I,6. The underlying set is Qe and the transformations are given by the elements s E S such that Qes c Qe. The partial function Qe + Qe produced by such an element s coincides with that given by ese. Further, if ese and es'e produce the same

70

I l l . Transformation Semigroups (continued)

partial function Q e

---f

Qe, then for each q

E

Q we have

qese = (qe)ese = (qe)es'e = qes'e so that ese

= es'e.

We thus obtain

This clearly is a tm with e producing the identity function Qe Qe. Henceforth we shall use the shorter notation X , instead of X Q e . A semigroup S is identified with the ts (S', S). If e is an idempotent of S, then S'e = Se. Thus --f

Se = ( S e , e S e ) Since S e is a subsemigroup of S, we also have the ts

Se

=

((Se)', S e )

which may be distinct from S,. T h e two coincide iff S e is a monoid, i.e. iff S e = eSe. PROPOSITION 5.1.

Let

X
be a covering. For each idempotent e of X there exists an idempotent e' of Y covering e. Further,

with y : Qye' Proof.

--j

Qxe deJined by Fe.

T h e set {t I t E S,, t covers e }

is a non-empty subsemigroup of S,- and thus, by Proposition 4.1, contains an idempotent e'. Since g,e c e'g, and Qyg, = Q x , we have

Q x e = Q Y Y ~c~Q y e ' ~c Qxe If s E Sx is covered by t

yese proving (5.1)

1

E

S y , then ese is covered by e'te'. Thus

= g,esee c

e'te'ye

=

(e'te')y

5. ldempotents in a ts COROLLARY 5.2.

71

Given a covering

x
x< ye v with v,o dejined by

pl

I

PROPOSITION 5.3. Let ( g , e ) f 0 be an idempotent of X e is an idempotent of Y and

(X 0 Proof.

Y)(g,e) c

X

0

o

Y . Then

Ye

Since (g,e)(g, e ) = ( g , e), it follows that ee = e. Clearly ( Q x X Q Y ) ( ~ , e ) c Q X X (QYe)

For any transformation ( f , s) of X

o

Y we have

( g , e ) ( f , s)(g, e>= (h,ese) for some function h: Q y - S X

Let

h': Q y e + S, be the restriction of h. Then (h', ese) covers ( g , e)( f , s)(g, e ) viewed as a transformation of (Xo Y),,,,, 1 COROLLARY 5.4.

Given a covering X< Y

0

2

c

where X is a tm, there exists an idempotent e in Z such that

X
where v,o is dejined by y . This follows from Corollary 5.2 and Proposition 5.3

I

72

Ill. Transformation Semigroups (continued)

Show that i f X is a transitive ts, then so is X , for any

EXERCISE 5.1.

idempotent e in X . Show that i f e is an idempotent in X , then

EXERCISE 5.2.

Thus X ,

-

X , c X I Qe

and

X I Qe c Xe

X I Qe.

Let X be a ts and 1its closure. If G # 1 is a group such that G < Sz (or G c SX),then G < SX (or G c SX). EXERCISE 5.3.

<

EXERCISE 5.4. Let X Y. In general, unless X is complete the inequality S, Sy need not hold. Show that if G is a group, then G S,

<

<

implies G < Sy. EXERCISE 5.5.

Show that 1' < X

EXERCISE 5.6.

Show that if X

and X

< Z.

isf Sx has an idempotent e # 0.

< l', then X < Y

o

Z holds ifl X


6. Localization

For any class X in TS the localization of X is defined by

LX = { X I X , E X for all idempotents e in X } PROPOSITION 6.1. For any class X, LX is again a class and LLX = LX. If X is weakly closed, then so is LX.

<

Proof. Let Y X and X E LX. For each idempotent e in Y there is, by Proposition 5.1, an idempotent e' in X such that Ye X,,. Since X,, E X and X is a class, it follows that Ye E X and thus Y E LX. If X is any ts and e an idempotent in X , then X , E LX iff X , E X. Thus LLX = LX. Let X be weakly closed and let X,Y E LX. An idempotent e in X X Y has the form e = ( e l , ez)where el is an idempotent of X and e, is an idempotent of Y.Further, ( X X Y ) e= X,,X Yes.Since X,,, YeaE X and X is weakly closed it follows that ( X X Y ) , E X and thus X X Y E LX I

<

73

6. Localization PROPOSITION 6.2. For any class X

(LX) n TM = X n TM and LX is the largest class with this property. Proof. T h e formula above is clear. Let Y be any class such that

Y n TM = X n TM. Let Y EY and let e be a idempotent in Y .

<

Since Ye Y it follows that Ye E Y. Since Y e is a tm, it follows that YeE X and thus Y E LX. This shows that Y c LX I Although it holds in important special cases, it is in general not known whether LX is closed when X is closed. EXAMPLE 6.1. Consider the closed class [O] consisting of the two ts's 0 and 0'. The ts X = (Q,S ) is then in L[O'] iff X,E [O'] for every idempotent e in S. This is equivalent with Qe = 0 for every idempotent e, and this in turn signifies that e = 8. Thus X E L[O'] iff 0 is the only idempotent in S or S = 0. Such a ts is called strongly nilpotent. The class L[O'] of all strongly nilpotent ts's is easily seen to be closed. In fact if X E L[O'], then Y o X E L[O'] for any ts Y . The class L[O'] will reappear in Chapters IV and VIII.

A class X in TS is called local if LX = X. PROPOSITION 6.3. For any weakly closed local class X in TS containing 2' and for any complete ts X , the following conditions are equivalent:

(i)

X E X.

(ii) S, E X. (iii) eSxe E X for each idempotent e in

Sx.

Proof. Since X is local and 1' E X, it follows that C E X. Thus the equivalence (i) o (ii) follows from Proposition 1.2. The equivalence (i) o (iii) follows from the equivalence (i) o (ii) applied to X, I EXERCISE 6.1.

Show that

[2]

= L[2] = L[1] = L[O] = L0

Deduce that any local class must contain 121. Thus 121 is the smallest local class.

74

111. Transformation Semigroups (continued)

7. Closed Classes Containing 2' THEOREM 7.1. Let X be a closed class in TS containing 2.. For any ts X the following conditions are equivalent :

(i) X E X. (ii) G E X for any group G in Sx. (iii) G E X for any simple group G Sx. Proof.

(i)

-

<

(ii). Let e be the unit element of G. Then (Qe, G)

X, < X . Thus ( X e , G) E X. Since ( X e , G) is complete. Proposition 1.1 implies G E X. (ii) 3 (iii). If G < Sx,then by Corollary 4.6, G < H where H is a group in Sx. Since H E X,it follows that G E X. c

(iii) * (i). This follows directly from the Krohn-Rhodes Decomposition Theorem I COROLLARY 7.2. If

X is a closed class in TS containing 2, then X

=

[Z', G,]

where Gx is the family of all simple groups belonging to X

I

Let X be a closed class in TS containing 2.. For each ts X the following conditions are equivalent: PROPOSITION 7.3.

(i) X E X. (ii) X c E X. (iii) X ' E X. (iv) X E X. (v) X,E X for all idempotents e in X . Proof. T h e equivalences (i) o (ii) o (iii) follow from the observation that S,, S,, , and S,. have the same non-trivial groups. T h e same holds for Sz in view of Exercise 5.3. Since each group in Sx is a group in eSxe for some idempotent e, the equivalence (i) o (v) follows I COROLLARY 7.4.

Every closed class containing 2. is local

I

PROPOSITION 7.5. Let X be a closed class in TS containing 2. If X contains a semigroup S , then it also contains the reversed semigroup Se.

7. Closed Classes Containing 2'

75

Proof. If G is a group, then the correspondence g +g-'

shows that G and Ge are isomorphic. Thus the conclusion follows from Theorem 7.1 I THEOREM 7.6. For any ts X , the following conditions are equivalent:

(i) x E [TI. (ii) There exists an integer n 2 0 such that snfl = s for all s E S,. (iii) All groups in S, are trivial.

-

(ii). Let A, be the family of all ts's such that (ii) holds Proof. (i) for a given integer n. We first show

(7.1) If X

< Y,X

is complete and Y E A , , then X E A,.

<

Indeed, since X is complete, we have Sx Sy.Thus there is a subsemigroup T of Sy and a surjective morphism T + S,. Since Sy E A , , it follows that T E A, and also that Sx E A,. Consequently X E A,,. Next we prove

(7.2) If X

E

A , , Y E&, then X

o

Y E A,,,.

Indeed, let ( f , S) E SXOY9 and let ( p , 9) E Qx x Q Y - If ( p , q)( f , s ) ~ # 0, then ( p , q)( f,s ) = ~ (p', q') with q' = qsm. Since qsmff = q' for all Y 2 0, it now follows by induction that ( p , q)( f , s),+~ = (p'(q'f ) r , 4'). Thus ( p , q)( f , s)"+, = ( p , q)(f , s),+,+l as required. Now assume that X E Then also X c E [?'I and thus X c ( p ) , for some k 2 0. Since all transformations of p are idempotent, we have 2. E A , . Consequently, by (7.2), E A, and by (7.1) xcE A,. Since Sx = Sxc it follows that X E A , . (ii) (iii). Let G be a group in Sx and let g E G. Since gn+l = gn, it follows that g is the unit element of G, and thus G is trivial. (i). This follows from Theorem 7.1 I (iii)

[z'].

<

(z'),

-

COROLLARY 7.7.

L[z']

=

[z.].

This follows from Corollary 7.4

[z']

I

The ts's in the class will be called aperiodic (the terms group-free and combinatorial also are used). Aperiodic ts's and semigroups play a very important role in the sequel.

111. Transformation Semigroups (continued)

76

8. The Derived t s and the Trace of a Covering

be a relational covering as defined in I,3. Aparametrization for ‘p consists of a finite alphabet l2 and a pair of morphisms a: 9 + + S , 6: 9 t - t T such that

(8.2) a is surjective; (8.3) ma is covered by 06 for all

w E

9.

The last condition implies

(8.4) wa is covered by wB for all w E 9+. Note that p always has at least one parametrization; it suffices to set

9 = {(s, t ) I s E S, t E T ,

t

covers s}

and define (s, t ) a = s, (s, t)P = t. The construction that follows does however depend upon the choice of the parametrization. Whenever a relational covering X dpY is established explicitly, the construction will usually indicate the “natural” (Le. the most useful) parametrization to be employed. EXAMPLE 8.1. Let p: S + T be a morphism of semigroups. I n 1,s the relational covering

S -p*-l 4T was established. A “natural” parametrization in this case is 9 = S, sa = s, SB = s’p. Given a parametrization of as above, it will be convenient to write q ( w 4 = qw,

P ( d ) = Pw

77

8. The Derived t s and the Trace of a Covering

for p

P, q E Q, w E Qf. In this way P and Q become Q-modules and (8.4) may be rewritten as E

(8.5 1

plw c w p

for all

w E Q+

The deriwed ts Q, of 9,relative to the given parametrization (Q, a,fi), will now be defined. The set of states of @ is simply the graph of q ~ i.e. ,

T o define S , we consider triples

(PI w9 P') with

p , p'

E

P,

w

E

pw

Qf,

c p'

Each such triple will be regarded as a partial function

( P , W , P'): Q,

(8.6)

-

Qe

by setting

Note that q E p"pl. Thus if p" = p then

qw c Pplw c PWpl c P'P, Thus if qw f 0 then (qw,p') E Q,, so that the partial function (8.6) is well defined. T h e ts Q, = (Q, , S,) is now defined with S , generated by all the partial functions (8.6). We note the following useful properties of So:

(8.7)

(p, w , p ' ) = 8

if pw

=

0

From these two properties we easily derive

(8.9)

Every transformation t of S, that is not 8 has the form t = (p, w, p ' ) with pw = p'. If further t is an idempotent, then p = p'.

78 THEOREM 8.1.

111. Transformation Semigroups (continued)

(Tilson). Let

be a relational covering equipped with a parametrization (9, a, p). QY # 0) then

(8.10)

y

X<@OY

where @ is the derived ts of p relative to the given parametrization. Proof. Define the partial function

Clearly e is surjective. Let s E S and let w E 52+ be such that s = wa. We assert that s is covered by the transformation (f,w p ) with

f : P+S@

Pf

as required. This proves (8.10)

=

(P9

WY

P')

I

We continue the analysis of the relational covering (8.1) with parametrization (52, a, B). For each p E P we define

9,+= ( w 1 w

E

D+,pw

c

p}

Clearly LIP+ is a subsemigroup o f 9+.Since ppw c pwp, it follows that

8. The Derived ts and the Trace of a Covering p p w c pp for all w

E Qp+.

79

Thus w defines a partial function

(8.11)

PP, -Pv

w:

There results a ts Tr,

=

(Pv,S,)

with S, composed of all the partial functions (8.11) for all w E Q,+. This ts is called the trace of p at p . T h e set of ts’s (Tr, 1 p E P} is called the trace of p. We emphasize again that these definitions are relative to a given parametrization. I t is useful to note that if X is complete and p p f 0 then Tr, is complete. EXAMPLE 8.2. Let p: G

--f

H be a surjective morphism of groups,

and let

G 4 H 6’

be the associated relational covering equipped with its natural parametrization as discussed in Example 8.1. Let N = ker p. Then for each h E H we have Tr, = (h’N, N ) where h’p = h. Thus it easily follows that Tr,

= ker p

for all h E H

We now compare the trace of p with the derived ts @ of p, both defined using the same parametrization of p. Given p E P we define the ts @, contained in @ as follows. T h e underlying set of @, consists of all pairs (4, p ) with q E p p , the transformations of @, are defined by the transformations of @ of the form ( p , w , p ) with w E Of, p w c p . If we identify the pair (q, p ) with q we then find Tr, w @,

(8.12)

THEOREM 8.2. (Tilson’s Trace Theorem).

Let

X 4 Y be a relational covering with

Qy

# 0, and let X be the least weakly closed

Ill. Transformation Semigroups (continued)

80

class in TS containing the trace of y relative to some parametrization. Then

X
<

(8.13)

Qie

c

Qip

with Qip defined above. Indeed if (4, p ' ) E Q, then (4,p')e is either empty or has the form (qw, p ) and qw E p y . Thus Qae c Qap. Every non-empty transformation of Qie must have the form ( p , u, p ) with u E 9+,pu = p . These transformations are present in Q i p . Since 0 c e the empty transformation of Qie also is contained in some transformation of Q i p . This proves (8.13). Formulas (8.12) and (8.13) jointly imply QeE

x.

Next assume e = 0. Then Qie = 0 . Since S, # 0 it follows from the construction of Qi that 9 f 0. Consequently there exists w E 9+such that wp is an idempotent. Since also Q y # 0 there exists p E Q y such that p w c p . Thus Tr, must contain some transformation and consequently 0' c Tr,. It follows that 0' E X and thus Qie E X I Q y = 0. Indeed if X = Y = 0' then Q y = 0 implies Qx = 0 and X 4 Y in this case already implies X < Y.

Note that the theorem fails if

Qi = 0. Also observe that

EXERCISE 8.1. Show that if Y = 1' then Qi M X . Show that if Sy = 0 then 12= 0 and S, = 0. Show that in all other cases 0 E S,. 9. The Delay Covering

Let S be a semigroup. We shall denote by 9 the alphabet whose letters are the elements of S. There results a morphism a: 9 + + S

81

9. The Delay Covering

which maps each letter of Q into itself. In order not to confuse a word w in 9+with its image w a in S, we shall write w = (sl, . . . , sn) if w has length n and si is its ith letter. Thus wa = s1 . . . sn. For brevity, we shall write E for wa. PROPOSITION 9.1. Let S be a non-empty semigroup. For each word w E Q+ such that I w I 2 card S, there exists an idempotent e E S and a factorization w = uu, I u 1 2 1 , such that iie = k. Proof. Let I w 1 = n 2 card S. For each 0 < i 5 n we have a factorization w = u p i with I uiI = i. If the elements iii are all distinct then n = card S and one of the elements iii must be an idempotent e of S. Thus Cie = iii. If iii = t i j for some 0 < i <j 5 n, then writing u j = uiz we have iiiI = iii. This implies z i i f m = iii. Since for some power m, Zm = e is an idempotent, we obtain iiie = iii I PROPOSITION 9.2. tents of S. Then

Let S be a semigroup and let E be the set of idempo-

S"

=

SES

for all n 2 card S. Proof. T h e conclusion is clear if S = 0. Suppose S f 0 and let s E Sn. Then s = 5 for some w E Q+, I w I = n. By Proposition 9.1

there is a factorization w = uv and an idempotent e of S such that iie = 1. If w # 1 then s = fie6 E SES. If 5 = 1, then s = zie = Gee ESES Let X = (Q, S) be a ts and let n 2 1. We consider the set Qn =

((4,

W)

For each s E S

I Q E Q, w E Q", I w I < n } u {W I w E Q", I w I =

D we define the transformation

by setting (4, w)s"=

{ [Z,:;

ws" = ( t z ,

. . . , tn, s)

i f IwI < n - 1 if

IwI=n-l

if w

=

(tl, ..

. , t,)

=

n}

111. Transformation Semigroups (continued)

82

+

In this definition (w, s) denotes the word of length I w I 1 obtained from w by concatenation with s. T h e functions s" convert Q, into a complete Q-module and define a complete ts X n = ( Q n , Sn) PROPOSITION 9.3. X , E

[z].

Proof. Consider an arbitrary non-empty finite set R. For each define the function

E

H

E

R.

7

P : Rn --t Rn (Y1,

. . . , 7,)P

= (r2,

. . . ,r,,

Y)

There results a cts Y, = (R", T,) with T, generated by P with Clearly Y , = R and also

Y,+l c Y ,

O

7

R

[z]

Consequently, it follows by induction that Y , E for all n 2 1. We now choose R to be the disjoint union R = Q v S v 70 and define 9: Q n - + R n (q, s l , . . . , ( ~ t1

Then 9: X ,

-+

.*

* 9

sn)9

=

(yo,

== (sly

. . . , 7 0 , 4, $ 1 , . . ,s i ) > sn)

* * *

Y , is an injective function and therefore X n < Yn

I

We note that X , is only vaguely related to X as in the definition of

X, , Q and S are used only as sets. The semigroup structure of S and the action of S on Q will however be used to define the nth delay covering

XdXn *tI

8,:

Qn+

Q l w l I n - 1

w8,=QE

if

lwl=n

The fact that the relation 8, is surjective is clear. T h e inclusion 6,s c $6,

9. The Delay Covering

83

is easily verified and shows that 6, is a relational covering. Observe that if S = 0 then 52 = 0, X , = Q, and 6, is the identity. There is a natural parametrization associated with this covering, namely, (52,a, /?) with Q and a as used above and with s/? = s^ for all s E 52. I t is with respect to this parametrization that we consider the trace of the covering 6,. First we consider the elements of X , of the form (q, w ) , q E Q, w E a*, 1 w I < n. There is no transformation t E S , such that (q, w ) t c (q, w ) . It follows that

Next we consider w = ( s l , . . . , s,) E Q,. Let t E S, be a transformation such that wt = w. Write t as a composition t = tAl . . . f, with t , , . . . , t , E S. From the definition of X , it follows that wt is the terminal segment of length n of the word (sl, . . . , s,, t , , . . . , t,) in Q+. Thus wt = w holds iff

Consequently if we define Rw =

{X

Ix

E

52+, w x E Q+w}

we find that the trace

Tr,

=

(!a,SW)

is represented by (Qg,R,). PROPOSITION 9.4.

If n 2 card S,

there exists an idempotent e E

then for each w E Qf,

I w I = n,

S such that

Proof. Since I w I 2 card S , Proposition 9.1 applies. From all the factorizations w = uv that satisfy the assertion of Proposition 9.1, choose the one with I v I minimal, and let e be the idempotent such that iie = Q.

Let x E R,. We show that v is a terminal segment of x so that xv-l is a word in 52". Indeed, since wx E Q+w, we have wx = yw = yuv for some y E Q+, so either v is a terminal segment of x or vice versa.

84

Assume 1 v I > I x

Ill. Transformation Semigroups (continued)

I. Then

v = zx for some z E

9+and

w x = yuzx

Cancellation of x yields w = yuz with j5ie = jjii and 1 z I < I v I, a contradiction to the minimality of I w I. Therefore v must be a terminal segment of x. Now w x = yuv implies wxv-1 = yu

(9.2) Since rie = zi, it follows that

with

v: "=

Qe-QZ

{?

if q E Q i i otherwise

Note that Qii = Qiie c Qe and that Qiiv = QE= QZ, so that v is well defined and surjective. We assert that each transfwmation x E R, of Tr, is covered by xq E eSe. Indeed if q6 E Qe then by (9.4) qiifpn = q w x = px-

= qzi(q)B

Since further by (9.3) and (9.2) qzi(xr]) = q i i a Z - l e = pTiZ5-1 = qJii E Qii

9. The Delay Covering

85

it follows that

qzqn = qC(xq)@ = qC(xq)q Thus q f c ( q ) qas required

I

Proposition 9.4 and formula (9.1) combine to give THEOREM 9.5. (Tilson's Trace-Delay Theorem). Let X = (Q, S ) be a ts with S # 0 and let X d8* Y be the nth delay covering of X with n 2 card S. Then for each q E Q y there exists an idempotent e E S such that Tr,<Xe I COROLLARY 9.6. Let X be a weakly closed class in TS containing 1, let X E LX and let X Qa, Y be the n-delay covering of X with n 2 card S, . Then Traced, c X I

As an application consider the case X = [l']. T h e conclusion above asserts that card(@,) 1 for all q E Q,. Thus d, is a covering rather than a relational covering. Thus X < X , and since, by Proposition 9.3, X, E it follows that X E We thus obtain

<

[z]

COROLLARY 9.7. L[1']

[z], c [z]

I

We shall see in Chapter IV that this inclusion is an equality. EXERCISE 9.1. Let X = (Q, S ) be a ts with S # 0 and let E be the set of idempotents of S. Show that

(9.5 1

QSn = QES

2 inf{card Q, card S}. Further, i f 1' < X show that (9.5) holds for all n 2 inf{card Q - 1, card S}. Hint: Consider the chain of subsets

for all n

QSi, 0 5 i. EXERCISE 9.2. Show that the covering

Tr,

<x,

of Proposition 9.5 is proper. Conclude that

ST,,,< , eSe.

86

Ill. Transformation Semigroups (continued)

References

The material of this chapter is mostly new or folklore. Tilson’s results of Sections 8 and 9 are hitherto unpublished. T h e delay covering of Section 9 was implicitly considered by Stiffler (1973) and explicitly and independently by Brzozowski and Simon (1971). T h e proof of Proposition 9.4 was inspired by an argument in Brzozowski-Simon. J. A. Brzozowski and Imre Simon, Characterization of locally testable events, IEEE Symp. Switching and Automata Theory 12th (1971). Price Stiffler, Jr., Extension of the fundamental theorem of finite semigroups, Advances Math. 11 (1973), 159-209.

CHAPTER

Iv Primes

Whenever one has a multiplication and (partial) order one can always define primes by saying that p is a prime whenever p 5 x y implies p 5 x or p 5 y . Further, one can define ( p ) = {x I p 5 x does not hold}. Applying this procedure to TS with order given by and wreath product as multiplication, we obtain the notion of a prime ts. I n this chapter the prime ts’s are enumerated, and the classes ( X ) , where X is a prime, are studied in some detail. These turn out to be some of the most “interesting” closed classes. By interesting we mean that these classes occur in various contexts and admit several seemingly unrelated descriptions.

<

1. The Exclusion Operator

For any ts X , the exclusion of X is defined as follows

<

<

If X It: Y and Y’ Y , then we also have X Y‘. Thus Y E ( X ) implies Y’ E ( X ) for all Y’ Y. Consequently ( X ) always is a class. Clearly X X‘ implies ( X ) = ( X ’ ) . Thus the class ( X ) depends only on the equivalence class of X and not on X itself.

-

<

PROPOSITION 1.l.

(X’)

=L(X)

Proof. By Corollary 111,5.2, the relation X‘ < Y is equivalent with X ‘ < Ye for all idempotents e in Y . This in turn is equivalent with X < Y e ,i.e. with Ye E ( X ) . Thus Y E ( X ’ ) iff Ye E ( X ) for all e I

IV. Primes

88

COROLLARY 1.2. PROPOSITION 1.3.

If X is a tm, then the class ( X )

is local

If the ts X has no sinks, then (X).

Indeed, assume Y O 4 (X), i.e. X and thus Y 4 (X) I

X
c

I (X).

< YO. Then, by Corollary 111,2.2,

Given any family X of ts’s we define

PROPOSITION 1.4. Let

TG’be the f a m i b of all tg’s Y = (P,G) with

G # 1. Then

(TG‘)= [TI Proof. Let X E [z’] and assume that Y < X with Y as above. Since Y is complete, it follows that G < Sx.However X is aperiodic and thus, by Theorem 111,7.6, all groups in Sx are trivial. Thus, since G is a quotient group of a group in Sx ,it follows that G = 1 contrary to assumption. Thus [TI c (TG‘). T o prove the opposite inclusion assume that X = (Q, S)is in (TG’). Let G be a group in 5’ with e as unit element. Then (Qe, G) is a tg and (Qe, G) < X . Consequently (Qe, G) E (TG’) and thus G = 1. It follows from Theorem III,7.6 that X E [TI I

2. Primes

A ts X is called a prime if it satisfies

(2.1) If X < Y

0 2 ,

then

X < Y or X < Z .

This is equivalent with requiring that X Y o 2, i.e. with

X
and

X

2 imply

(2.2) (X) is a closed class. If X and X’ are equivalent (i.e. if X <X’ and X’ < X ) , then (X) = (X’). Thus X is a prime iff X’is one. THEOREM 2.1.

Up to equivalence, the prime ts’s are the following:

(2.3) all divisors of 2, (2.4) the cyclic groups Z , of prime order p > 1.

2. Primes

89

The proof is deferred to Section 3. For the purpose of the proof and also for convenience in later work, it will be useful to enumerate the divisors of ? (up to equivalence there are 22 of them!) and arrange them in some ordered manner. The following seven ts's will be called the main divisors of 2 :

0, 1, 2, E, C, F,

2

The first three have no transformations at all: E, C, and F are as foIlows

E: 1 5 0

c: 1

2 0 3 .

F: 1 s O Thus C = (2, 0). The tm divisors of 2 are obtained from the main divisors by the dot-operation. They are

0 , l', 2', E , c',F', 2 The remaining eight divisors of 2 are called intermediate and are as follows (1, el, (2, el, 1 v I*, 1' v 1-

We observe the following inclusions

2

E

c

1

(1, e)

1-

(2, e)

1 1-

1' v 1-

2-

i =0,l Ei c E,, c E , c c c, c C'

IV. Primes

90

We recall that we only enumerate the divisors of 2. up to equivalence. It is this that makes the list moderately short. For instance, the ts

is equivalent with 2 since a7 = 1. I n Section 4-7, we shall calculate the closed classes ( P ) for the main divisors P of 2. and also for some (but not all!) of the other divisors of Finding (2,) where p is a prime is easy. Since Z , T, we have 2' E (2,). If G is a group, then Z , G iff p divides the order of G. Thus Corollary III,7.2 yields

<

<

(2,)

=

[T, G

z.

simple of order relatively prime to p ]

Since every simple group of odd order is cyclic (Thompson-Feit Theorem), we have (zZ)= [T,Z , for p prime, p > 21 PROPOSITION 2.2. If P is one of the primes F, F', each ts X the following conditions are equivalent :

2, T , Z , ,

then for

x

0) E (P) (ii) XO E ( P ) (iii) Xc E ( P ) (iv) sx E ( P ) Proof. (i) 3 (ii). This follows from Proposition 1.3, since P has no sinks. (ii) * (iii). Obvious. (iii) * (iv). This follows from the inequality Sx = S,, Xc(k). (iv) e-(i). Since 2' and C are in ( P ) , the conclusion follows from Proposition III,1.2

<

3. Proof of Theorem 2.1

We begin with the following very useful PROPOSITION 3.1. (Lifting Lemma).

x
Let X and Y be ts's such that

3. Proof of Theorem 2.1

91

and

x
or

z,,

X=

p prime

Then there exists a ts X' such that

X ' C Y,

X"X

Proof. First assume X = Z , wherep is a prime. Since Z p is complete S y . By Corollary III,4.6 there the inequality Z p Y implies Z, exists a group G in S y and a surjective morphism G + 2,. Consequently, the order of G is a multiple of p and by general fact from group theory, G contains an element g or order exactly p. Let e be the unit element of G. Replacing Y by Yewe may assume that e is the identity transformation of Y. Since g # e and gp = e, it follows that g: Q y + Q y is a function other than the identity. There exists then a state q E Q y such that q f qg f 0. Then the states q, qg, . . . , qgp-l must be all distinct because p is a prime. These states together with the transformations e, g , g2, . . . , gp-I form the required ts X ' c Y . Next we must consider the case X p . The proof has to be conducted case by case covering all the possibilities for X (up to equivalence). We shall limit ourselves to the explicit consideration of the cases X = El, and X = 2.. These two are typical in that they represent all the peculiarities. The remaining cases are treated quite similarly. Let X
<

<

<

where 0, 1 are the two states of X . We first consider the case X = El,. Thus X is the ts

We choose transformations ij, I, 8 of Y covering q, E , a. Since 7 and E are idempotents, we may choose i j and F to be idempotents. Let d = ?jag. Choosing a E A arbitrarily, the diagram

gives the required X ' since uij E A, a d E B.

IV. Primes

92

Next we consider X = 2. Thus X is the tm given by

with transformations u, t,and 1. We choose transformations 5, i, and i of Y covering u, z,and 1. Further we may choose 5, i, and i to be idempotent. Replacing 5 and t by iai and iii, we may assume

ia =

5=

ai,

i t = i = Ti

Choose n > 0 so that 2 = (3). is idempotent. Since 57 covers ur = z, it follows that 2 covers t. Further 52 = 3. For any a E A, consider

Since a2i= a?, a2rii = a25, a2 E A, a25 E B, it follows that we have a copy of Z. in Y We can now prove one of the assertions of Theorem 2.1, namely that all divisors of 2 are primes. Let then X 2 and X Y o 2. In view of Proposition 3.1 we may assume that X c Y 02.Leaving out the trivial case when card Qx < 2 we thus have

<

QX = ( ( P o ,

qo),

(PI

9

<

q1))

with p o 9 P 1 E Q Y , q o , q1 E QZ and ( p o ,qo) # ( p , , q l ) . For each s E SX we have a transformation (fa, f ) of Y o 2 such that s c (fs ,5). If qo # q l , then the states qo, q1 together with the transformations i yield a sub-ts X ' of 2 equivalent with X.Thus X 2. We may thus assume qo = q1 = ql and consequently Po #pl . Then the states po ,p 1 with the transformations qfs yield a ts X" c Y such that X - X". Thus X Y as required. The next step is to prove that 2, , with p a prime, is a prime ts. Assume Y o 2. By Proposition 3.1 we may assumq that 2, N X, then 2, X c Y 02.Let then

<

<

<

be the p states of X and let (f,s) be a transformation of Y o 2 such that

(3.1)

(4i 9 y i ) ( f 1

S) =

(qi+lp

~i+l)

3. Proof of Theorem 2.1

93

for 0 5 i < p . From (3.1) we deduce (34

YiS = T i f l

for 0 5 i < p . If the states y o , . . . , Y ~ -of~ 2 are all distinct, then we obtain 2, c 2. If r o , . . . , Y ~ are - ~ not all distinct, then (3.2) and the fact that p is a prime easily imply y o = yl = . . . - Y ~ - ~ Let . Y = yo a n d l e t h = rf E Sy.From(3.1)andthefactthatr = r i f o r a l 1 0 silp we deduce qih = qi+l

i < p . Since the states ( q i ,ii) = ( q i , Y) are all distinct for for 0 I 0 5 i < p , it follows that qo, . . . , qp--l are distinct. We thus obtain 2, c Y. Consequently Z, is a prime ts. We now conclude the proof of Theorem 2.1 by showing that there are no other prime ts’s than those listed there. Let then X = (8,S) be a prime ts. Applying the Krohn-Rhodes decomposition we have X < X I o . . . o X , with X i = or X i a simple group dividing S . Since X is a prime, we have X X i for some 1 5 i 5 n. If X i = 2, we are done since then X ?. There is left the case when X i = G S and G is a simple group. Thus X G S and G is a simple group. By Corollary 111,4.6, there exists a group H in S and a surjective morphism y : H -+ G. Let e be the unit element of H. Then (Qe, H ) is a tg and (Qe, H ) c (Qe; eSe) = (Q, S ) G

<

<

z.

<

< <

<

<

Since (Qe, H ) is complete, Proposition I,5.5 implies H G. Consequently, card H = card G and y is an isomorphism. Thus we may assume G = H c S . Choose g E G, g # e. There exists then a state qo E Qe such that qo # qog f 0. Consequently, card qoG > 1. Since G is simple, the action of G on qoG is faithful and therefore (qoG,G) is a transitive tg. This yields the inequalities

-

<

Since (Q,S ) is by assumption a prime, we obtain (Q, S ) (qoG,G). We have thus shown that ( Q , S ) (qoG,G). Thus we may assume that X = ( Q , G) is a transitive tg with G simple. We are now in the situation described in Exercises 142.1 and 2.2 and thus we may assume X = (G/H,G ) where G is a simple group, H is a subgroup of G, and H # G.

IV. Primes

94

Let K c G, K f G be any subgroup, and consider the tg Y = ( G / K , G). Thus G Y ( k for ) some integer k. Since X G, it follows that X Y ( k )Since . X is a prime, we deduce that

<

<

<

(GI% G )

<(G/KG)

By Exercise II,2.4, this implies (3.3)

card K

divides card H

Now let c = p;1

...pF

the prime decomposition of c = card G. Assume that k > 1, i.e. that c is not a prime power. Let Si be a pi-Sylow subgroup of G for 1 5 i 5 k. Then Si f G and card Si = pqi. I t follows from (3.3) that p;i divides card H and thus card H = c. This proves H = G, contrary to assumption. It follows that k = 1, i.e. G is a p-group for some prime p. Since G is simple, it follows that G w 2, and H = 1. Thus X w 2, I 4. The Low Primes

We begin with formulas (4.1) (4.2) (4.3 1 (4.4) T h e first three equations are obvious. T o verify (4.4) consider a ts X = (Q, S). Then X E ( E ) holds iff qs c p for all s E S and q E Q. This is equivalent with s c 1Q . Thus X E ( E ) iff X c Q.. This proves (4.4). Next we consider the dotted primes 0', l', 2', and E'. T h e formula (4.5 1

(0')

= L(0) =

[2]

is obvious. T h e class

(1')

= L(l) = L[O']

95

4. The Low Primes

was considered in Example III,6.1 and is the class Nil of all strongly nilpotent ts's. Thus

(1')

(4.6)

= Nil

Observe that this class contains no complete ts's. Next we consider the class

(2')

= L(2) = L[l']

From Corollary III,9.7 we have L[1'] c c (2'). Thus

2 E (2') it follows that [2] (4-7)

(2') =

Let X = (Q, S ) be a ts. Then potents e in S. We thus obtain

Since, however,

[2]

X E L[l'] iff card Q e 5 1 for all idem-

(2') iff all idempotents of X are subconstants.

(4.8)

X

(4.9)

A complete ts X = potents e in S.

E

[z].

(0, S ) is

in (2') iff Se = e for all idem-

(4.10) A semigroup S is in (2') iff Se

=e

for all idempotents e in S.

Next we consider the class = L(E) =

(E') We have X E L[2'] iff X , subidentity. This implies

E

[2'], i.e. iff each transformation of X , is a

(4.11)

X

(4.12)

A complete ts X potents e in S.

(4.13)

A semigroup S is in ( E ' ) iff eSe

E

( E ' ) iff ese c e for all s E S, and all idempotents e.

EXERCISE 4.1.

=

(Q, S ) is in ( E ' ) iff eSe = e for all idem-

Show that a ts X

;sf s1

for any sl, . . . , s, E S where n

IIL9.1.

L[2']

=

=

e for all idempotents e in S.

(Q, 5') is in the class (1')

= Nil

. . . s,=e

2 inf(card Q, card S ) . Hint: Use Exercise

IV. Primes

96

EXERCISE 4.2. Given an integer k >_ 1 let Nilk be the set of all ts's X such that either S , = 0 or SXk= 0. Establish the following facts:

(i)

u Nilk

= Nil.

k21

x

E N i l k , then Y o (ii) If (iii) Nil, # Nilk+,.

x E Nilk for

Deduce from the above that the class (1') EXERCISE 4.3.

all Y.

is not finitely generated.

Establish the following facts:

(i) ((1,O)) = [O'I u (0.). (ii) ((2,O)) = [l'] u (0.). (iii) (1 v 1') = [l'] u (1'). (iv) X E (1' v 1') 8 either Qx e c @ f o r all idempotirnts e.

=

0 or there exists q E Qx such that

5. The Primes C and C'

The study of the prime C is carried out using a construction which is of independent interest. Given a ts X = (9,S ) and given q l , q2 E Q , we define q1 2 q2 if either q1 = q2 or q2 E qlS. I f both q1 2 q2 and q2 2 q l , then we write q1 qz. Note that q1 q2 signifies that q1 and qa are in the same transitivity class of X.We shall write q1 > q2 if q1 2 q2 but q1 + q2. Given a ts X = ( Q , S ) , consider the set R of all sequences

-

-

q = (41, * . 91,.**,qkEQ,

9

4k),

21

Q1>Q2>**.

>qk

For each s E S define the partial function

S: R - + R

("

qs"= (411 * * * qkt qks) (419 * * * 9 qk-19 qks) f

Consider the ts 2 = ( R , T)with with s E S. We have

with 9,: R

-

if qks = 0 if Qk > 4ks if qk qks

-

T generated by the transformations S

x - 0 P

Q defined by qq = qk . Clearly S covers s.

5. The Primes C and C'

X

PROPOSITION 5.1.

0 # qt

97 E

(C)

=

if and

q't

only

implies

if

the t s

2 is injective, i.e.

q = q'

for all q, qf E R, t E T. cannot contain C which is not injecProof. If 3 is injective, then tive. Thus 2 E (C) and thus X 2 implies X E (C). Now assume that X E (C) and that 8 is not injective. Then for some s E S , the transformation f : R + R is not injective. Let

<

Clearly qks = pls. If qk > qp and p 1 > pls, then it follows immediately that p = q. Thus we may assume that p1 - p p , i.e. p 1 = plst for some t E s u 1,. Consequently qgt = plst = p l . Thus we obtain

Since

X E (C) it follows that

qk =

p l and thus qk = qkSt. Consequently

Since qf = p f and q k = p l ,it follows that p

= q contrary

to assumption

1

For any ts X the following conditions are equivalent:

PROPOSITION 5.2.

(i) X E ( C ) . Y for some injective ts Y . (ii) X X Y for some tg Y . (iii)

< <

<

(i) * (ii). Follows from Proposition 5.1 since X 2. (ii) 3 (iii). Let X Y with Y = (R, T ) injective. Then Y c (R,G) where G is the group of permutations of R. Thus X (R, G) and (R,G) is a tg. (iii) (i). Let X Y with Y a tg. Since Y cannot contain a copy of C we have Y E (C) and thus X E (C) I Proof.

-

<

<

<

IV. Primes

98

Proposition 5.2 implies that (C) = [TG]. However [TG] = [GI (Exercise II1,l.l) and thus

(5.1)

(C)

PROPOSITION 5.3.

=

[GI

X E ( C ) i f and only

if every idempotent

of X is

a subidentity. Proof. Let X = (Q, S ) . Assume X E (C) and let e E S be an idempotent. Let q E Q be such that qe # 0. Then in X we have

Since X E (C) it follows that q = qe and thus e c 1,. Conversely, assume that e c 1, for all idempotents e in S. If X $ (C), then C X and, by Proposition 3.1, X contains a configuration

<

with

p # q. Then for all n 2 1, we have

Since s" is idempotent for some n

2 1, a contradiction results I

COROLLARY 5.4. A complete ts X is in ( C ) i f and only i f X is a tg or S, = 0.

Indeed, if X E (C), then 1, is the only idempotent in S , and thus SAY is a group by Proposition III,4.3 I PROPOSITION 5.5. X E ( C ' ) if and only if for any idempotents e' and e in X , the relation e' E eSxe implies e' c e.

This follows from Proposition 5.3 since (C')

= L(C)

COROLLARY 5.6. A complete ts X is in ( C ' ) if and only for all idempotents e in X .

This follows from Corollary 5.4

I

I if X e is a

tg

5. The Primes C and C’

99

PROPOSITION 5.7.

(54

(C’)

=

[G, Z]

Proof. T h e inclusions [G, 21 c (C’) follow from 2 E ( C ’ ) and G c (C) c (C). T o prove the opposite inclusion, consider X = (Q, S ) E (c‘)= L(C). Let X -3, Y be the nth delay covering of X with n L card S. From Tilson’s Trace-Delay Theorem 111,9.5, we deduce

Trace p c (C) Let @ be the derived ts of p. Then by Theorem III,8.1,

X<@o Y Since by Proposition 9.3, Y E

[z],

it suffices to prove @ E (C). Let then

be a diagram in @. Utilizing the notation of 111,8, we then have

with q, qr E Q, p , p’ E Q y . Since rt = r’ it follows that t must have the form t = ( p , w , p ‘ ) with p w = p‘, qw = 4’. Since also r‘t = Y ’ , it follows that t = ( p ’ , w , p ’ ) with p‘w = p‘, q‘w = 4’. Thus p = p ‘ and pw = p . This yields the diagram

in the trace Tr,. Since, however, Tr, E (C) it follows that q Thus r = r’, proving that @ E (C) I EXERCISE 5.1.

Show that

=

q’.

X is aperiodic $7 2 is.

EXERCISE 5.2. Show that X tive and aperiodic Y .

E

(C)

n [FJ 2 3X

< Y for some injec-

IV. Primes

100

x

EXERCISE 5.3. 0

Show that

if X

and Y are injective ts's, then so is

Y.

6. The Primes

F, Z, F , and p

We now come to the study of the primes

P = F,

(6.1)

Z, F', Z.

The principal results consist of the formulae

We consider the primes

P

= 2,

c, 2',

C'

depending upon the choice of P in (6.1). We already know that

(6.6)

( P ) = [I], [GI, [Zl,

[G, 21

Thus formulae (6.2)-(6.5) can be replaced by the single formula

(6.71 Since P that

(P)

=

[C, (P)1

< P it follows that ( P ) c (P). Since also C E (P) it follows [C, (

0 1 = (P)

(P)

= [C, ( 0 1

Thus only the inclusion

(6.8)

needs to be established. From (6.6) it follows that ( P ) is completely generated and therefore [C, ( P ) ] also is completely generated. Since further C E [C, (p)], Theorem III,3.1, may be applied. It shows that X E [C, (p)] iff X I R E [C, (p)] for each transitivity class R in X. Consequently (6.8) follows from

6. The Primes F,

3, F', and 3.

PROPOSITION 6.1.

in

101

A transitive ts X is in ( P ) if and only if it is

(P).

Proof. T h e inclusion need to prove

P

XE(P)

c

P implies


c (P). Thus we only

implies

XE(P)

implies

P

or equivalently

P

cX

c

x

We now consider the four cases for P separately. If P = F, P = 2, and 2 c X , then X contains the distinct states p , q. Since X is transitive, X contains the ts

and thus F c X . If P = 2, = C , and C c X , then X contains the configuration

with p # q. Since X is transitive, we have qt Thus X contains the configuration

=p

for some t

E

Sx.

so that 2 c X . This takes care of the cases P = F, 2 with P = 2, C. T h e dotted cases P' = F',2 with = 2', C' now follow easily, Indeed, assume that X is transitive and X E (P). This implies X , E ( P ) for each idempotent e in X . Since X , is transitive (Exercise 111,5.1), it follows that X , E { P ) and thus X E (P) I COROLLARY 6.2. X tivity class R of X I

es

E

( P ) if and only

if X I R E Y for

EXERCISE 6.1. Show that a semigroup S is in = e for all e, s E S, e2 = e.

each transi-

(2) isf ese = e implies

102

IV. Primes

Derive (6.5) from (6.3) using the technique employed in the proof of Proposition 5.7. Show that (6.4) cannot similarly be derived from (6.2). EXERCISE 6.2.

7. Switching Rules

Our objective here is to establish the following equations:

[X,Z] = [XI [Z] [c', GI = [C']o [GI

(7.1)

0

(74

where X is any family of tm's. T h e wreath product of classes was defined in II1,l. A consequence is the equation

[C',G,Z] = [C']0 [GI0 [Z]

(7.3)

The equations above are easy consequences of the following "switching rules"

Z x <x

(7.1')

0

G o C'

(7.2')

0

Z(k)

< Cz' o G(2)

where X is any tm, k is an integer depending on X , G is a group, and 1 is an integer depending on G. Rule (7.1') is a consequence of PROPOSITION 7.1. (Stiffler).

If X = (Q, S ) is a tm, then

ZoX<XoB where R is the set 2 Q c xS. Proof.

Define

v: Q x R - + ~ x Q k f ,s)v = (qf, 4s) where q E Q, f E 2Qc,s E S. Given a transformation v = ( g , t ) of with t E S , g E 2Qc define the transformation k of X o R by setting

( q , f , s)k

=

(qs, g , t )

oX

7. Switching Rules

Thus k covers

ZI =

103

(g, t )

I

T h e proof of (7.2') requires some preliminaries. Given any (finite) set fJ consider the monoid Q union) with multiplication qq'

ql

= 4,

=

u 1 = Lo (disjoint

11 = 1

q = lq,

If card Q = n, then we also may write L, instead of L,. Note that L, is the reversal of the monoid i' which is usually denoted by U,,. Also note that L,, = K,I with K , as defined in Example II,9.3. PROPOSITION 7.2.

L,

E

[c'].

This follows from the observation that L ,

E

(F)

I

Switching rule (7.2') follows from Proposition 7.2 and the following PROPOSITION 7.3. (Stiffler).

For any group G

G o C' < L,

o

G'2'

104

IV. Primes

We define the transformation k of L,

Thus K covers (f,s)

o

G'2' by setting

I

We have seen that c' w U,. Since Proposition I,9.6 implies

is monogenic and complete,

2<2
Show that

> 1. What happens if n = 1 ? Explain why Un was dejined as 5' rather than ii'.

for n

EXERCISE 7.2.

Show that for any classes X and Y in TS

L(x OY) = L(X OLY) Derive the ruh

[X, any famiiy

~ O Y

Z]

=

[XI

0

[Z]

c

L([X]

x of tm's. Hint: Use [Z]

0

[Z])

= L[X]

= L[I'].

Show that i f X is monogenic, then X X E [ZJ.Hint: Use Exercise I,9.5. EXERCISE 7.3.

E

[Z', 21 implies

8. Summary and Open Problems EXERCISE 7.4.

105

Use Exercise 7.3 to show thut the ts

8. Summary and Open Problems

Among the 22 (up to equivalence) primes that divide seven were called the main series

0, 1, 2, E, C, F,

2. the following

Z

The prime tm's that divide ? are obtained from the main series by the dot operation, i.e. by adjoining the identity transformation. We tabulate the results for the exclusion classes:

(0) = 0 (1) = r0.1

(0') = 121 (1') = Nil

(2) = l1.1 ( E ) = [2'1

(2') = [ZI (E') [2', Z] = [2] 0 [Z]

(c')= [G,21 = [GI0 [Z] (F') = [C', Z] = [c']0 [Z] ( F ) = [c'l (Z) = [C',GI = [c']o [GI (T)= [C', G,Z] = [C']o [GIo [Z] (C) = [GI

We remark that, with the exception of 0 , all the generators listed in the brackets [ 3 are complete ts's and that in the left column they are complete tm's. We also note that inside the square brackets we can replace the ts's C' and 2 by the semigroups U, and 2. The left column in the table is complete. The right column contains two exceptional entries. The class (1.) consists of all the strongly nilpotent tds. We have seen that this class is not finitely generated (Exercise 4.2). It is also not completely generated since the only complete ts's that it contains are n for n > 0. The class (E') consists of all ts's X such that ese c e for all s E Sx and all idempotents e in S,. The inclusion [2, 21 c ( E ' ) holds because

106

IV. Primes

2 , 2 E ( E m ) .T h e opposite inclusion fails even if we restrict ourselves to complete ts's (Exercise 7.4). Most likely the class ( E ) is not finitely generated and also not completely generated ; however, these are open questions. In addition to the 14 primes considered above there are eight intermediate primes (among the divisors of 2'). Each such prime contains P and is contained in P' where P is one of the main primes. There is one prime between 1 and 1' and three between 2 and 2'. Their exclusion classes are considered in Exercise 4.3. In addition there are three primes between E and E and one prime between C and C'. We have no information concerning their exclusion classes ; they are likely to be very complicated. In Section 6 we have shown that the primes

P = F,

Z, F', T

are paired off with the primes

P = 2, c, 2',

C'

in the following sense: a ts X is in (P) iff X I R E ( p ) for each transitivity class R in X . In addition to the primes dividing 2' there are the primes 2, where p is a prime integer. The exclusion classes (2,) are (2,)

For p

=

[T, G

simple of order relatively prime to p ]

= 2,

(2,) = [Z., 2, for p prime, p > 21 T h e situation is somewhat reminiscent of the classification of simple compact Lie groups where there are four main series and five exceptional groups. EXERCISE 8.1. For n 2 1 consider the ts

and let En denote the same ts without the curved arrow. Show that if X where X is En,En*, Fn, or Fn*, then there exists a ts X' such that

X ' C Y,

X'MX


8. Summary and Open Problems

107

Assume that

EXERCISE 8.2.

X
X = En

and

Z E [2'] n ( C )

= En'

and

Z E [TI n ( C ' )

X = F,

and

Z E [?I n (2)

X

and

Z E [2'] n (2)

or

X or or

= F,'

Prove that

X
or

x
Hint: Use Exercise 8.1 followed by the projection n: Y o Z -+ Z ; examine all possibilities for X'n. EXERCISE 8.3. Let X be a closed class in TS.A ts X in X is said to be prime in X provided

X
YEX,

implies

X
or

ZEX

x
Show that if X iffinitely generated as a closed class, then X contains only a finite number of primes (up to isomorphisms). Use this to prove that the closed classes

(c), [TIn (Z), [TI

I 1

[Z.] n ( C ) [TI n (2.)

are not finitely generated. Show that [TI n ( C ' ) is the localization of n ( T ) is the localization of [TI n (2).

EXERCISE 8.4.

and that

[TI

[2']

n( C )

Show that if X E [TI n (2) or X E [TI n ( p ) , then the same holds for Xc. Conclude that the closed classes [TI n (2) and [PI n (T) are completely generated. EXERCISE 8.5.

IV. Primes

108

EXERCISE 8.6. For each integm n 2 0, define the classes

P, and R,

in TS as follows:

for all q E (i) (ii) (iii) (iv) (v) (vi)

X

E

P,

X

E

R,

iff qsn+l # 0

implies qs = q i . qsn+lt = q implies qs = q

Qx, s, t E Sx . Establish the following facts:

Po= ( E ) = [TI,R, = (F) = [c‘]. P, and R, are closed classes for all n 2 0. P, c P,+l and R, c R,+l for all n 2 0. UP, = [2‘] n (C),UR, = [TI n (2). En E P, - Pn-,and F, E R, - Rn-l for n 2 1. If X E P, - Pn.-l then En < X and if X E R, Fn < X.

-

R,-l then

References

The results of this chapter have been obtained jointly by the author and Tilson. In the proof of Theorem 2.1, the argument that 2, ( p prime) are the only prime tg’s is due to Jerrold R. Goodwin (unpublished). T h e calculation of the classes (T), ( U,), and (F’) (for complete ts’s only) appears in Stiffler (1973), .as do the switching rules of Section 7. Exercises 8.1 and 8.2 were inspired by Lallement (1969). Exercise 8.6 is due to Vijay Aggarval.

(z),

Gerard Lallement, Sur l’irrkductibilitk de certains monoides finis. C. R. Acad. Sci. Paris 268 (1969), 1312-1315. Price Stiffler, Jr., Extension of the fundamental theorem of finite semigroups, Advances Math. 11 (1973), 159-209.

CHAPTER

v Semigroups and Varieties

So far we have treated semigroups and monoids within the larger context of transformation semigroups and transformation monoids. There are situations when one is led to look at finite semigroups and monoids in their own right. This chapter contains the minimum needed in the subsequent chapters. 1. Varieties of Semigroups and Monoids

Let V be a family of semigroups. The usual definition in universal algebra says that V is a variety if it satisfies the following conditions:

(1.1) If S

E

V and T is a subsemigroup of S, then T E V.

(1.2) If S

E

V and T is a quotient semigroup of S , then T E V.

(1.3) The direct product of any family in V is in V. By the well-known theorem of Birkhoff such a variety is defined by equations that must hold for all elements of semigroups in V. Thus equations of the form xy = yx, XR = x, xyx = yxx, etc. give rise to varieties. Since we shall be concerned only with finite semigroups, it will be natural to restrict (1.3) to finite direct products. Such a family V is sometimes called a pseudovariety. Since these are our main interests here, we shall call them S-varieties. Thus an S-variety V is a family of semigroups that satisfies the following 109

V. Semigroups and Varieties

110

two conditions:

< S imply

(1.4)

S E V and T

T E V.

(1.5)

S, , S, E V implies S, x S,

E

V.

For any family X of semigroups we denote by (X),the least S-variety containing X.Clearly S E (X),iff S S, x . . . x S, with S , , . . . ,S, E X. We call (X), the S-variety generated by X. The largest possible variety of semigroups is the variety S of all semigroups. T h e smallest one is the empty variety. This one is not to be confused with the variety consisting of the empty semigroup. An M-variety is defined by the same conditions (1.4) and (1.5) applied to monoids only. For any family X of monoids

<

(XI,

(1.6)

=

(X),n M

is the least M-variety containing X, or the M-variety generated by X. The family G of all (finite) groups is an M-variety. Any M-variety contained in G will be called a G-variety. PROPOSITION 1.1. For any M-variety V either

.(i)

U, E V

or

(ii)

V is a G-variety.

Proof. If S is a monoid that is not a group, then by Proposition 111,4.3, S contains an idempotent e other than 1. Then the elements I, e form a submonoid of S isomorphic to U,. Thus U , S and U, E V I

<

Given any S-variety V, clearly V n M is an M-variety. Starting with an M-variety V, there are two procedures which lead to an S-variety V’ such that

(1.7) V’ n M = V The first procedure assigns to the M-variety V, the S-variety generated by V. Thus

Vs = {S I S E S, S

< T for some

T E V}

V,

1. Varieties of Semigroups and Monoids

111

Clearly Vs is the least solution of the equation (1.7). If V is a non-empty G-variety, then Vs is obtained from V by adjoining the empty semigroup to V. We say that Vs is an extended G-variety. S-varieties of the form Vs are called monoidal. PROPOSITION 1.2. For any S-variety

v

the following conditions are

equivalent : (i) V is monoidal. (ii) V = (V n M)s. (iii) V = (X), where X is a family of monoids. (iv) S E V implies s' E V. T h e proof is omitted

1

For instance, the variety Corn of commutative semigroups is monoidal. The variety N of nilpotent semigroups is not monoidal. A semigroup S is said to be nilpotent if either S = 0 or S has a zero 0 and Sn= 0 for some exponent n 2 1. T h e second procedure of passing from an M-variety to an S-variety is localization. For any M-variety V we define the class of semigroups (1.8)

LV = {SI eSe E V for all idempotents e E S }

If S < T , then by Proposition III,5.1, S, < Te,for some idempotent e' E T . Therefore eSe < e'Te'. Thus T E LV implies S E LV. Since LV is closed under direct product, it follows that it is an S-variety. We call LV the localization of V. Clearly LV n M = V and LV is the maximal solution of equation (1.7). Clearly Vs c LV. If V is an S-variety, then LV may again be defined by (1.8). Clearly LV = L(V n M). T h e operation L is idempotent, i.e. LLV = LV. An S-variety V is called local if LV = V. Although the difference between the notions of a semigroup and a monoid seems to be minor, the distinction is quite important in the applications. This will become amply apparent in Chapters VI and VII. In general, we shall give priority to M-varieties. Thus if an S-variety V is monoidal, we shall usually treat the M-variety V n M instead. PROPOSITION 1.3. Let V be an M-variety of commutative monoids and let X be the family of all cyclic monoids in V. Then V = (X)M.

112

V. Semigroups and Varieties

Proof. Let S E V and let T be the direct product of all the cyclic submonoids of S. Define the morphism p: T + S mapping each element of T into the product in S of its coordinates. T h e p is surjective and thus

a

sE(X)M

EXERCISE 1.l.Given S-varieties (or M-varieties) V and W,denote by V v W the least S-variety (or M-variety) containing both V and W.Show that if 1 E V and 1 E W then S E V v W ifl S < V x W with V E V and W E W.Show that ZjrV and W are M-varieties then

v s v w s = (VVW), I n the mixed case when V is an S-variety while W is an M-variety dejne V v W to be the least S-variety containing both V and W. Show that

v vw

=vvws.

EXERCISE 1.2.

Given any S-variety (or M-variety) V,dejne

V@ =

{solS E V }

Show that Veis an S-variety (or M-variety). Show that ZjrV is a G-variety, then V = Ve. EXERCISE 1.3.

that SI

E

Let V be a monoidal S-variety containing U , . Show

V ;f S E V.

2. Varieties Defined by Equations

Let

E* be the free monoid generated by the infinite sequence of letters

. . . , x,, . . . . Given

u, v that S satisjes the equation

xl,

E

8"and given a monoid S we shall say u=v

(or that the equation u = z, holds in S ) if u p = vp for every morphism q: SIC+ S of monoids. For a fixed pair (a,v), let V(u, v ) be the family of all monoids satisfying the equation u = v. Clearly if S, T E V(u, v ) , then Sx T E V(u,v). Next assume T S and S E V(u,v ) . There is then a submonoid S' of S and a surjective morphism y : S' T . Let p: 6" -+ T be any morphism. There exists then a morphism y : 5" + S' such that yy = p. Regarding y as a morphism 8" + S we obtain

<

---f

2. Varieties Defined by Equations

113

uy = vy. Consequently uq~= vg, and thus T E V(u, w ) . We have thus shown that V(u,v) is an M-variety. Clearly 1 E V(u, w ) . Given a sequence of pairs (Ui,V i ) E

8 + x 8",

i>l

we may consider the two M-varieties

n V(Ui, un m

V' =

Vi)

i=l

DOm

V"

=

V(Ui, Vi)

k = l i-k

A monoid S is inV' if it satisfies all the equations ui = wi . We shall say that V' is defined by the equations ui = vi. A monoid S is in V" if it satisfies the equations ui= wi for all i sufficiently large. We shall say that V" is ultimately defined by the equations ui = v i , i L 1. THEOREM 2.1. Every non-empty M-variety V is ultimately dejined by a sequence of equations.

-

The proof will utilize congruences in Z". We recall that a congruence in Z* is said to be finite provided the monoid PI- is finite, i.e., if there is only a finite number of congruence classes. The following two propositions of independent interest play a key role in the proof of Theorem 2.1.

-

PROPOSITION 2.2. Let Z be a j n i t e alphubet and

-

a j n i t e congruence in Z+. Then isjinitelygenerated, i.e., there exists afinite set W c Z xX Z* such that u v for all (u, v ) E W and that is the smallest congruence with this property.

-

N

-

is finite, there exists an integer k > 0 Proof. Since the congruence such that each congruence class contains an element w such that I w I < k. Define v, I u I I k I 'u I < k) {(u, v )

w=

1

-

Let = be the congruence generated by W. Clearly 2c = v implies u We shall now prove the opposite implication. We first prove

(2.1) For each w w 3 w'.

E

P,there exists w'

E Z" such that

I w' I < k

-

w.

and

V. Semigroups and Varieties

114

If I w I < k, then take w’ = w . If I w I 2 k , write w = uE with 1 u 1 = k. By the definition of k we have then (u, v) E Wfor some v E Z*, I v I < k . Consequently w = vE with I VE I < 1 w 1. Thus (2.1) follows by iteration. Now assume u

Since u‘ required

-

-

v. By (2.1) we have

v=v’,

u=u’,

v’, we have (u’, v‘)

)v‘I
IU’I
E

W and thus u’

E

v’. Thus u = v , as

PROPOSITION 2.3. Let S be a j n i t e monoid and Z a j n i t e alphabet.

Dejne u-sv E Z*whenever the equation u = v holds in S. Then -8 is a j n i t e for some congruence in Z* and Z ~ * / N is~ isomorphic to a submonoid of Sk’ jinite exponent k .

for u, v

-,

be the congruence in Proof. For each morphism q ~ :Z* --* S let Z* defined by u mIPv whenever ug, = vq~.Clearly Z*/Np% z*v c

and thus

s

-,is a finite congruence. Since

-S=n-, m

the intersection extending over all morphisms 9: Z* + S, and, since this family of morphisms is finite, it follows that mS is finite. Further Z*/.-&yc

nc*/-, I c

We can now prove Theorem 2.1. Let

s, < s, < . ’ < s,< .. *

be a sequence in

V such that

(2.2) For each S

E

V there exists n > 0 such that S < S,.

T o construct such a “cofinal” sequence in V it suffices to write a sequence T I , T,, . . . , T , , . . . which contains all the elements of V up to an isomorphism and then define S , = T , x . . . x T,.

2. Varieties Defined by Equations

115

B be the countable infinite alphabet composed of the letters let c", be the subalphabet {xl, . . . , x,}. For each integer n 2 1 consider the congruence w n= defined as in ProposiLet

xl,

. . . , x,, . . . and

mS,

tion 2.3. This congruence is finite, and thus by Proposition 2.2 it is finitely generated. Let W, c Enxx En* be a finite generating set for w n . We now regard En* as a subset of By and define

w=u

W,

n2l

We assert that W is a countable set of equations ultimately defining V . Let (u, v ) E W,, let y : 6" + S,, be a morphism, and let y : .Fey--+ 5, be the restriction of y. Since u mil v it follows that u wBZI and thus uy = v y . This implies uy = v y . Consequently S, satisfies the equations u = v for all (u, v ) E W,. For any S E V we have S S, for all n sufficiently large. Therefore, S satisfies the equations u = v for all (u, v ) E W , provided n is sufficiently large. Conversely, assume that S is a finite monoid satisfying the equations u = v for all (u, v ) E W, provided n 2 no. Choose n so that n 2 no and n 2 card S and choose a surjective morphism y : Eny-+ S. Thus (u, v ) E W, implies uy = v y . Since W , has generated -,, we obtain

<

u -,v

implies

uy = v y

It follows that y admits a factorization

and that y is surjective. Since, by Proposition 2.3, By/-, is a submonoid of SAk' for some finite exponent k, it follows that is in V and thus SEV I

%*I-,

COROLLARY 2.4.

Every M-variety generated by

Q

single monoid is

equational. and let ( v i ,ui), i 2 1 be a sequence Proof. Assume V = (So)M, ultimately defining V. There exists then an integer K 2 1 such that So satisfies all the equations vi = ui for i 2 k. Thus the same holds for all S E V. Assume now that S is a finite monoid satisfying the equations vi = ui for all i 2 k. Then S E V. Thus V is defined by the equations ui = v i , i 2 k I

V. Semigroups and Varieties

116

T h e arguments above apply equally well to S-varieties. We only need to replace Zwby Z+ throughout. We thus obtain THEOREM 2.1’. E v u y S-variety V containing the semigroup 1 is ulti-

mately defined by a sequence of equations

I

T h e M-variety 0 and the S-varieties 0 and {0} cannot be defined by equations. It should be observed that Theorem 2.1’ implies Theorem 2.1. Indeed, if V is the S-variety ultimately defined by the equations ui v i , i 2 1, then the same equations ultimately define the M-variety V n M. If ui = v i , i 2 1, are equations ultimately defining an M-variety V, then for some i we may have vi = 1 so that the corresponding equation is ui = 1. Such an equation may always be replaced by either or both of the equations uiy = y , yui = y where y is a letter not appearing in u i . With equations of the form ui = 1 eliminated, the equations can be used ultimately to define an S-variety V’. Clearly V’ n M = V. However, the S-variety V‘ will depend upon the choice of equations used to describe V. = 1

EXERCISE 2.1. Show that if the equations u i= vi ultimately dejine the M-variety (or S-variety) V, then the equations uiQ= vie ultimately define VQ. EXERCISE 2.2. In Proposition 2.3 assume that Z is countably inJinite. Show that Z”/-s is isomorphic with a submonoid of a product of a countable number of copies of S. 3. Examples of Ultimately Equational Varieties

In treating specific M-varieties or S-varieties, we shall not use Theorem 2.1 to obtain the ultimately equational description. We shall use whatever information we have about the variety to obtain a “nice” set of equations. T h e trivial M-variety consists of the monoid 1 only and is defined by the single equation x = 1. Also, the single equation x = -v can be used. T h e same equation can be used for the S-variety (0, l}. T h e equation x = x describes the varieties S and M. Some power of every element of a finite semigroup is an idempotent. EXAMPLE 3.1.

3. Examples of Ultimately Equational Varieties

117

Define an exponent of S to be any positive integer k such that sk is an idempotent for each s E S. For each n 2 1, let Z denote the least common multiple of 1, . . ., n. If K is an exponent of S, then s" is an idempotent for all n 2 k and for all s E S. EXAMPLE 3.2. T h e M-variety G of (finite) groups is ultimately defined by the equations xs=1, n>l

Indeed, if k is an exponent of a group G, then g" = 1 for all g E G and n 2 k. Conversely, any monoid satisfying xs = 1 for some n is a group. T h e equations above may be replaced by x5y

= y = yx",

n21

When this is done, we can use them as ultimate equations of an S-variety which is easily seen to be G s = {0} u G . If we use the equations y=yx5, n 21 to ultimately define G, then we obtain another S-variety that we shall denote by Q.This variety Q consists of all semigroups S satisfying the right cancellation law t,s = t,s 3 t , = t , for all t , ) t , , s E S. Since S is finite this is equivalent with the condition

ss = s

for all s E S . Such semigroups are called left simple (since the only nonempty left ideal in S is S itself). A detailed study of the variety Q will be made in Section 10. The S-variey LG is ultimately defined by the equations (yl'xyl'y = yl',

n 21

Indeed, given any S E L G , let k be the exponent of S. Then for all n 2 k and y E S, ya is idempotent and ~ " x Y "E Gut

118

V. Semigroups and Varieties

It follows that

(yfixyfi)"=y"

for all x , y E S

and n

2K

Conversely, if S satisfies the above equation for some n, then choosing y to be idempotent leads to the equation

(exe)" = e which implies S

E

for all idempotents e in

S

LG.

We recall that a monoid S is said to be aperiodic if all groups in S are trivial. T h e variety of all aperiodic monoids is denoted by A and plays an important part in the sequel. An ultimately equational description of A is supplied by Theorem III,7.6 and is EXAMPLE 3.3.

Xn+l = Xn

,

If we denote by Anthe M-variety defined by the single equation xn+l = xn, then A, c A, c . . . c A, c . . . and

A=

u An

n21

This gives one of the many hierarchies that can be found in A. This hierarchy is particularly interesting since it is invariant under reversal. T h e same equations can be used as an ultimately equational description of the S-variety As. Note that A, is the M-variety of idempotent monoids. T h e M-variety R consists of all monoids S such that

EXAMPLE 3.4.

the configuration

is present in S only when s

=

t. Thus

(3.1) S E R iff sxy = s implies sx = s. An equivalent statement is

(3.2) S E R iff sS = t S implies s = t. Indeed sS

=

tS is equivalent with sx

=

t and ty = s for some

y E S. This is equivalent with sxy = s where t = sx.

X,

3. Examples of Ultimately Equational Varieties

119

Define

R,

=

R n A,

Then

m

R= URn n=l

We assert that R, is defined by the equation

(3.3)

(XY)"X

= (XY)"

Indeed, suppose S E R,. Then (xy)"+' = (xy)"for all x, y E S. Thus (xy),xy = (xy)" and (3.1) implies (3.3). Conversely, assume that S is a monoid satisfying (3.3). Setting y = 1 yields xn+l = xn so that S E A,. If sxy = s, then s = s(xy)" = s(xy)nx = sx and thus S E R by (3.2). Thus S E R,. Since R = UR, it follows that equations (3.3) with n 2 1 ultimately define R. EXAMPLE 3.5. Define the M-varieties Jn =

R, n R,,@ = R n Re n A,

J=RnRe=

u Jn

n 21

Equivalently J, may be defined by the pair of equations

(3.4)

(XY)"X =

(XY)" = Y(XY)"

These equations considered for all n >_ 1 constitute an ultimately equational description of J. We now show that Jn may also be defined by the pair of equations

(3.5)

Xn+l

= Xn 3

(XY)" = (YxY

Indeed, assume (3.4). Setting y = 1 yields xn+l = xn. Further, (xy)" = y(xy)nx = (yx)n+l= (yx)".Conversely, assuming (3.5), we have

(xy)"= (xy)"+' = (yx)n+' = y(xy)"x This implies

(xy)"= yn(xy)nxn= y,(xy)nxn+1= (xY)". Similarly (xy)" = y ( x y p .

V. Semigroups and Varieties

120

It follows from (3.5) that

J1

is defined by the equations

x2 = x ,

xy

=y x

Thus J1 is the M-variety of idempotent and commutative monoids. Since (up to isomorphism) U , is the only cyclic monoid in J,, Proposition 1.3 yields

Ji = (uih Next we show

(3.6) S E J iff SsS = StS implies

s = t.

Indeed, assume that the condition stated in (3.6) holds for the monoid S. Since sS = t S implies SsS = StS and thus implies s = t, it follows from (3.1) that S E R. Similarly S E Re and thus S E R n RQ= J. Next assume S E J and SsS = StS. Thus atd = s, bsc = t for some a, b, c, d E S. It follows that abscd = s and thus s = (ab)=s(cd)"for all n 2 1. For sufficiently large n we then have by (3.4)

This proves (3.6). I n the language of the Green relations (see XI,l), J is the variety of monoids in which the 3-relation is trivial. Similarly, R consists of monoids that are *-trivial while Re consists of monoids which areP-trivial. EXAMPLE 3.6.

Consider the S-variety

V consisting

of all semigroups

S satisfying the condition ese = e +-es = e

(3.7)

for all idempotents e in S and all s E S. Exercise IV,6.1, asserts that this is exactly the variety (2) n S. Choose an exponent n for S. We assert that S satisfies (3.7) iff S satisfies the equation

(3.8)

(x"y)nx"

=

(xny)n

Indeed, assume (3.7). Then xnxn = xn

and

(xny)* = (xny)*(xny)n = ( ~ ~ y ) ~ x ~ ( x ? ) n

3. Examples of Ultimately Equational Varieties

121

Thus (3.7) implies (3.8). Conversely if (3.8) holds then ese

=e

implies

eses = es and thus es = (es)n = (ens)" = (ensyen = (es)ne = ese = e

Consequently (3.7) holds. We conclude that V is ultimately defined by the equations n 21

(xfiyy)"= (xfiy)"x",

The variety

V is monoidal because if S satisfies (3.7) then

so does 9.

EXAMPLE 3.7. The S-variety D consists of all semigroups S such that S e = e for all idempotents e in S. We show that D is ultimately defined by the equations (3.9)

yx" = xn,

n 21

Indeed, assume S satisfies (3.9) for some n 2 1. Then se = sen = en = e for any s E S and any idempotent e. Thus, S E D. Conversely, assume S E D. Let k be an exponent of S. Then (3.9) holds for n = K and thus also for all n 2 K. Notice that 5 is not required in these equations. Another ultimately equational description of D is (3.10)

yxl

. . . x,

=XI

. . . x,,

n 21

Clearly (3.10) implies (3.9). Conversely, assume S E D, S # 0 and let n 2 card S. Then by Proposition 111,9.2, S" = SES where E is the set of all idempotents in S. Thus x1 . . . x , = uev for some u, v E S, e E E. Consequently yx,

. . . x,

= yuev = ev = uev = x1

. . . x,

as required by (3.10). The equation (3.9) taken singly defines an S-variety D,. Clearly

D , c D , c ... C D , C UD, = D

...

(z),

The variety D, defined by the equation y x = x is the S-variety which consists of all semigroups of the form where Q is any finite set. Note that D is not monoidal, in fact D n M = (1).

8

V. Semigroups and Varieties

122

D consists of

all semigroups S satisfying eSe = e for all idempotents e. Using the same procedures as in Example 3.7 we may obtain the following ultimately equational descriptions of D EXAMPLE 3.8. The S-variety

(3.11)

xfiyxfi = X I ,

n 21

or

(3.12)

x1

. . . xjiyx, . . . xji = x1 . . . x,

Also note that D is not monoidal. EXERCISE 3.1. Show by an explicit example of a monoid with three elements that R, # J 1 . EXERCISE 3.2. Show that the equations

dejine the S-variety R,,. Show that the second equation alone dejnes an S-variety in V, such that

and that these inclusions are proper. Conclude that the S-variety R, is ultimately dejined by the equations (3.3). Hint: With Z = {u, T} utilize the semigroups Zf/.ZkZffor suitable integers k. EXERCISE 3.3. Show that the equation (3.5) dejines the S-variety J,,, and that these equations considered for all n 2 1 ultimately dejne the S-variety Js . EXERCISE 3.4. Let V be an M-variety ultimately defined by the equations u, = v, , n 2 l . Let

Replace the equation u,

(3.13)

= v,

by the equation

xfiy,x" . . . xfiy*,xfi = X f i Z 1 X f i

. . . xfiz,*xfi

where x is a letter diflerent from y l , . . . , y p n ,xl, . . . , zqn ( i f v,, = 1, then v, is to be replaced by x"). Show that the sequence of equations (3.13)

4. Semidirect Products

123

ultimately defines the S-variety LV. Show that the same conclusion is obtained by replacing xfi by x1 . . . x, where the x’s dtfer from all the y’s and X’S.

EXERCISE 3.5. Show that the S-variety N of nilpotent semigroups is ultimately defined by the sequence Xn

EXERCISE 3.6.

n21

=y”,

Show that the sequence of equations X1+A

=x

,

n 2 1

is the ultimately equational description of the M-variety of all monoids S such that S is the union of groups in S.

<

EXERCISE 3.7. Let S E J 1 . Show that S Uik’with 1 Hint: Consider the function v: S + 2s given by sp = sS.

+ k = card S.

4. Semidirect Products

Let S and V be semigroups. T o decongest the notation, it will be convenient to write V additively, without however assuming that V is commutative. A left action of S on V is a function SXV-v (s, v ) + sv satisfying the following conditions

for all s, s’ E S , v, v l , v, E V . Condition (4.1) expresses the fact that each element s E S determines an endomorphism of V. Thus a function S -+ End( V) is obtained. Condition (4.2) expresses the fact that the function Se + End(V) is a morphism of semigroups. Note that S has been replaced by the opposite monoid Se. This is due to the fact that, functions being written on the right, the composition in End(V) is opposite to that implied by (4.2).

V. Semigroups and Varieties

124

Given a left action of S on V we define the semidirect product V I S as follows. T h e elements of V I S are pairs (v, s) with v E V , s E S. Multiplication is given by the formula

(4.3)

(v, s)(v‘, s‘) = (v

+ sv’, ss‘)

We have

+ sv’, sd)(v”, s”) = (v + so‘ + (ss”v”, ss’s”) = ( v , s)(v’ + s’“‘, s‘”’) = (v + s(v‘ + s‘”‘), ss’srr)

[(v, s)(v‘, s’)](v‘’, 3’’) = (v

(v, s)[(v’,s’)(“‘,

s”)]

From (4.1) and (4.2) we deduce that the two results are equal and thus the multiplication in V * S is associative. Thus V I S is a semigroup. Note if the-action of S on V is trivial (i.e. if sv = a for all s E S, v E V ) , then V I S coincides with the direct product V X S. The semidirect product is not associative (see however Exercise 4.8) and therefore when dealing with iterated semidirect products, parentheses are needed. Given semidirect products V iI Si,i = 1, 2, we define a left action of S,XS, on V1x V 2 by setting (s1 I s2)(v1 9 v2)

=

(WI

9

s2vz)

With this definition we have the useful isomorphism

If S is a monoid and the action of S on V satisfies (4.5 1

lv

=v

for all a

E

V

then we say that the action is left unitary. If V is a monoid (with 0 as unit element), and the action of S on V satisfies

(4.6)

SO

=

0

for all s E S

then we say that the action is right unitary. If both S and V are monoids and the action of S on V is both left and right unitary, then we say that the action is unitary. I n this case V I S

4. Semidirect Products

125

is a monoid with (0, 1 ) as unit element. Indeed

(v, s)(O, 1 )

=

(0, l)(g, s)

=

+ so, s) (v,s) (0 + 1% = (v, (v

=

s)

s)

If V and S are groups, and the left action of S on V is unitary, then V # S is a group. T h e inverse of (v,s) is (TI, s)-1 =

(-s-lv,

s-1)

If V and S are monoids and the left action of S on V is unitary, then the following morphisms are defined: V / V # S I S B

vcf

= (0,l),

(v,s ) n = s,

s p = (0, s)

with a z = 1 and /?z= identity. This implies

V
S
Observe that n can also be defined in the semigroup case and the inequality S V S still holds provided V # 0.

<

EXAMPLE 4.1.

Let G be a group and K and H subgroups of G

satisfying the following conditions :

(4.7) K is an invariant subgroup of G. (4.8) K H = G. (4.9) K n H g

=

1.

This implies that each element g E G admits a unique factorization with k E K and h E H. Next we note the identity

= kh

khk’h’ = k(hk’h-’)hh’ Thus if we write ( k , h ) instead of kh and use additive notation in K , then

( k , h)(k’, h’) = ( k

+ hk’h-’, hh’)

This implies

G=K+H

V. Semigroups and Varieties

126

where the left action of H on K is given by

h *g

= hgh-l

This is the original motivation of the definition of the semidirect product. There are close ties between the semidirect product of semigroups and the wreath product of ts's which we shall now examine. Let X = (Q,S ) , Y = (P, 7') be complete ts's. Then x o

Y

=

(QXP, W )

where

W=SPxT

.>

and the multiplication in W is given by

(f,t x g , with h

E

S p given by

Ph

=

=

(h, t u )

P f f (Ptk

where multiplication in S is written additively. If we define the left action of T on S p by setting P ( t * g ) = (Peg for all g

E

S p and all p

E

P we find that h=f+t

*g

and thus

W=SP*T We have thus proved PROPOSITION 4.1.

For any complete ts's X and Y S'yay =

with k = card

sy

0

s y

Qy I

COROLLARY 4.2. For any semigroups U and V

suov - U'k' *r

with k

= card

V' I

v

4. Semidirect Products

127

For any ts's X and Y

COROLLARY 4.3.

s*yoy

with k

=

< sy x sy

card Qyc.

<

This follows from Proposition 4.1 and the remark that Sxoy SXcoYe1 Observe that if X is a tm, then the action of SI-on Slyk) is right unitary, while if Y is a tm, the action is left unitary. For any semigroup V we consider the complete ts V# = (V', V ) where Vz = V u 1 is obtained from V by adjoining a unit element (regardless whether V has a unit element or not). Clearly

v c V# c V' and V

< W implies V# < W#.

PROPOSITION 4.4. of S on V

For any semlgroups V and S and any left action

v x s <(V S)# < V# v s < S,,'*SP< *:

0

S#

S"#OS#

i*

If S is a monoid and the action is left unitary, then S# may be replaced by S. If V is a monoid and the action is right unitary, then V# may be replaced by V . If both V and S are monoids and the action is unitary, then

v+s
Proof. We shall only prove the first part since the remaining three

parts are proved quite similarly. Note also that the second line follows from the first, since all the ts's involved are complete. Thus we only need to prove ( V x S)# V# o S#. By identifying 1 with (1, 1) we may regard ( V XS ) u 1 as a subset of (V u 1)x (S u 1). With this convention we prove ( V x S ) # c V# 0 S#

<

For each

TI E

V consider the function

f v : S' + v

V. Semigroups and Varieties

128

We then verify that for each transformation ( 0 , s) of (V t S)# we have (0, s) c (f u , s) where (f v , s) is a transformation of V# o S# I EXERCISE 4.1.

Show that

wts=o=vw0, Further, show that V t 1 l v = v , for all v E V ) .

=

l*S=S

V if the action of 1 on V is left unitary (i.e.,

EXERCISE 4.2. Assume that the semidirect product V t S is a monoid with unit element (0, 1). Show that V and S are monoids with 0 and 1 as unit elements and that the action of S on V is unitary. Hint: Show that the function v -+ l v is injective and idempotent and hence is the identity. EXERCISE 4.3. For any monoid V ,show that both inequalities V < V I1 and V I1 < V fail i f the action of 1 on V is not left unitary (e.g., when V # 1 and the action is given by l v = 1 for all v E V ) . Deduce that the inequality V t S < V o S for monoids V and S may fail i f the action is not unitary. For a detailed discussion of V t 1 with non-unitary action see Section 10. EXERCISE 4.4.

Show that i f V is a group, then any action of S on V is

right unitary. EXERCISE 4.5. Show that any left action of S on V can be extended to a lejt unitary action of Sx on V so that V t S c V t S'. Similarly, show that the action can be extended to a right unitary action of S on VI and a unitary action of S' on VI. EXERCISE 4.6.

Show that with unitary action Z , t Z , is either

Z,x Z , or Z, depending on whether the action of Z , on Z , is trivial or not.

-

EXERCISE 4.7. Given semigroups S and V consider the free product V 0 S. In this product consider the least congruence such that s o v (sv) 0 s where sv is the result of a given left action of S on V. Show that V o Sl- w V t S.

-

EXERCISE 4.8. Given a semidirect product V I( T IS ) of monoids with unitary actions, define a lejt action of T on V and a left action of S on

5. Varieties V x W

129

V x T by setting

where 1 is the unit element of S and 0 is the unit element of T. Show that with the actions so defined

V * ( T * S ) = ( V # T )# S 5. Varieties V

*W

We shall now relate the notion of an S-variety with that of a semidirect product. Given S-varieties V and W, we denote by V x W the S-variety generated by all semidirect products V x W with V E V, W E W and with any left action of W on V. T h e isomorphism (4.4)implies that R E V x W iff R ' < V x W f o r some V E V , W E W . I n the same manner we define the M-variety V x W when V and W are M-varieties. I n this case one only considers semidirect products V x W in which the action of W on V is unitary. Similarly we shall have occasion to consider the S-variety V II; W in the mixed case when V is an M-variety while W is an S-variety (or vice versa). In this case one only considers semidirect products V x W in which the action of W on V is right (left) unitary. Contrary to what happens with the semidirect product of semigroups, the semidirect product V * W of varieties is associative. This is based on the examination of the connection between the semidirect and the wreath product. PROPOSITION 5.1.

Let V and W be S-varieties and let X and Y be

ts's. If SxeV then

SyeW

and

sxoyE v

x

w

If V is an M-variety, then X is assumed to be a tm, while M-variety, then Y is assumed to be a tm. Proof, By Corollary 4.3, Sx0y

< Skk)* SY I

if W

is an

V. Semigroups and Varieties

130

Let V and W be S-varieties and let X be a complete ts. Then Sx E V + W if and only ;f X < Y o Z where Y and Z are complete ts's such that S yE V and Sz E W . If V is an M-variety, then Y should be a complete tm. If W is an M variety, then Z should be a complete tm. If both V and W are M-varieties, then X , Y , and Z should be complete tm's. PROPOSITION 5.2.

Proof. We shall only consider the case when V and W are S-varieties. If X < Y o Z and S , E V,S, E W, then since X is complete, Sx < SFOz and by Proposition 5.1, SyOzE V # W. Therefore Sx E V # W. Assume Sx E V + W. Then Sx < V y: Wfor some V E V and W EW. We first note the inequality

X

< Qx*x sx#

given by the surjective function

(4, S)V

=

if s E Sx if s = l

qs q

Next, Proposition 4.4 yields

S,# Therefore

X

< (V

b

W).

< V#

< Q x ' x S x # < Qx'

o

0

W#

V# o W#

PROPOSITION 5.3. The associative law

(U +V) + w

=

u

*;

(V b W)

holds where U,V, W each could be an S-variety or an M-variety. Proof. There are eight cases to be considered. We shall only consider the case when U and V are M-varieties while W is an S-variety. Assume T E (U # V) lit W. Applying Proposition 5.2, we obtain T R o Z where R is a complete tm, Z a complete ts, and SR E U + V while Sz E W. Applying Proposition 5.2 again to R we obtain

<

T<XoYoZ

5. Varieties V

where

#

W

131

X and Y are complete tm’s, 2 a complete ts, and S,UE u,

Sy

EV,

SZE

w

I t follows that

T

< SXOYOZ = SXO(YOZ,

From Proposition 5.1 it now follows that Syoz E V * W.Applying Proposition 5.1 again we find S X o ( y o zE) U * (V * W). Thus T E U * (V * W). This proves the inclusion (U *V) iy W c U * (V * W).T h e opposite inclusion is proved similarly I PROPOSITION 5.4.

Let V be an M-variety containing U,. Then

w “V, = (W x v ) s ,

v, * w = (v * W),

for any M-variety W and

w # V S= w *v, vs * w = v x w for any S-variety W.

W y: Vs = (W x V), where W * Vs we have (W * V), c W * V,. T o prove the opposite inclusion consider W * V with W E W, V E Vs and with a right unitary action of V on W. Since V, is monoidal and U, E V it follows from Exercise 1.3 that VI E V. T h e action of V on W Proof. We shall only prove the equality

W is an M-variety. Since W * V

c

may now be extended to a unitary action of V’ on W (see Exercise 4.5), so we have W*VcW * V ~ E W * V

COROLLARY 5.5.

If V and W

aye M-varieties containing U,, then

PROPOSITION 5.6. Let V be an S-variety or an M-uaviety and let be an M-variety. Then

v # L W c L(V x W)

W

132

V. Semigroups and Varieties

Proof. We shall only treat the case when V is an S-variety. Let

R E V # LW. Then by Proposition 5.2 we have

R<XoY where X and Y are complete ts’s such that

S,EV,

SyELW

For each idempotent e in R we then have eRe 111, 5.4,implies eRe X o Yd

<X

o

Y and Corollary

<

for some idempotent d in S y . Since S y E LW it follows that Sy4 = dSyd

Thus by Proposition 5.2, eRe E V

#

E

W

W. Consequently R

E

L(V * W)

I

Show that in Proposition 5.4 the assertions Vs * W (V u: W), and Vs # W = V * W are valid without the assumption U , E V. Similarly in Corollary 5.5. Hint: Use Proposition 1.1 and Exercise 4.4. EXERCISE 5.1.

=

6. Varieties vs. Weakly Closed Classes

There are close connections between the notions of an S-variety and that of a weakly closed class in TS. PROPOSITION 6.1. If X is a weakly closed class in TS, then X n S is an S-variety and X n M is an M-variety. If further X is generated by tm’s, then X n S is a monoidal S-variety, i.e.

X n S = (X n M)s Proof. T h e subtle point of the proof is the discrepancy between the definition of the product in TS and in S. If S, T E X n S, the product in TS is

(S’xT’, S x T ) which covers

( ( S xT)’, S x T )

6. Varieties vs. Weakly Closed Classes

133

The latter ts is the ts corresponding to the semigroup S X T . Thus SX T E X n S. If X is generated by tm's and S E X n S then S < X for some tm X in X. Since S' X' = X it follows that S' E X n S. Thus X n S is monoidal I

<

For any family X of semigroups

PROPOSITION 6.2.

Iffurther X c M then

Proof. We recall that (X) is the weakly closed class in TS generated by X, while (X), is the S-variety generated by X. Set V = (X)s. Then clearly (V) = (X). T h e equality V = (V) n S is trivial. T h e second half follows similarly I PROPOSITION 6.3.

If X is a weakly closed class in TS,then LX n S c L(X n M)

If, further, 2'

E

X, then equality holds.

Proof. Let S E LX n S. Then S, = ( S e , eSe) E X for all idem-

potents e in S. Since eSe c S , , we have eSe E X n M. Thus S E L(X n M). Next assume 2' E X and let S E L(X n M). Thus eSe E X for all idempotents. Corollary I,9.7 now implies that 8,= (Se, eSe) E X and thus S E L X n S I T h e following example shows that the inclusion in the proposition above may be proper. Let X = [l']. If S E LX n S, then S, = (Se, eSe) [l']for all idempotents e E S. This holds iff Se = e for all idempotents. This variety is denoted by D. On the other hand, X n M contains EXAMPLE 6.1.

E

only the trivial monoid, and thus L(X n M) consists of all semigroups S such that eSe = e for all idempotents. This variety is denoted by b. Clearly D is a proper subvariety of b.

134

V. Semigroups and Varieties

PROPOSITION 6.4. Let

X and Y be weakly closed ctasses in TS. Then

(X n M)t (Y n M) c (XoY)n M If,further, X and Y both contain C, then

(Xn S) (Y n S) c (X Y) n S Q

Proof.

0

Let S E X n M, T

E

Y n M.Then by Proposition 4.4

St T <SOTEXOY Thus S t T E (XoY) n M. Next assume S E X n S, T

S

x

E

Y n S. Then by Proposition 4.4

T

< S# o T#

Since C E X and C E Y,Proposition I, 9.1, implies S# E X and T# E Y. Thus S x T E (X oY)n S I PROPOSITION 6.5. Let

X and Y be completely generated weakly closed

classes in TS. Then

(XoY) n S c (X n S) t (Y n S) If X is generated by tm's, then

(X oY) n S c (Xn M)t (Y n S) If Y is generated by tm's, then

(X oY) n S c (X n S) t (Yn M) If both X and Y are generated by tm's then

(X oY) n M

c

(Xn M)t (Y n M)

Proof. We prove only the second assertion as the other proofs are quite similar. Let T E (X o Y) n S. Then T X o Y with X E X and Y E Y. We further may assume that X and Y are complete and that X is a tm. Proposition I, 9.8, now implies Sx E X n M and SpE Y n S. Thus by Proposition 5.1

<

T

< S x o y E (Xn M)* (Yn S) I

7. Closed Varieties

135

Let X and Y be completely generated weakly closed classes in TS.If X and Y are generated by tm's, then COROLLARY 6.6.

(X n M ) c ( Y n M) = (XoY) n M If X and Y both contain C , then

(X n S) #(Y n S) = (X oY)n S

I

7. Closed Varieties

An S-variety V is said to be closed if T , S E V imply T any left action of S on T. Equivalently, V is closed iff

*SE V

for

V#VCV

(7.1)

If X is any family of semigroups, then [XI, will denote the least closed S-variety containing X. T h e notion of a closed M-variety is also defined by condition (7.1) For any family X of monoids, we denote by [XI, the least closed Mvariety containing X. If V is a closed S-variety, then it is clear that V n M is a closed M-variety. PROPOSITION 7.1. is a closed S-variety. Proof.

If V is a closed M-variety containing U , , then Vs

From Corollary 5.5 we have

vs " V S = (v "V), and since V is closed, we have

Proposition 7.1 discriminates against G-varieties. We show here that if V is any G-variety, V # 0, V f { 1}, then the S-variety Vs is not closed. Indeed let V E V be a group of cardinality >1. Consider a non-unitary left action of 1 on V (e.g. set l v = 1 for all v E V ) .Then, by Exercise 4.2, V # 1 is not a monoid. Consequently V * 1 is not in vs = {0} u v. EXAMPLE 7.1.

V. Semigroups and Varieties

136

PROPOSITION 7.2. If X is any closed class in

TS,then X n M is a

closed M-variety. If,further, X contains C then X n S is a closed S-variety. Proof, Proposition 6.4 yields

(X n M)# (X n M ) c (X ox) n M c X n M The second assertion follows similarly

I

PROPOSITION 7.3. Let X be any family of semigroups and let [XI be the least closed class in TS containing X. Then

[XI n s

= [XIS

and equality holds whenever U , E X. IfX c M then [XI n M = [XlM Proof. Assume T E [XI n S. Then T

<

<X , o .. .

o

X , with X , ,

. . . , X,, E X. Consequently T Sxlo...ox,. Since Sx,= XiE X we have Sx,E [XI,. An iterated use of Proposition 5.1 then shows that SX1O...OXn E [XI,. Thus T E [XIs and [XI n S c [XI,. If U , E X then C E [XI and, by Proposition 7.2, [XI n S is a closed S-variety. This implies [XI, c [XI n S.The proof of [XI n M = [XI, is similar, but the condition U , E X is not needed I PROPOSITION 7.4.

Let V be an S-variety such that ~ E V , R*VcV

For each ts X = (Q, S ) , the following conditions are equivalent:

(i) S E V (ii) X E (V). Further, V is a closed S-variety Proof.

(7.3 1

if and only if (V) is a closed class in TS.

(i) e-(ii). From Corollary I,9.4 we have

X

< Q' x Sx C

The conditions on V imply R c V and thus U , E V. Therefore C E (V) and 2' E (V) and thus also Q' E (V). Thus (7.3) implies X E (V).

7. Closed Varieties

137

<

(ii) e- (i). First assume that X is complete. Then S X k )for k = card Q and thus S E (V). Thus S E (V) n S = V, as required. If X i s not complete, then since (V) is completely generated, we have X Y for some complete Y E (V). By what we just proved, it then follows that S y E V. Further, Proposition II1,2.3 implies X" c' o Y . Since S,. = U , E R and Sy E V, Proposition 5.2 implies

<

<

S = S ~ E R # V C V Next, assume that (V) is a closed class. Since (V) n S = V it follows from Proposition 7.2 that V is a closed S-variety. Finally assume that V is a closed S-variety, and let X , Y E (V). Since (ii) * (i), it follows that S,-, Sy E V. Thus by Proposition 5.1, SxovE V x V c V. Consequently X o Y E (V) since (i) e- (ii). Thus (V) is a closed class I COROLLARY 7.5. Let V be a closed S-variety containing U , . Then (V) is a closed class in TS and

(V)= { X I S x E V l

I

Let V and W be closed S-varieties. Show that the SW is closed i . W ry V c V x W. Similarly for M-varieties.

EXERCISE 7.1.

variety V

x

EXERCISE 7.2.

Show that for any S-variety V

[V],

=

fi V"

n=l

where Vn is the n-fold product V EXERCISE 7.3.

x

. . . x V.

Similarly for M-varieties.

Let X be a closed class in TS containing c'.Show that X= [XnS]

and that

if X

is generated by tm's, then X= [ X n M ]

EXERCISE 7.4. Let X be a family of semigroups that contains a nontrivial semigroup that is not a group, i.e., X $ G s . Show that

[XI ns

=

[XIS

138 EXERCISE 7.5.

V. Semigroups and Varieties

Show that

V = [V] n S for every closed S-variety V Open problem.

Does the inclusion

LV XLW c L(V XW) hold for any M-varieties V and W ? In particular, if V is a closed M-variety, does it follow that LV is a closed S-variety? 8. Examples of Closed Varieties

The M-varieties 0, {l}, G , and M are closed. T h e Svarieties 0, {0}, (0, l}, and S are closed. T h e S-variety G s is not closed (see Example 7.1). EXAMPLE 8.1.

EXAMPLE 8.2. The S-variety Q of left simple (i.e., right cancellable) semigroups is closed. Indeed, suppose that V , W E Q and assume that in V x W we have (v1, W d V ,

w ) = ( v z , wz)(v,w )

Then

w1w

= WZW,

v,

+

WlV =

v,

+ w,v

The first equation implies w1 = w, and the second then implies vl = v 2 . EXAMPLE 8.3.

For the M-variety A it follows from Theorem 111,

7.6, that A

=

[z']

nM

Consequently, Proposition 7.2 implies that A is closed. Furthermore, since 2 is complete and monogenic, we have [PI = [ U,] and thus

A

=

[U,] nM

Proposition 7.3 now implies

(8.1)

A

=

[U,lh.l

8. Examples of Closed Varieties

139

The formula above leads to a hierarchy

This is one of the many hierarchies in A that can be considered. Another way to see that A is closed is to consider the M-varieties A, defined by the equation xnfl = x n . An easy calculation (following closely the argument used in the proof of Theorem 111,7.6) then shows

Proposition 6.1 implies that [ U , ] n S is a monoidal S-variety. Thus = [U , ] n S. Proposition 7.3 now yields

As

As =

(8.3)

W2lS

so that A, also is closed. Formula (8.2) remains valid with by Ans.

A, replaced

EXAMPLE 8.4. We now consider the M-variety R defined in Example 3.4. We recall that S E R iff the configuration s+t

is present in S only when s = t . This condition asserts that no copy of the ts F is present in S.. Thus

R=(F)nM Consequently R is a closed M-variety. Since ( F ) = [C'] follows from Proposition 7.3 that (8-4)

R

=

[ U J n M = [U&

For the M-varieties Rn = R n A, we have

Since Rp * R, c R it follows that

Rp * R, c R n A,,,

=

R,+,

=

[Ul], it

140

V. Semigroups and Varieties

EXAMPLE 8.5. We shall now show that the M-varieties J and Re are not closed. Since J = R n Re and R is closed, it suffices to show that J is not closed. We observe that U , E R and since U , = Ule we have U, E J. Next we note that F U, and thus U, 4 ( F ) . Thus U , 4 R and UZQ 4 RB. Consequently, U Z p4 J. It now suffices to show that U,P U, x U, for some unitary action of U, on U,. Let U , = (1, u ) and let the action be given by ax = 1 for all x E U,. Then in U , # U , we have

<

<

cx, a)(%

for all x, y

E

a) = (x

+ cy,

0)=

(x

+ 1, ).

= (x, 0)

U,. It follows that the elements (1, 11, (1, 01,

(0, ).

form a copy of U Z pin U , x U,. EXAMPLE 8.6. The S-varieties D, De, and ( f , e ) be an idempotent in V W. Then

f+

+ ef

ef

let

ee = e

ef = f ,

This implies

b are closed. Indeed,

= ef

so that e is an idempotent in W while ef is an idempotent in V. Assume now that V , W EI).Then (U,

.)(f, e ) = (.

+ v f , we)

Since W ED we have we = e. Further u

+ v j = u + wf+

(we)f = u

+ vf + ef

= ef

since V E D. Thus (u, w)(f, e ) = ( f , e ) and thus V x W ED. Next assume V , W E DP. Then (f,

). = (f+ a,e v )

8. Examples of Closed Varieties

Thus ev

141

and

=e

f

+ eu =f + ef + eu =f + ef =f

Consequently ( f , e)(u, v ) = ( f , e ) and thus V Finally assume V , W E b. Then

( f , e)(u, v ) ( f ,e ) = ( f

r*

W E DO.

+ eu + ev!, eve)

Thus eve = e and

f

+ eu + evf = f + ef + eu + evf + evef =f

+ ef + eu + evf + ef = f + ef = f

Consequently ( f , e ) ( u , v ) ( f , e ) = ( f , e ) and this proves that V # W E 0. Another way to see that D and 0 (but not DQ)are closed is to note that by IV, (4.10), and IV, (4.13),

D=(2')nS,

D=(E)nS

Since C E (2') and C E (IT), Proposition 7.2 yields the result. Further, since (2') = [Z] = 121, Proposition 7.3 implies

D But

=

[2] n S c [Z],

2 E D, so D

(8.6)

=

[21s

EXERCISE 8.1. Show that the S-variety QQof left cancellable (i.e. right simple) semigroups is not closed. EXERCISE 8.2.

Show that the S-variety DnQdeJned by the equation x,

. . . x,y

= XI

. . . x,

is closed for all n 2 1. EXERCISE 8.3. Show that for any semipoup S f 0 the following conditions are equivalent :

(i) (ii) (iii)

S is nilpotent. S has a zero which is the only idempotent in S. e S = e = Se f m all idempotents e in S.

V. Semigroups and Varieties

142 EXERCISE 8.4.

Show that the variety N of nilpotent sern&roups is

closed. EXERCISE 8.5. Let S be a semigroup with a zero 0. In the ts S = (S, S), 0 is then a sink state. Show that S E N zfl the t s S - 0 is strongly nilpotent, i e . , S - 0 E Nil.

9. Triple Products

There is a dual notion of a right action o f a semigroup T on a semigroup V in which conditions (4.1)-(4.2) are reversed end to end. T h e reversed semidirect product T +e V is then defined using the multiplication

(9.1)

(t', v')(t,v) = (t't, v't

+v)

Comparing this formula with (4.3) we find the identity

(9.2)

T #e V = (Ve* Te)e

where Veand T e are the semigroups V and T reversed, and with the left action of Te on Ve obtained from the given right action of T on V by reversal (i.e., tv defined as vt). We shall also need a product that combines the semidirect and the reversed semidirect products. This is the triple product ( T , V , S) in which T , V , and S are semigroups (with V written additively). A left action of S and a right action of T on V are given and it is assumed that the two actions commute, i.e. that

(9.31

(sv)t = s(vt)

The triple product ( T , V , S) is then composed of all triples (t , v, s) with multiplication given by

(9.4)

(t, v, s)(t', v', s')

=

(tt', vt'

+ so', ss')

The product ( t, v, s)(t', v', s')("', v'', s") calculated in two ways gives the same result, namely,

(tt't'') vt""

+ sv't'' + ss'v'',

Thus ( T , V , S) is a semigroup.

SS'S")

9. Triple Products

143

The triple product ( T , V , S ) may be obtained from the semidirect products as follows. Consider W = V x S written additively and define the right action of T on W by setting (v,s ) t = (vt, s)

Then setting w

=

(v, s), w’

=

(v’,s‘), formula (9.4) becomes

(t, w ) ( t ’ ,w’) = (tt’, wt’

Indeed

wt‘

+ w’)

+ w’ = (vt’,s ) + (v’,s’) =

(vt’

+

SV’,

ss’)

The formula above shows that

( T , V ,S ) = T * , ( V x S ) Dually we also have

( T , V , S ) = (T*,V)x:S for the left action of S on T *, V defined by s ( t , v ) = ( t ,sv). PROPOSITION 9.1. Let G be a group in ( T , V ,S). There exists then an invariant subgroup H of G such that H is isomorphic with a group in V while GIH is isomorphic with a group in T X S . Proof. Let (c, d, e ) be the unit element of G. Then c is an idempotent in T and e is an idempotent in S. Define

H

-

= {(c, v, e ) I (c,

v, ).

E

GI

Clearly H is a subgroup of G and is the kernel of the morphism v: G T XS given by (s, v, t)q = (s, t). Thus GIH is isomorphic with a group in T x S. T o show that H is isomorphic with a group in V we define y : H V by setting

-

(c, v, e ) y = evc

We have [(c,

+ ev’, e ) y e(ac + e d ) c evc + ev’c v, e)y + v‘, e ) y

v, e>(c, v‘, e ) ] y = (c, ~c = =

=

(c,

(c,

144

V. Semigroups and Varieties

so that y is a morphism. Since (c,

0, e )

= (c, = (c,

d, e)(c, v, e)(c, d, e ) dc evc ed, e )

+

+

it follows that y is injective. Thus y maps H isomorphically onto a group inV I COROLLARY 9.2. Let G be a group in V # S. There exists then an invariant subgroup H of G such that H is isomorphic with a group in V , while G / H is isomorphic with a group in S I PROPOSITION 9.3. Let G be a group such that G exists then an invariant subgroup H of G such that

H
< ( T , V, S). There

G/H
Proof. By Corollary 111,4.6, there exists a group G‘ in ( T , V, S) and a surjective morphism p,: G’ -+G. By Proposition 9.1, there exists an invariant subgroup H’ of G’ such that H‘ V , G’IH’ T x S . Since p, is surjective, H = H’p, is an invariant subgroup of G and GIH G‘/H‘. Thus H V and GIH T x S I

<

<

<

<

COROLLARY 9.4. Let G be a group such that G exists then an invariant subgroup H of G such that

H<

<

v,

GIH<S


1~

S. There

I

The definitions and results of this section apply equally well if T , V , or S are assumed to be monoids rather than semigroups, provided that appropriate “unitary” conditions on the actions are imposed. EXERCISE 9.1. Show that if S, V , T are monoids and if the actions of S and T on V are unitary, then ( S , V , T ) is a monoid with (1, 0, 1) as

unit element. 10. G-Varieties PROPOSITION 10.1. Let V and W be G-varieties. Then V * W is the G-variety consisting of all groups G that have an invariant subgroup H such that GIH E W H E V,

10. G-Varieties

145

Proof. Assume that H is such an invariant subgroup. By Corollary

II,2.3,

G


0

(GIH)

Consequently Proposition 5.2 implies G E V 8 W. The converse follows from Corollary 9.4 I

Let V, , . . . ,vk be G-varieties. Then the G-variety consists of all groups G having subgroups

COROLLARY 10.2.

Vk

8

. . . 8 V,

G = Gk 3 GkPl 3 ...

r>

G,

3

Go= 1

such that Gipl is an invariant subgroup of Gi and GJG,-,

for l s i s k

E

Vi

I

PROPOSITION 10.3. A G-variety V is closed

if and only if for each

group G and each invariant subgroup H , the conditions H, G / H E V imply G EV. This follows from Proposition 10.1

I

Let V be a variety of groups. We define =

{ S I S E M, G E V for all groups G in S )

PROPOSITION 10.4. For any G-variety V, the largest M-variety such that n G = V.

v

< T with

v is an M-variety, and is <

T E V. If G is a group in S, then G T and there exists a group H in T such that G H . Since T E V, we have H E V and thus G E V. Consequently S E If S, T E 9, then clearly S X T E ~ . Clearly n G = V. Assume now that W is any M-variety such that W n G = V. Let S E W and let G be a group in S. Then G E W and thus G E W n G = V. Consequently S E and thus W c 1 Proof. Let S

<

v.

PROPOSITION 10.5.

For any G-variety V

ft = VQ

146

V. Semigroups and Varieties

Proof. This follows from the observation that each group G is isomorphic with its reversal GQ,the isomorphism being given by g +g-I I PROPOSITION 10.6. For any G-variety V

LV = and

vs

vs is local.

Proof. If V = 0, then both sides are 0. Thus we may assume 1 EV. Let S E Lv and let G be a group in S, with unit element e. Then G is a group in eSe. Since eSe E it follows that G E V. Since further 1 E V, it follows that each group G in s' is in V and consequently S' E Thus S E 9, and this implies Lv c T h e opposite inclusion is obvious I

v.

vs.

PROPOSITION 10.7. Let V and W be G-varieties and let ( T , V , S ) be a unitary triple product of monoids with V E and T , S E Then ( T , v, S ) E v * w.

v

-

w.

Proof. Let G be a group in ( T , V , S). By Proposition 9.1, there exists an invariant subgroup H of G such that H is isomorphic with a group in V while GIH is isomorphic with a group in T x S. It follows that H E V and G/H E W. Thus, by Proposition 10.1, G E V * W. Consequently

( T , V , S)

E

v w I wr

PROPOSITION 10.8. For any G-varieties V and W

V*WCVKw If W = 0 both sides are 0. Assume W # 0 and let V S E wr # with V E S E #. Taking T = 1 in Proposition 10.7 we obtain V*S€V*W I Proof.

- v,

PROPOSITION 10.9. If V is a G-variety, then the M-variety J is closed is closed. Further, IJ V is closed and non-empty, then

if and only i f V

147

10. G-Varieties

v n G. Next, assume that V

is closed, then so is V = Proof. If is closed. Then, by Proposition 10.8,

V*VCv*:CP so that ft is closed. Let S E pi. Since all groups in S are in V, it follows from the Krohn-Rhodes Decomposition Theorem that S E [T, V]. Since 3 U , , this implies S E [U , , V] n M . From Proposition 7.3, it then fOllOWS that E [ u, ,VIM. Thus c [ u, ,VIM. Since v c E and is closed, the opposite inclusion also holds I

<

v, u, v,

s

COROLLARY 10.11.

If V is a closed M-variety containing U , , then

-

V=VnG=[U,,VnG], COROLLARY 10.12. Let S be a monoid and let G,, simple groups dividing S. Then

sE [U,, G l ,

*

'

* 9

. . . , Gn be

the

GnlM

This follows from Proposition 10.9 applied to the G-variety * * * 9 Gn]M I

v=

COROLLARY 10.13. Let S be a semigroup and let G,, simple groups dividing S. Then

sE [ U , , G l , .

*

. . . , Gn be

the

. G,lS 9

Indeed, since S and S are divided by the same simple groups, we have

Corollary 10.12 is a monoidal version of the Krohn-Rhodes Decomposition Theorem, while Corollary 10.13 is a semigroup version. Taking V = (1) we have = A. Thus Proposition 10.9 yields A = [u&, a fact we already obtained in Example 8.3. We have seen in Proposition 7.1 that if V is a closed M-variety that is not a G-variety, then Vs is a closed S-variety. In Example 7.1 we have shown that this is not true for V = G. Thus

148

V. Semigroups and Varieties

An interesting question thus arises, namely, what is the S-variety [GI, or, more generally, what is [V], for any closed G-variety V? The answer is given by the next proposition. It involves the closed S-variety (1)s = (0,1 ) the S-variety Q of left simple semigroups, and the S-variety D14 of all semigroups defined by the equation xy = x. PROPOSITION 10.14. Let

Then

Qn

V be a G-oariety, V

+ ld and V f

(1).

vs = DIPvV= V * (l)s

If, further, V is closed, then

yV],

(10.1)

=

DIPvV.

Proof. We begin with the study of a semigroup S E Q. By the definition of Q, each element s E S yields a bijection P: S -+ S defined by t4 = ts. There results a tg (S, 3) and a surjective morphism p: S -+ s defined by sy, = 4. Let K be the kernel of y, i.e.,

K

=

{k I k E S, uk = u for all

u E S}

Clearly K E DIP.We assert that

SwKxS

(10.2)

To prove this we define

KxS-+S (k, P)pl = k4= ks pl:

Since

(I,

f)pl(k, P)pl = Ztks =

Its

[(I fp, , 4)]pl = ( I , (ts)")pl = Its

morphism. Given s E S, define k to be the unique element satisfying KF = s. Then for u E S we have uks = us and thus uk = k. It follows that k E K and s = (k,4)pl. This proves that pl is surjective. If (k', f)pl = s then k't = s and us = uk't = ut for all u E S.Thus 4 = f and k's = s. This shows that for a given s E S, the pair (k,4) is unique and thus pl is injective. We have thus proven (10.2). pl is a

10. G-Varieties

If S

E

Qn

149

vs then

E

< S. Thus (10.2) implies

V since Q

nv,

c DIevV

T o establish the inclusion

DIBV V c V

(10.3) we first note that to show that

V

c

V

L

L

(l)s

(l)s and that DIe = (Ze),. Thus it suffices

2QE v

L

Since V f 0 and V f (1 >,there is a group G E V with card G 2 2, Consider K = G L 1 with action given by lg = 1. Then

(g, l ) ( h , 1) = (g

+ Ih, 1) = (g, 1)

for all (g, l), (h, 1) E G rw: 1. It follows that c K , and (10.3) is established. Since both V and (I), are contained in Q and Q is closed, it follows that V L (l), c Q. Let S E V L (l)s and let G c S be a group. Corollary 9.4 then implies that G E V. Therefore S E and V * c Thus the final inclusion V L (l)s c Q n

vs vs

vs.

is established. Now assume that V is closed. By Proposition 10.9, 'iiis closed and thus is also closed. Since Q is closed, it follows that by Proposition 7.1, Q n is closed. T h e inclusion V c Q n thus implies [V], c Q n The inclusions V c [V], and (l)s c [V], imply V L (l), c [V],. Having proved the inclusions

vs vs.

vs

vs

V * (1)s

c

[V]s

c

Q

0s

equation (10.1) follows COROLLARY 10.15.

[G]s=Q=D,evG==Grw:(l)s

I

If in Proposition 10.14 we assume V = {l}, the proof of the inclusion Qn c Dle v V remains valid. Since = A we obtain Q n As c Dle. The opposite inclusion being obvious we obtain

vs

V. Semigroups and Varieties

150 PROPOSITION 10.16.

I

Q n A s = DIe 11. Primes

A semigroup S is said to be prime if the inequality

where T , and T , are semigroups with T , acting on the left on T , , implies

S

< TI

< T,

S

or

As in IV,l we consider the class ( S ) of all ts’s X such that S does not hold. It follows that S is a prime semigroup iff

<S)

<X

*s

is a closed variety of semigroups.

The prime semigroups are

THEOREM 11.I.(Krohn-Rhodes).

as

fol-

lows : (i) all divisors of U,, (ii) all simple groups. Proof. Let S be a prime semigroup not encompassed by (i) and (ii). From Corollary 10.13 we have

s E [U,, G l ,

* * * 9

G,ls

<

where G, , . . . , G, is the family of all simple groups Gi such that Gi S. Since by assumption G, F+ S we have S Gi , i.e. Gi E ( S ) . Similarly since S < U, we have U , E (S). Since ( S ) n S is a closed S-variety, it follows that

<

[UZ,GI,

* * *

, Gnls = <S)

This yields S E (S), a contradiction. This proves that there are no more prime semigroups than listed in (i) and (ii). Next assume that S is a simple group and let S T1 o T , . By Corollary 9.4, there is then an invariant subgroup H of S such that H T , and

<

<

11. Primes

<

151

S/H T,. Since S is simple, we have either H = S or SIH = S. Thus either S TI or S T , . Consequently S is a prime. Next assume that S U , . This gives the following five possibilities for S :

<

< <

u,, 2,

0, 1,

u 2

Clearly

(0) n S = 0,

(1) n S = {0}

and consequently 0 and 1 are prime semigroups. There remain the following three cases

s= u,, 2,

u 2

Correspondingly we consider the prime cts's

P = C', 2, 2 Then

s=s p Since P is monogenic we have, by Proposition I,9.9,

p<s

(11.1) We assert that (11.2)

( S ) n S = (P) n S

<

<

Indeed, let T be a semigroup. If S T,then (11.1) implies P T. If P T,then S = S, ST = T . Thus (11.2) holds. Since (P) is a closed class in TS containing C, it follows from Proposition 7.2 that ( P ) n S is a closed S-variety. Thus S is a prime semigroup I

<

<

The notion of a prime monoid is defined quite similarly. Theorem 11.1 remains valid, but the prime semigroups 0 and 2 which are not monoids must be removed from the list. Thus the only prime monoids are 1, U , , U , , and simple groups. EXERCISE 11.1. Show that $ S and T are semigroups, S < T and S c U , , then T contains a subsemigroup isomorphic with S.

152

V. Semigroups and Varieties

EXERCISE 11.2. A semigroup S is called critical (resp. hypercritical) if S is not in the S-variety (resp. the closed S-variety) generated by the semigroups T such that T S and T # Show that any S-variety (resp. closed S-variety ) is generated by critical (resp. hypercritical) semigroups.

<

s.

Formulate similar notions for monoids and M-varieties. EXERCISE 11.3. Show that a semigroup S containing U , is hypercritical tff

s= u,.

V be a closed S-variety and let V E V. Say that Vis prime inV if V < V , # V , and V,, V , E V imply V < V , or V < V , . Show that V is prime in V iff V n ( V ) is a closed S-variety. Show that i f V is prime in V, then V is hypercritical. EXERCISE 11.4. Let

EXERCISE 11.5. Consider the closed class

x = [TI n (2) in TS and the closed S-variety

Use Exercise IV,8.3 and Exercise 7.3 to prove that V is not finitely generated as a closed S-variety. 12. A Tabulation

We shall tabulate, in a convenient form, results concerning the varieties (P) n M and ( P ) n S, where P is one of the fourteen prime ts’s contained in for which (P) was calculated in Chapter IV. The fourteen primes are arranged into two series: the main series (12.1)

P = 0, 1, 2, E, C, F,

Z

and the “dotted” series (12.2)

p’=

0 , l’, 2 , E , c‘,F’, 2.

We first note the relation

( P ) nM

= (F) n M

12. A Tabulation

153

expressing the fact that, for a monoid S, conditions P are equivalent. The tabulation for (P) n M is as follows:

(0) n M = (0') n M

(12.3)

< S and P' < S

=0

(1) n M

=

(2) n M

= (2')

n M = (1)

(E) n M

= (E')

nM


=

(1') n M = (1) n M = 0

=

{ 1}

(c')n M = (U,)n M = G

< F ) n M = (F') n M = R

(2) n M = (2.) n M =
= R rc

G

All these are closed M-varieties. The first six formulae are restatements of facts already known. The last one requires a proof. Next we pass to the S-varieties (P) n S with P given by (12.1). The table in IV,8, shows that the class (P) is generated by tm's. Thus it follows from Proposition 6.1 that (P) n S is a monoidal S-variety. Consequently

(0 n s = ((P) n MIS This yields the following table:

(0) n S = (0) n S = 0 (1) n S = (1) n S = {0} (2)nS(12.4)

{0,1)

( E ) n S = (0,1>

(C) n S = Gs ( F ) n S = RS

(2) n S = (2) n S = Rs r G For the last formula we use the equality (R *: G ) s = Rs rc G implied by Proposition 5.4. All these S-varieties with the exception of G s are closed. For the first four this is clear, while for the last two this follows from Proposition 7.1 since R and R # G are closed M-varieties containing U,.

154

V. Semigroups and Varieties

Next come the S-varieties (P) n S, where P is one of the seven dotted primes. The table is ( 0 ) n s =: 0

(1') n S

=

{@}

(2') n S = D (12.5)

( E ) n s = Z,

(c')n S = ( U , ) n S = G r D (F') nS = R

D

(F) n S = ( U , ) n S = R

#

G

#

D

All these S-varieties are closed. For the first two this is clear. The third and the fourth we treated in Example 8.6. For the last three this follows from Proposition 7.2 since (P) is a closed class containing C. The first four formulae are restatements of earlier results. T h e last three require proofs. We are thus left with the task of proving the last formula of (12.3) and the last three formulae in (12.5). Since the proofs are quite similar, we shall only prove the most involved one, namely the last formula in (12.5). From IV,8, we know

(2')

=

[C']

0

[GI

0

[Z]

Thus Proposition 6.5 applied twice yields

(2.) n S c ([C'] n M)

#

([GI n M)

#

([Z] n S)

Using the earlier results in the tables above this implies

(2.) n S c R # G r D T o prove the opposite inclusion we observe that (2) n S is a closed S-variety. Thus it suffices to prove that (2.) contains R, G, and D. This, however, follows from the fact that (2) contains (F), (C), and

(2.)We note that the tables above contain the varieties (S) n S for the five prime semigroups S and the varieties ( S ) n M for the three prime monoids S.

12. A Tabulation

155

T o the formulae above we add three more formulae dealing with localization G+D=LG

(12.6)

R+D=LR R+G+D=L(R+G)

Indeed, settingV = G, R, or R x G, the tables show thatV = (P) n M where P is C or F o r 2 depending on what V is. Since 2 E (P) and since (P.)= L(P), Proposition 6.3 implies

(P)n S = L(P) n S = L((P) n M ) = LV Equalities (12.6) thus follow by looking up formulae (12.5). It should be observed that if V is any M-variety, then V + (1) and thus by Proposition 5.6

V r L ( 1 ) c L(V+ (1)) Since L ( 1 )

=

b

(12.7)

= V,

= LV

and D c b we obtain V + D cVWZ) c L V

Equations (12.6) thus supply three cases in which the inclusions in (12.7) are equalities. I n VIII,8 we shall prove that (12.8)

JI

+

D = LJ,

No doubt there are other cases in which equality in (12.7) holds. It is an open and interesting question to find criteria for equality in (12.7). It should be noted that ultimately equational descriptions of most of the varieties discussed above are contained in Section 3. EXERCISE 12.1.

Let P be a simple group. Show that

(P) n M =

v

where

V

=
156

V. Semigroups and Varieties

where P ranges over all simple groups. State analogs of the above for Svarieties. EXERCISE 12.2. Establish the equation

[ U , , 21, = R

x

D

References

The notion of a semidirect product is folklore. The definition of the triple product has been extracted from Schutzenberger (1965). The ultimately equational approach to varieties is from Eilenberg-Schutzenberger (1975). Theorem 11.1 is by Krohn-Rhodes (1965). Samuel Eilenberg and M. P. Schiitzenberger, On pseudo-varieties of finite monoids (to appear). Kenneth Krohn and John L. Rhodes, Algebraic theory of machines, I. Prime decomposition theorem for finite semigroups and machines, Trans. Amer. Math. SOC. 116 (1965), 450-464.

M. P. Schiitzenberger, On finite monoids having only trivial subgroups, Information and Control 8 (1965), 190-194.

CHAPTER

VI

Decomposition of Sequential Functions

The purpose of this chapter is to establish a bridge between the theory of sequential functions and that of transformation semigroups. Once this bridge is established, the results of Chapters I-V can be, in a wholesale manner, interpreted for sequential functions yieIding various decomposition theorems. The reader should at this stage review A,XI and A,XII. 1. Syntactic Invariants of Sequential Functions

Let

A=

(8, i, A ) : 2-r

be a sequential machine as defined in A,XI,l. In particular, Q is a right Z*-module and the action of Zf on Q represents a ts

TSA=

(Q, $4)

where S4 is the semigroup of all partial functions f: Q Q of the form qf = qs for some s E Z+. There results a surjective morphism ---f

6:,

(1.1)

Z+ +

s,

If we replace Z+ by Z*, we obtain the tm TM-

=

(Q, M-)

Clearly

TM-=

s,

c

M-= 157

TS-'

s,

v 1,

VI. Decomposition of Sequential Functions

158

If .A is the minimal sequential machine of a sequential partial function f : Z* -d*, then the subscript A? is replaced by f and we obtain T S f , Sf, TMf, Mf which are called the syntactic invariants off. O f these, the syntactic ts T S , is the strongest since it determines the syntactic semigroup S f , the syntactic tm TM,, and the syntactic monoid M f . The difference in strength between the various syntactic invariants is illustrated by the obvious PROPOSITION 1.I. Let f: Z* +P be a sequential partial function. The following conditions are then equivalent:

(i) f is a function. (ii) TSf is a complete ts. (iii) T M f is a complete tm

I

On the other hand, it is impossible to determine whether or not f is a function from the knowledge of S, or M,. PROPOSITION 1.2. Let

Z+r A ’= (Q’, i f ,A’): Z+r A= ( Q , i , A ) :

be sequential machines and let Q]:

A?+&’

be a state mapping. If Q]: Q -,Qr is surjective, then

TS& < T s , Proof. We recall that (as defined in A,XII,l) q? is a partial function q?: Q + Q‘ satisfying lrp = i‘

(qQ])a= ( Q 4 P

kq?,a)A’ = (q, 4 Thus if Q] is surjective the first inclusion above shows that (T viewed as a transformation of Q covers (T viewed as a transformation of Q’. This yields the conclusion I

1. Syntactic Invariants of Sequential Functions PROPOSITION 1.3.

f : Z*

+r*.Then

159

Let A:Z+T be a sequential machine with result TSf


Proof. Let Af be the minimal sequential machine o f f and let A* be the accessible part of A. There is then a state mapping A+ Aa which is the inverse of inclusion, and a state mapping Aa Af given by reduction. There results a state mapping A+ A f ,which is surjective. Thus the conclusion follows from Proposition 1.2 1 ---f

PROPOSITION 1.4. For any sequential partial function f : Z*

+

r*,

the tm T M f is monogenic. Thus TMf

< Mf

and $f is a function, then

TMf

< Mf < (TMj)(k)

for some integer k 2 1. Proof. Let A= (Q, i, A ) be the minimal sequential machine o f f . Since &is accessible we have iZ* = Q. Thus i M f = Q and consequently T M , is monogenic. The remaining conclusions follow from Proposition

I,9.9

I

Note that Proposition 1.4 may fail for TS,. PROPOSITION 1.5. For any sp-function f: Z* +T*, the following conditions are equivalent

<

(i) TSf 1'. (ii) f is a very j n e morphism with domain C,* for some subset XI of Z. Proof. Indeed, both conditions signify that the minimal s-machine A, has a single state 1

Let (1.2)

A'= (Q, i, A): Z - + r

be a sequential machine. We recall (see A,XI,5) that A is said to be state dependent if there exists a function

8: Q-r

VI. Decomposition of Sequential Functions

160

such that (! 012 I,= ( P ) B

for all q E Q, o E Z. Proposition A,XI,5.1 constructs for each sequential machine A! a state dependent sequential machine A?’with the same result as A. The explicit construction of A!’ is as follows

A‘= ( r x Q, ( 7 0 , i ) , 2’) (Y,410 = ((Q, o)A 40) ((Y,q), 4’ = (Q, a ) l

r

with yo E arbitrary. From this construction, it.is clear that

T S d ,
o

TS,

Letting A be the minimal sequential machine of f: Zw-+ obtain PROPOSITION 1.6. Each sequential partial function f : Z* -+ result of a state dependent sequential machine A! such that

rXwe

r+is the

If we consider the case when f : Zw-+ r*is a very fine morphism, then the machine (1.2) may be chosen to have a single state. The state dependent machine will then be

so that I f f is surjective, then TSM, = f,and it is easy to see that the machine 4‘ is accessible and is “minimal” when compared with other state dependent machines. This shows that the result asserted in Proposition 1.6 is in some sense best possible. The above discussion shows that even though the replacement of sequential machines by state dependent sequential machines is harmless as far as describing sequential functions is concerned, it leads to considerable complications for the kind of problems that are discussed in this chapter.

1, Syntactic Invariants of Sequential Functions EXAMPLE 1.1. Let f: Z*

+ Z*

161

be the function defined by

If= 1 =

UWf

bow

where u E C,w E C*,and cro is a fixed letter in Z. Assuming that Z has more than one letter, the minimal machine off is

with one straight edge and one loop for each letter of 2. It follows that

T M f = C'

TSf = C ,

M- = u,

Sf = 1, EXAMPLE 1.2.

Let f: Z# -+ Z* be the partial function defined by

The minimal machine for f is ala

ala

-+o-o---co

and TSf is the strongly nilpotent ts I

I

O---+O+O

This shows that even though f is a subidentity, its syntactic invariants may be more complicated than that of the identity. EXERCISE 1.1. Let f: Z* + I'* be an sp-function, and let s E Z* be such that sf # 0. Consider the sp-functiong =f o s: Z* -+ P dejined by

Prove

where R is an ideal in TS,.

VI. Decomposition of Sequential Functions

162

f: Z*+r" be a sequential partial function. Show that there exists a sequential function g : 2" + P such that EXERCISE 1.2. Let

f c g TM, TS, < TS,",

< TM,"

M, < M f

S,<Sf, 2. Composition PROPOSITION 2.1.

If

are sequential partial functions, then

TSfLf.a

TSf2 O TSfI

Proof. Let

J f 1 = ( Q 1 , i l , A 1 ) :Z + Q

-4= (Qz,

i,,A2):

Q

+r

be the minimal sequential machines of f l and f 2 . Let

A?=

Jfz=

(QzxQ1,(G,i1),4:

Z-F

is the composition fifi . Thus

By A,XII, (8.4)the result of


TSfd2

Formula (2.1) shows that the transformation a of TSA is in the wreath product TSd8 o TSdl. Thus TSd c TSd2 o T S d I .Consequently TSfl,
Consider

3. Decomposition

163

where f is a sequential partial function and p and ly are very Jine morphisms. Then TSdV TS,

<

Indeed, Propositions 2.1 and 1.5 yield

< TS,

TSpf,

TSJ 0 TS,

0

< 1'

0

TS, 0 1'

=

Ts,

Corollary 2.2 becomes more clear if we adopt the following conventions: A function p: C, Z will be identified with the very fine morphism p: Zlw + Cr that it defines. We shall also identify p with the sequential machine Z, + C with a single state i and with output (i, a,)A = a,pl for all c1 E 2,. Then for each sequential machine -+

A = (Q, i, A): C - + r and any functions p:

z,-+z,

y:

r+rl

the composite machine

YAY = (Q, i) 21):

2 1

+Ti

satisfies 401

k'

=

4(w)

=

(q, a1Pvly

It is then clear that

TS,..nV = T S I it follows Thus taking to be the minimal machine of f: Zr -+P that TSpfV TS,. What is most important is that Proposition 2.1 may be inverted to yield decompositions of sequential partial functions. This will be done in the next section.

<

3. Decomposition

THEOREM 3.1. Let f: Z* +.Pbe a sequential partial function and let X , , X , be ts's such that

Ts,< x, XI

VI. Decomposition of Sequential Functions

164

There exist then sequential partial functions

such that

f =f i f i TSf, < X j ,

j = 1, 2

If, further, Xi and X , are complete, then f l and f z may be chosen to be functions. Proof. Let off. Thus

A = (9,i, A ) :

r

Z + be the minimal sequential machine

T s , = (Q, S ) and a surjective morphism [: Z + + S

is given such that

qs = q w for all s E Z+. Let

X j = ( Q j , Sj), j TSf

= 1, 2

< x2 x, 0

9

Thus V; Q z X Q i - + Q

is a surjective partial function. Since q~ is a covering, we may choose for each u E Z an element t, E Sl and a function fa:

Qi

+

S2

such that the transformation (f u , tu) of Qzx Q1 covers the transformation u = uE of Q. Thus

(42 4 1 ) v 9

= M 4 1 f 1, 41tO)V

Since g~ is surjective, we may choose (i2 >

4 ) E Qa x Qi

3. Decomposition

165

We now consider the alphabet !2=Q1XZXSz

The sequential machine

is defined by setting

41“ (41 “111 9

= 41tU =

(91 > u>4 1 f u )

The sequential machine

is defined by setting %(41 > ((42

(42,41,

“9

s2)Az

“>

$2)

= 42%

if ( 4 2 9 4119 f 0 f if qzs2 = 0 otherwise

, 4l>P,“)A

=

y arbitrary,

The inequalities

TS,

c

are then clear. If

are the results of

and dz, then

is the result of the composite machine

Xj

42s2

VI. Decomposition of Sequential Functions

166

In d1A2 we have

We now assert that rp: Q2xQ1 -+ Q is a state mapping 9: A14

-

A,

Indeed, we have

( i 2 ,i,)V (42 9 4 1 ) P

= (42

9

=i

41)(fo , t u k

= (42

9

41)ag,

We also must verify the inclusion

= (42

( d 4 2 9 41)%

9

417 c)l’

or equivalently

(3.1)

((42

9

= (42

41)94

9

417

c, 4lfu)I

If (q2, ql)v = 0, then the left-hand side is 0 and there is nothing to verify. If q2(4,f u ) = 0, then ( q 2 , ql)( f u , t,) = 0 and therefore (q2, ql)pa = 0. This implies ( ( q 2 , ql)v, e)A = 0 and again the inclusion (3.1) holds. If ( q Z , ql)p f 0 # q2(q1f c ) , then (3.1) is an equality by the definition

of

a!.

Having proved that p7: A,&2+ Afis a state mapping, the inclusion f cflf2follows from A,XII, (1.4). If X I and X 2 are complete, then the machines A, and A2as constructed above are complete and thus fl and f 2 are functions I COROLLARY 3.2. Let f: Z* -+P be a sequential partial function

and let

TS,<X,o

... O X , ,

n>l

where X , , . . . , X , are ts’s. There exist then sequential partial functions fj: Qj-l

-Qj,

1 cj I n

3. Decomposition

167

such that

l f XI, . . . , X , are complete, then f i , functions.

. . . ,f n

may be chosen to be

For n > 2, the corollary follows from Theorem 3.2 by induction. For n = 1, the corollary follows from Theorem 3.2 by taking X,= 1’ COROLLARY 3.3. Let f: Z* -+Pbe a sequential function and let X,, . . . , X,, n 2 1 be complete ts’s. The following conditions are then

equivalent : (i) (ii)

TS, < X, o . . . o X , There exist sequential functions f i : Q.j!-, Q,

1 5j 5 n

+Qj*,

=

z,

Q,

=

r

such that .*.fn

f = f i

TS,,

<X j

for

1 5j 5 n

The implication (i) * (ii) follows from Corollary 3.2. The implication (ii) * (i) follows from Proposition 2.1 I Both in this section and in Section 2 we utilized the syntactic invariant TS,. The reason for this is that the results formulated using TS, imply those for the invariant TM,. As an example and also for future reference, we shall give the M, version of Corollary 3.3.

Let f : Z* + r* be a sequential function and let X , , . . . , X , , n 2 1, be complete tm’s. The following conditions are then equivalent : COROLLARY 3.4.

(i) (ii)

T M f <X , o .. . o X , There exist sequential functions f i : Qj?, - + Q j * , Q~=Z,

1 5j I n

~,=r

VI. Decomposition of Sequential Functions

168

such that

f=f1 . . . f a TMf, < X i Proof. Since X,,

. . . ,X ,

for

1Cj 5n

are tm's, condition (i) is equivalent with

TSf -< X ,

o

... o X ,

<

Similarly, condition TM,, Xi is equivalent with TS,, the result follows from Corollary 3.3 I

< X j . Thus

4. Parallel Composition

Given two finite alphabets Zl, Za we shall consider the product Z, x Z2. A word in (Z, x Za)* may be interpreted as a pair (sl, s2) where s1 E Zl*, s2 E Z2*are words of equal length: 1 s1 I = I s2 1. In this way (Z1xZ2)* becomes a submonoid of Zl*x Z2*. Given length-preserving partial functions

fj: Zj* -frj*, j = 1, 2 the partial function

f =f1xf2:

(Zlxw-(rlxr2)*

is defined by setting ($1

9

szlf = (Slfl, S Z f 2 )

The right-hand side is interpreted as 0 if either s1 f, = 0 or s2f 2 = 0. If, further, Zl= Za = Z, then we define f1

4:Z*

---+

(r l x r2 ) *

by setting 4fi ~ f 2 = ) ($1

$a)

for s E Z*. The two operations can be expressed one in terms of the other as follows :

fl A f 2 = S ( f l X f 2 ) where

6: Z* -+ (ZXZ)"

4. Parallel Composition

169

is the diagonal mapping s 6 = (s, s). Conversely, if f j : Zj# -+I'j#,then

f i xf 2

= Zlfl A %f2

where nj: (Zl x Z2)# Zj# --f

is the projection (sl, s2)nj= sj, j = 1, 2. These operations were considered in A,XII,10 (the notation v was used there instead of the more appropriate notation A used here). I t was shown that if f l and f 2 are sequential, then so is the result of the operation f i x f 2 or f l A f 2 . This was done by defining the operations x and A for sequential machines. Because the two operations x and A define each other, we shall restrict our attention to the operation A. PROPOSITION 4.1.

Let

be sequential partial functions. Then

is a sequential partial function, and

be the minimal machine of f j . Then f is the result of the machine

VI. Decomposition of Sequential Functions

170

This is the analog of Proposition 2.1. We now formulate the analog of Theorem 3.1. THEOREM 4.2. Let f : C+ + P be a sequential partial function and let X , , X 2 be ts’s such that

TSf

< XI x x2

There exist then sequential partial functions

fj: C+ + Qj+,

j = 1,2

and a function

p:

Q1XQ2+F

such that

f = (fl AfiM TS,,

<X i ,

j = 1,2

If,further, X l and X , are complete, then f l and f i may be chosen to be functions.

Proof. Let A?= (Q, i, A ) : C +F be the minimal sequential machine o f f , Thus TSf = (Q, S ) and a surjective morphism

E : C++S is given such that

9s = s(sE) for all s E C+. Let

X j = (Qj, Sj), TSf

j = 1,2

< Xl x x2 0

Thus

V : QixQz+Q is a surjective partial function. Since pl is a covering, we may choose for

4. Parallel Composition

171

each a E C elements

j

tju E S j ,

=

1, 2

such that (41

9

4 2 ) v

c (Qltl, 9

42tzu)v

Since cp is surjective, we may choose

(4

9

i2)

E

Qi

x Q2

so that

(il, iZ)v= i We now consider the alphabet Qj

= QjxC,

j

=

1,2

Let fj be the sequential partial function defined by d j .Then

TS,,

< TSdj c Xj,

The sequential partial function

is the result of the machine

with

j

=

1,2

VI. Decomposition of Sequential Functions

172

We assert that rp: A+ A, is a state mapping. Indeed, we have (il, i,)v = i. Further (41 9

= (q1t1, 42t2,)rp = (41 42)arp 4 = (41 0, .)B = (41 a)l’

42)w

((41, 421%

Y

9

42

9

9

9

42

9

This implies

f = ( f l .fdB as required. If X l and X , are complete, then kland k2 as defined above are complete and thus f l andf, are functions I COROLLARY 4.3.

and let

Let f : Zw+ P be a sequential partial function

TS, < X 1 X * .. X X n

where X , ,

. . . ,X ,

are ts’s. There exist then sequential partial functions f j : Zx+Qj+,

1l j l n

and a function

p: Q,x.. .X Q , + ~ such that

f

c

Afn)B

( f iA

1l j l n

TSfj<Xj,

If tions

X,,

I

. . . ,X ,

are complete, then f l ,

. . . ,f n may be chosen to be func-

COROLLARY 4.4. Let f : Z# --+ P be a sequential function and let X , , . . . ,X , be complete ts’s. The following conditions are then equivalent:

(i) TS, < X l x . . . x X , . (ii) There exist sequential functions

f j : Zx-+Qj*, and a function

p:

Q1X

1 <j
. . . XQ,-+P

such that

f

=

(fi

Tsjj<Xj,

* *

A.fn)B

1l j L n

173

4. Parallel Composition

-

T h e implication (i) s (ii) follows from Corollary 4.3. T h e implication (ii) (i) follows from Proposition 4.1 I

As in the case treated in Section 3, the results for the syntactic invariant TS, imply similar results for the syntactic invariant TM,. We state the analog of Corollary 3.4 without proof. COROLLARY 4.5. Let f: L'* +T* be a sequential function and let X , , , . . , X , be complete tm's. The following conditions are then equivalent:

(i) (ii)

<

TM, X , x . . . xX,,n 2 1. There exist sequential functions

COROLLARY 4.6. Let f: Z* +T* be a sequential function and let X , , . . . , X , be monoids. The following conditions are then equivalent:

<

X , x . . . x X , , n 2 1. (i) M, (ii) There exist sequential functions

<

Indeed, condition (i) is equivalent with TM, X , X . . . x X , since TM, is complete and monogenic. Similarly M,, < X j is equivalent with TMfj <Xi I

VI. Decomposition of Sequential Functions

174

be a sequential partial function PROPOSITION 4.7. Let f : Z* + rW and let k = card r. There exist then sequential partial functions

and a function such that

for all 1 5 j 5 n. Proof. Let I' = { y l , function of y j . Thus

. . . , yk}, and let xj: r -+ 2 be the characteristic

Then

x

xfl

lr. Define f j = f x i . Then

=

is injective and there exists a function

(fi

... IZfk)B = ( f x l A

=fh A

* * *

* * *

p : 2k -+I' such that

.fXk)B A

X d B =fxB = f I

5. Examples of Decompositions

We shall now illustrate on some examples how Theorems 3.1 and 4.2 (or more exactly their proofs) can be used to decompose sequential machines and sequential functions. EXAMPLE 5.1.

Consider the sequential function

f: u * - + 2 * which is the mod 4 counter. Thus

uy = (ooo1)po'

5. Examples of Decompositions

175

T h e minimal machine d of f is

1

and

T h e decomposition theory for groups (see II,2) gives

(i,j ) y = 2i + j for i, j E 2. With this notation a is the generator of 2, and also the generator of 2,. T h e generator 0 of 2, is covered by (f,a) with I f = a, Of = a2 = e. Applying the construction in the proof of Theorem 3.1, we obtain the following factorization of A : 0

.Acl

Ma 2x 2, d 2

with dland A, as follows

Since dldoes not have Oa and le as outputs, these letters may be elim-

VI. Decomposition of Sequential Functions

176

inated. Replacing Oe and l a by 0 and 1, we find the factorization of &?

with dl and d..as follows

Indeed, if we compose dland d2we obtain the sequential machine

01

which is the original machine A with the states renamed using p-'. The function fl computed by dlis a 2-counter and converts a string of a's into 0101010101. . . The second function leaves the zeros alone but converts the first, third, fifth, etc., 1 into a zero. Thus the string above becomes

0001000100... as it should be for a mod 4 counter EXAMPLE 5.2. We consider the function

f : a*-+22" which is the mod 6 counter. Thus anf = ( 0 5 1 ) p O r

if

n=6p+r,

O(r<6

5. Examples of Decompositions

177

The minimal machine A f o r f is

\ -0

3

Applying the construction of the proof of Theorem 4.2 we obtain the sequential machines

A2:1 - + 3

A1;1 - 2 , a10

1 u

+ 0-

-0

')all

L

a/ 1

2

a12

The parallel product A1t,A2; 1-2x3 is the machine all1

11 -02

+

k

10

00 12

aiol

/lo

01

If we now relabel the states using the bijection Z,xZ, (a, a ) -+ 6, we obtain the machine all1

1-2

s-4 a101

-

Z6 given by

VI. Decomposition of Sequential Functions

178

Comparing this with the original machine A we find that the function ,!I: 2x 3 -+ 2 must be chosen to map 12 into 1 and all other letters into 0. Thus 4= (Al A &.),!I.

6. The Function

I n this section and the next one we shall only consider sequential functions and complete sequential machines. Let S be a monoid. We define the sequential machine As= ( S , 1, n):

s+ s

n: S x S - + S

(s, s')n

=s

The action SX S -+ S is given by multiplication in S. This machine clearly is accessible since 1s = s. If s1 # s2, then (sl,s)n f (s2, s ) n and thus the states sl and sg are not equivalent. Thus A8also is reduced. Consequently Asis the minimal machine of a function

s: S" -.+ S" The monoid S plays here a dual role; in addition to being a monoid, it also is the base of a free monoid S". T o avoid confusion, we shall write a word of length n in S" as an n-tuple (sl, . . . , sn). T h e 0-tuple A = ( ) is then the unit element of S", not to be confused with the unit element in S. With these conventions, the function 9 is given by (s1,

. . . , s,)S

=

(1,

$1,

SlSZ,

. . . , s1 . . . q - 1 )

From the minimal machine JH~we can read off the syntactic invariants of S. We have

TMg = Mg = T S g = Sg The key result concerning the function THEOREM 6.1.

Let

f: P

-+r+

=

is

S

6. The Function

s

179

be a sequential function and let S be a monoid. The ,following conditions are then equivalent:

<

( 9 M, s. (ii) There exist functions a:

z js,

8: sx2-r

such that

f = (OlSA 1 Z ) B We recall that a stands also for the very fine morphism Z x defined by a. Similarly for /? and 1,. Proof.

(i)

* (ii).

--f

Sx

Let

A?= (Q, i, A): Z - - f F be the minimal sequential machine o f f and let

We thus have a surjective morphism p:

z* ’ M ,

such that 90 = 4.P)

for all q E Q, 0 E 2: Since M, S, there exists a submonoid T of S and a surjective morphism v: T + M, of monoids. We shall regard v as a partial function S + M, with T as domain. Since p is surjective, we can find a function

<

a : Z-FS

such that the triangle

commutes. Thus

(6.1)

VI. Decomposition of Sequential Functions

180

Define

(s, 0 ) B =

(i(sq),u)A

{ arbitrary

if sq # 13 if sp = 0

Consider the machine

Define the partial function y: S-Q

sy = i ( s q )

Clearly ly

=i

Further by (6.1) (sy)u = i ( s q ) ( u a q )= i(s(ua))q = i(su)q = (so)y

Since (SY

, u)A = (i(sv),u)A

it follows that if sy f 0 then (sy, u)A = (s, u)B. These facts show that y is a state mapping J”- kf. Thus the two complete machines have the same result, i.e., (ii) holds. (ii) = (i). The syntactic monoid o f f = as^ lZ)B is dominated by that of a s A 1, (Corollary 1.7). Since the identity mapping 2* Z* has a syntactic monoid 1, it follows from Proposition 4.1 that the syntactic monoid of as^ 1, is the same as that of a s . Again, by Corollary 1.7, the syntactic monoid of a s is dominated by that of 3. Thus M, Ms

-

<

=s

EXERCISE 6.1.

Given a monoid S consider the sequential function

s: s*+s* (sl, . . . ,Sn)S = (sl, s,s,,

. . . ,s1 . . . s,)

7. Varieties of Sequential Functions

Show that (sl, .

Show that

. . , s,)S

=

s= p:

181

( 1 , s1,

( S h

. . . ,s,-l)s

1,)p

sxs+s

(s, t ) p = st

Show that 3 and S have the same syntactic invariants. Given a finite alphabet Z and a distinguished letter 2 define the function d: Z* + Z*

EXERCISE 6.2.

a,

E

Id= 1 (wu)d = U,W for all w E Z* and u E Z. This function is called the delay of Z relative to a,. Show that d is a sequential function and that

TM, = ZShow that the functions 9, 15: S --+ S are related as follows

S where d : S*

+ S*

= dS

i s the delay relative to 1.

7. Varieties of Sequential Functions

Let V be an M-variety. We shall denote by the family of all sequential functions f: Z*+r*for all possible finite alphabets Z and F, such that

M,EV Any class of sequential functions defined in this manner will be called a variety of sequential functions. THEOREM 7.1.

Let V be an M-variety generated by a non-empty family

X of monoids

v=

or),

VI. Decomposition of Sequential Functions

182

v

Then is the smallest class F of sequential functions satisfring the following conditions :

1,: Z w - + Z *is in F. If f: Z* + P is in F and tp: Zl# -+Z* and w: F* -d,# are very Jine morphisms, then tpfy E F. (iii) If f i : Z* -+Tix, i = 1, 2, are in F then fiA fi E F. (iv) If S E X, then E F. (i) (ii)

s

v

Proof. We first verify that satisfies conditions (i)-(iv). Since V # 0 we have 1 E V and therefore 1, E 'fi. Since Mpfw M f (Corollary 2.2), it follows that f E 'fi implies tpfy E 'fi. Since MfIAfp Mflx Mfa(Corollary 4.6), it follows that f l ,fi e implies f i A fi E Finally, if S E X c V, since Ms = S E V. then E Next assume that F is any class of sequential functions satisfying (i)-(iv) and let f : Z* + be in We shall prove that f E F. Since f E 'fi and V is generated by X,we have

v.

s

v.

r*

Mf < X , X . . for some X I , . . . , X , E

< <

. xx,

X. From

Corollary 4.6 we then deduce that

f = (fi

.

A

* *

Afn)Y

for some sequential functions

and some very fine morphism

Thus, by conditions (i)-(iii), it suffices to show that f i E F for 1 5 i 5 n. Since Mf, X i , Theorem 6.1 implies

<

fi =

(.xi

Thus conditions (i)-(iv) imply fi PROPOSITION 7.2.

function

Let V and

E

A

1z)B

F. Consequently

vcF 1

W be M-varieties. For any sequential

f: Z* +r*

7. Varieties of Sequential Functions

183

the following conditions are equivalent (i) f E (V x W)' . (ii) f is the composition

with

Proof. (i)

* (ii). Since f E (V x W)"

we have M j E V x W and there-

fore

Mf
<

<

< w,

Corollary 3.4 then yields a factorization f = gh with TM, W, TM, v. It follows that M, and Mh Thus M, E Mh E v or equivalently g E W, h E (ii) * (i). Let X = T M h , Y = TM,. Corollary 3.4 then implies

<w

<

v.

< v.

TMf<XoY Since sx = Mh E v and sy = iWgE w, Proposition V,5.1 implies SxoyE V x W. Thus Mf E V x W and consequently f E (VP W)-. THEOREM 7.3. A n M-variety V is closed if and only sequential functions is closed under composition. Proof. Assume that

if

the class

v of

V is a closed M-variety and let g : Zx +a*,

v.

h : a* + P be sequential functions in From Proposition 7.2 we then deduce gh E (V x V).. Since V x V c V, it follows that (V x V). c Vv and thus gh E Conversely assume that is closed under composition, and let S E V x V. Then E (V x V). and Proposition 7.2 implies s = gh with g, h E Since is closed under composition, it follows that 9 E fl and thus S = M, E V. Thus S E V proving V x V c V 1

v.

v.

s

v

v

VI. Decomposition of Sequential Functions

184

EXERCISE 7.1. Show that condition (iii) in Theorem 7.1 may be replaced bY (GI) If f i : Zj* +ri*,i = 1, 2, are in F, then f i x f i E F. EXERCISE 7.2. Explain why Proposition 7.2 may be restated as

(V*W)"=W EXERCISE 7.3.

Show that a class F of sequential functions is a variety (i)-(iii) of Theorem 7.1 and also the following

zy it satisjes conditions condition (v) I f f

E

F and S

EXERCISE 7.4.

= M,,

then

s E F.

In Sections 6 and 7 the syntactic invariant M, and the

discussion utilized monoids throughout. Examine in detail the arguments to see where the use of the syntactic invariant S, and of semigroups in general will prove inadequate.

CHAPTER

VII Varieties of Sets

With each recognizable subset A of Zf we associate syntactic invariants such as the syntactic semigroup S, or the syntactic ts TSA . T h e objective of this and the following three chapters is to relate the properties of the set A with algebraic properties of its syntactic invariants. A key theorem in this effort is Theorem 3.2, which establishes a 1-1 correspondence between S-varieties and families of sets that we shall call +-varieties. 1. Syntactic Semigroups

We shall give here a rapid review of the basic facts about syntactic semigroups. Our approach in this section will be purely algebraic. T h e tie-up with the theory of automata will be made in Sections 2 and 6. Let S be any semigroup and A a subset of S. We define a congruence -A in S as follows: $1 -A

$2

holds iff uslv E A o us2v E

A

for all u, z, E s'.Thus s1 - A s2 signifies that s1 can be replaced by s2, and vice versa, within any context without affecting membership in A. T h e congruence -A is called the syntactic congruence of A, and the quotient semigroup S / - * , which we may denote by S / / A is called the syntactic semigroup of A. T h e natural factorization mapping

S

+

S//A

is called the syntactic movphism of A . ias

186

VII. Varieties of Sets

The subset A of S is saturated for the congruence and s1 E A implies s2 E A. This means that

i.e. s1

s,

AeAe2 = A or equivalently

A

= A' e A -1

for some subset A' of SIIA, notably A'

= AeA.

S -+ T be a surjective morphism of semigroups and let A = Bp-' for some B c T. There exists then an isomorphism PROPOSITION 1.1.

Let

ip:

y: S / / A-+ T / / B

such that eAY

= ipeo

Proof. We may assume that y is extended to a morphism s' + T' of monoids. Let sl, s, E S. T h e condition usiv E A for u, v E s' is equivalent with (uv)(siip)(vv) E B. Since i p : s'+ T' is surjective, it follows that s1 -A s2 holds iff slip -B s z p This defines the isomorphism y as required I PROPOSITION 1.2. Let y : S + T be a morphism of semigroups and let A = Bq-lfor some B c T. There exists then a subsemigroup T' of T and a surjective morphism y : T'+S//A Proof. Let T' = Sip, B' = Aip, and let y : S / / A m T'//B' be the isomorphism given by Proposition 1.1. Then the composition

T'

QB'

--+

T'//B'

S//A

yields the required surjective morphism. The conclusion of Proposition 1.2 may also be recorded as

SIIA PROPOSITION 1.3.

ment of S, and f: T


Let S be a semigroup, A, A' subsets of S , s an elea morphism. Then:

+S

1. Syntactic Sernigroups

187

S / / A= S / / ( S- A). (i) n A’) ( S / / A ) x(S//A’). (ii) (iii) S//(s-’A)and S//(As-’) are quotient sema&oups of SIIA. (i.1 TI/(& - l ) VIA.

<

<

We recall the definitions

KIA= { u ~ u E SS ,U E A } ,

As-’= { u ~ u E SU, S E A }

(i) follows from the observation that the syntactic congruences of A and S - A coincide. T o prove (ii) consider the morphism Proof.

f?J: 5-

+

( S / / 4x (SIIA’)

whose coordinates are the syntactic morphisms

Let B c S / / A and B’ c SIIA‘ be such that A = Bell, A’ = B‘el?. Then A n A’ = ( Bx B’)qrl. Thus the conclusion follows from Proposition 1.2. T o prove (iii) set u = s @ A E S//A. If A = Atell, then

s-’A

=

(u -1A’)el1,

AS-’

=

(A’u-’)ell

Thus the conclusion follows from Proposition 1.1 applied to the surjective morphism P A : S + S//A. T o prove (iv) observe that if A = A’eal with A’ c S / / A , then Af -I = A’eA’f -l = A’(feA>-’. Thus the conclusion follows from Proposition 1.2 applied to the morphism q~ = f @ A : T S//A -+

A subset A of a semigroup S will be called rigid if s1 -A s, implies s1 = s2, or equivalently if P A : S S / / A is an isomorphism. A semigroup S that has a rigid subset will be called syntactic. Clearly for any semigroup S and any subset A of S, the semigroup S / / A is syntactic because A@,is a rigid subset of S//A. -+

PROPOSITION 1.4.

If A is a rigid subset of

This follows from the fact that

-A

and

S, then so is S - A. coincide.

PROPOSITION 1.5. If {Si , i E I } is a family of syntactic semigroups, then so is their product S = Si.

n

VII. Varieties of Sets

188

If Si = 0 for some i E I , there is nothing to prove. Thus we may assume that Si# 0 for all i E I. Let Ai be a rigid subset of Si. In view of Proposition 1.4, we may assume that A if 0 for all i E I. Let A = A i and let zi:S + Sibe the projection morphisms. If tni. This implies sni= tni for all s, t E S and s -A t then sni i E I and thus s = t. Consequently A is rigid and S in syntactic I Proof.

n

PROPOSITION 1.6. Let S be a semigroup. For each s E S consider the

syntactic morphism

es: s-+ Slls

and let y:

s

+

n Slis

8 ES

be the morphism dejned by

{e,, s E S } . Then y

is injective.

Proof. Let s, t E S and sy = ty. But then sen = tea and s -d implies s = t. Thus y is injective I

t. This

COROLLARY 1.7. Every semigroup (or monoid) S is isomorphic with a subsemigroup (or submonoid) of a syntactic semigroup (or monoid) T. If S is jinite, then T may be chosen to be jinite 1 COROLLARY 1.8. Each S-variety (or M-variety) V is generated by the syntactic semigroups (or monoids) that it contains.

Indeed S

E

V implies Sl/s E V and thus also

n Slls E V

I

EXERCISE 1.1. Let G be ajinite group and A a subset of G. Show that A is rigid isf K A f A for all non-trivial invariant subgroups K of G. Deduce that all singletons in G are rigid. EXERCISE 1.2. Show that the monoids

Show that semigroup.

uk (k > 2 ) with

uk with k > 2 are not syntactic.

the unit element removed is not a syntactic

2. Syntactic Semigroupr and Recognizable Sets

T h e considerations of Section 1 were purely algebraic and involved no finiteness conditions of any kind. We shall now apply them in two cases which are of utmost importance for the study of recognizable sets.

2. Syntactic Sernigroups and Recognizable Sets

189

In the first case we set S = Z* where Z is a finite alphabet. For any subset A of Z*, the semigroup Z*llA is a monoid denoted by M p and called the syntactic monoid of A. The syntactic morphism P A is denoted by p.4: Z*

+ MA

T h e set A is recognizable iff M A is a finite monoid. This could be taken as a definition, but is a theorem if recognizability is defined using automata. PROPOSITION 2.1. For any subset A of ,Z* and any monoid M the following conditions are equivalent:

There exists a morphism p: Z* + M such that A B c M. (ii) M A M (i)

-

= Bv-l

for some

<

(i) 3 (ii). Proposition 1.2. (ii) (i). Since MA M, there exists a submonoid M' of M and a surjective morphism y : M' + Ma*.Since Z* is free, there is a morphism p': 2* -+ M' such that p A = ~ ' y Let . p: ,Z* + M be the morphism defined by p'. Define B = ApAv-'. Then Proof.

<

A

= ApApl' = ApAy-lp'--l =

Bp'-'

zz

Bp-1

I

PROPOSITION 2.2. For all subsets A , A' of Z* the following hold: (i) MA = Mz*-A. MA x MA,. (ii) M A n A t (iii) Mo-IA< M A and MAo-l MA,for all (T E Z. MA for all monoid morphisms f : r*-+ Z*. (iv)

< <

<

This follows from Proposition 1.3

I

The second case that will be of interest to us is the case S = Z+where Z is a finite alphabet. For any subset A of Z+ the syntactic semigroup Z+llA is denoted by SA and the syntactic morphism P A is denoted by EA:

Zf

+ s A

Since A may also be regarded as subset of Z*, we also have PA:

Z'

-+

MA

VII.

190

Varieties of Sets

The syntactic congruence in Z+ agrees with that defined by A on Z* and therefore the morphism p A agrees with EA on Z+. Consequently

where 1 is the unit element of M A . Thus SA is finite iff MA is finite, i.e. iff A is recognizable.

For any subset A of Z+ and any semigroup S the following conditions are equivalent: PROPOSITION 2.1s.

(i)

There exists a morphism p: Z++ S such that A

= Btp-l

for some

B c S. (ii) S A

< S.

The proof is a replica of that of Proposition 2.1 PROPOSITION 2.2s.

1

For all subsets A, A' of Z+the following hold:

(i) S A = S,p-A. SA x S A t . (ii) S A n A t SA for all a E Z. (iii) So-IA SA and S,,-, -+ Z+. (iv) SAr-I SA for all semigroup morphisms f : r+

< < <

<

This follows from Proposition 1.3

I

It should be noted that in (iii) the sets a-'A and Aa-' are defined relative to Z+ as the ambient semigroup thus

a-'A

= {U

Iu E

Z+, uu

E

A}

and similarly for Aa-l, while in Proposition 2.2 we have

a-'A

=

{u 1 u

E

Z", uu E

A)

An important point is to compare statements (iv) in Propositions 2.1 and 2.1s. In the latter, the morphism f: r+.--, Z+admits an extension f: P -+ Z# by setting If = 1. The resulting morphism r*+ 2 7 is non-erasing, i.e. satisfies yf E Z+ for all y E I: In (iv) of Proposition 2.2 arbitrary monoid morphisms f: r*-+ Z* can be used. In this section and in the preceding one, we emphasized the purely algebraic nature of arguments by making no assumptions of recognizability. In the rest of this chapter (and also in the exercises below) we

2. Syntactic Semigroups and Recognizable Sets

191

shall assume that all subsets A, A' of C* or Cf considered are recognizable and that all monoids and semigroups, except for C* and its subsemigroups, are finite. PROPOSITION 2.3. A jnite monoid (or semigroup) S is syntactic if and only if there exists a Jinite alphabet 2 and a recognizable subset A of C* (or of 2.) such that S % M A (or S M SA).

Proof. Let A be a recognizable subset of C*.Then M A = P / / A is syntactic. Thus S M M Aimplies that S is syntactic. Conversely assume that S is syntactic with a rigid subset B. Choose C so that there exists a surjective morphism q ~ :C* --+ S and define A = By-'. Then by Proposition 1.1, MA M S / / B = S. T o treat the semigroup case replace Z* by Z+ I

In A.III,8 a large number of examples of syntactic monoids and semigroups are computed using minimal automata. EXERCISE 2.1. Let Z = u and let A = Z+ c C x .Show that SA = 1 while M A = U , = SA1. Thus S, # MA even though SA is a monoid. EXERCISE 2.2. Let y : C* + G be a surjective morphism, where G is a jinite group. Show that G M MBp1-' for all g E G. EXERCISE 2.3. Show that an element m of a (finite) monoid M is invertible isf mm' = 1 for some m' E M . Hint: Show that the transformation x + xm is a permutation. EXERCISE 2.4. Deduce from Exercise 2.3 that a (finite) monoid M is a group ;ff MmM = M for all m E M . Give an example showing that this condition is not sujicient when M is infinite. EXERCISE 2.5. Given a (recognizable) subset A of 2* and the syntactic morphism pA4 : 2* -+M A ,denote by GA the group of all invertible elements of M z 4 . Dejine A' = GAp3' A D= lp-il,

Show that each initial and each terminal segment of a word in AG again is in AG.Conclude that AG= r*for some subset r or C. Show that the following conditions are equivalent:

VII. Varieties of Sets

192

(i) GA = 1. (ii) A@= A D . (iii) AD = for some

r*

r c 2.

Show that the EXERCISE 2.6. Assume in Exercise 2.3 that A c C+,. following conditions are equivalent : (i) SA = MA - 1. (ii) A D = 1. (iii) A@= 1.

Conclude that

if GA # 1,

=

MA.

Show that the following conditions are equivalent for

EXERCISE 2.7.

A

then S,

c Z*:

(i) M A is a group. (ii) A D n sZ*# 0 for all s E Z*. (iii) A D n C+s # 0 for all s E C*. (iv) A D n C*sC* # 0 for all s E C*. (v) There exists an integer k 2 1 such that sk E A D for all s

E

Z*.

3. Varieties of Sets

Suppose that for each finite alphabet C a family C * Y of recognizable subsets of Z* is given. We shall then say that

Y ={ P F } is a +-class of sets. Given such a *-class T we define the M-variety

V

=

(MAI A

E

Z * Y for some Z)M

Thus V is the M-variety generated by the syntactic monoids of all the sets in F. We shall use the notation

to indicate that V is defined by F i n the above manner. Given an M-variety V we may define the *-class T by setting

Z * Y = { A I A c C*,MA E V)

3. Varieties of Sets

193

We shall write

V * Y to indicate that Y is defined by V in the above manner. THEOREM 3.1. For any M-variety V , the relation

V6.Y implies

Y-V Proof. Let Y - V ’ . For any A E Z + Y we have MA E V by the definition of V s T. However the definition of Y - V ‘ shows that V’ is generated by the monoids MA; thus V’ c V. T o prove the opposite inclusion V c V’ we consider a monoid S E V and assume that S is syntactic. By Proposition 2.3, there exists a finite alphabet Z and a recognizable subset A of Z+ such that MA w S. Then MA E V and it follows that A E Z + Y . Thus MA E V‘ and consequently S E V’. Since, by Corollary 1.8, V is generated by its syntactic monoids, it follows that V c V’ I THEOREM 3.2.

Let Y be a +-class and let

Y-=+V*Y Then Y c p.The equality Y = holds satisjies the following four conditions :

if

and only

if

the +-class

Y

(3.1) If A E Z + T , then Z+ - A E Z + T . (3.2) If A, B E Z + T , then A n B E Z + Y . (3.3) If A E Z + T and a E Z, then the sets a-’A = {w I w Aa-’ =

{W

E

Iw E

Z+) ow

E

Z+, w a E

A} A}

are in Z+Y-. (3.4) If f : P -+ Z+ is a morphism of monoids and A f -l E F + Y .

if A

The proof is deferred to the next section. We note that conditions (3.1) and (3.2) jointly assert

(3.5) Z + Y is closed under boolean operations.

E

Z + Y , then

VII. Varieties of Sets

194

In view of Theorem 3.1, we shall say that a +-class Y satisfying conditions (3.1)-(3.4) (or equivalently conditions (3.3)-(3.5)) is a +-variety. COROLLARY 3.3. If V is an M-variety and V e-Y , then Y is a

+-variety. Indeed we have by Theorem 3.1

V=Y*V This implies

Y=>V*Y and thus Y i s a +-variety, by Theorem 3.2 The results above add up to the following THEOREM 3.4.

M-varieties and +-varieties are in a 1-1-correspondence

given by

v-Y

Y*V*

The notion of a +-class is defined exactly like that of a +-class but with Z+ replaced by Z+. Thus a +-class

r== {Z+T} consists of a family 2+Y o f recognizable subsets of Z+defined for every finite alphabet Z. For each such +-class Ywe define the S-variety

V

=

(S, I A

E

Z+ Y for some Z)s

-

generated by the syntactic semigroups of sets in Y. Notation : Y For each S-variety V we define the +-class Y (notation: V defined by Z + Y = { A I A c Z+, s, E v> THEOREM 3.1s.

V.

Y)

For every S-variety V, the relation

V * Y implies

Y-V The proof is a replica of that of Theorem 3.1 and is omitted

1

3. Varieties of Sets THEOREM 3.2s.

195

Let

F be

a +-class and let

Then Yholds if and only if the +-class Y c Y7 The equality Y = satis-es the following four conditions :

(3.1s) r f A E Z+U, then C+- A E Z + F . (3.2s) If A, B E Z+Y, then A n B E Z + Y . (3.3s) If A E Z + Y and u E Z, then the sets

I w E Z+, uw E A } A u - l = {w I w E Z+, w u E A } a-’A

= {W

are in Z + Y . (3.4s) iff: r+ Z+ is a morphism of semigroups and Af -* E T + F . -+

if A E Z + Y , then

For the proof, see the next section. Conditions (3.1s) and (3.2s) may be replaced by

(3.5s) Z + Y is closed under boolean operations. T h e +-class Y is called a +-variety if it satisfies (3.1~)-(3.4s) (or equivalently if it satisfies (3.3~)-(3.5s)). COROLLARY 3.3s.

+-variety

I

THEOREM 3.4s.

dence given by

If V is an S-variety and V * Y , then F is a

S-varieties and +-varieties are in a l-l-correspon-

T-v,

V=.Y

In comparing the axioms defining a +-variety with that of a +-variety, we observe that conditions (3.1~)-(3.4s) are simple adaptations of conditions (3.1)-(3.4) obtained by passing from 2 3 to Z+. However, in some sense, axiom (3.4) is much “stronger” than (3.4s). Indeed a morphism f:Tf Z+ may be extended to a morphism f: T y -+ Z* by setting If = 1. The resulting morphism is non-erasing, i.e. satisfies yf # 1 for all y E T. Moreover all non-erasing morphisms can be thus obtained. Axiom (3.4) is asserted for all morphisms f: re .ZCx and not just the non-erasing ones. This is an important difference between +-varieties and +-varieties. --f

----f

VII. Varieties of Sets

196

Let 59 be the #-variety corresponding to the variety G of groups. Let Z = u and let A = (c2))*. Then M A = 2, E G and thus A E Z * Y . Now consider the augmented alphabet f3 obtained by adding a new letter t to Z. Viewed as a subset of f+ the set A has as syntactic monoid Zzo,i.e. the group 2, with a zero added. This is no longer a group and thus A 4 LkV. This shows that the informal notation A E Y is dangerous. It also shows that the inclusion morphism Z+ + f+ does not carry z#Pinto 2*P. EXAMPLE 3.1.

EXERCISE 3.1. Given any +-class F d e f i n e the +-variety v b y

T&-V=>F

Show that is the least +-variety containing F and that y = F. Similarly for #-classes. EXERCISE 3.2. Show that the empty S-variety corresponds to the empty +-variety. Similarly the empty M-variety corresponds to the empty 1c-variety. EXERCISE 3.3. Assume that V is the S-variety consisting of the empty semigroup 0 alone. Show that the corresponding +-variety Y contains only one set, namely, the empty subset of Z+for the empty alphabet Z. I f Z # 0, Z + F is empty.

V is the S-variety consisting of the semigroups 0 and 1. Show that in the corresponding +-variety Y, the family Z + F consists of the sets Z+ and 0. EXERCISE 3.4. Assume that

EXERCISE 3.5. Assume that V is the M-variety consisting of the trivial monoid 1 only. Show that in the corresponding +-variety Y, the family Z*Tconsists of the sets .Z* and 0. EXERCISE 3.6. Given anyfinite alphabet Z, define the augmented alphabet f3 = Z v t where t is a letter not in Z. Define the morphism n: + .Z* such that an = a, tn = 1. For each A c Z* &fine A = An-'. Prove that for each recognizable subset A of Z+

z#

4. Proof of Theorems 3.2 and 3.2s

197

EXERCISE 3.7. Show that axiom (3.4) for a *-variety is equivalent to the conjunction of the following two axioms:

I f f : r++ Z* is a non-erasing morphism of monoids (i.e. if If and A E Z * Y , then Af -l E r*Y. (ii) If A E Z * F , then A E t*Y.

(i)

--I

=

1)

r*

Given A c E*, B c where C and I' are non-overlapping alphabets, consider the external shuffle product A LLI B c (Zv r)* as defined in A,II,3. Show that if Y i s any *-variety, then A E Z * F a n d BEPTimplyAwBE(Zur)+F. EXERCISE 3.8.

EXERCISE 3.9.

Show that MAUL3

-

MA x MR

4. Proof of Theorems 3.2 and 3.2s

We shall only conduct the proof of Theorem 3.2s. We ask the reader to verify at each step that the arguments remain valid for +-classes and M-varieties. We consider the situation

F-=+V*Fwhere Y is a given +-class. Let A E Z + Y . Then by definition of V, we have S, E V, and thus by definition of p,A E Z+P. Consequently Y c

. 7 .

Next we observe that Proposition 2.2s implies that p satisfies conditions (3.1~)-(3.4s). Thus if F = p,then F a l s o satisfies (3.1~)-(3.4s). Thus the conditions of Theorem 3.2s are necessary. Next we assume that Y satisfies (3.1~)-(3.4s). We must prove that c Y. For this we consider a set A E Z+p. From the definition of V p i t then follows that SA E V. By definition of V, there exist then sets A iE Zi+F-, 15i5n

-

r*

such that SA

The syntactic morphisms

<s A , x

* * *

XSA,

VII. Varieties of Sets

198

define a surjective morphism Y = ~ ~ X . . . X ~W , :4

where

T

w = Z 1 + X . . . xzn+ T

Since SA morphism

< T, it

r=

. .

S A ~ X. X

SA,,

follows from Proposition 2.1s that there exists a cp: Z+-+T

and a subset B of T such that

A The morphism

= Bcp-’

v admits a factorization Z+&WY’T

Let

yi: Z+ -fZi+,

1 < i
pi: Z+* s A i ,

1
and let

be the coordinates of y and cp. Then

pli = y i f i . We

have

Since Y is closed under boolean operations, it suffices to show that tv-l E 2+Yfor all t E T. If t = ( s l , . . . , s,) E T, then

n n

tp-l =

sicp;l

i-1

Thus by (3.2s) it suffices to show

Since s& that

= (s&l)y;l,

it follows from (3.4s) that it suffices to prove s&1

for all si E

SAI

and all 1 5 i 5 n.

€

zi+r

5. Operations on Varieties

199

Thus to complete the proof it suffices to establish

(4.1) If A

E

Z+Tand

5: Zt * S A is the syntactic morphism of A, then for all s

E

SA

sE-' E Z + Y

This, of course, under the assumption that Y satisfies (3.1~)-(3.4s). Note that until now we have not used condition (3.3s) and it is expected that it will have to be used in proving (4.1). We define for each w E Z+

Rw = {(u, v ) 1 u, v E Z", uwv = ((u, v ) I u, 2, E

P, w E

E

u-1Av-'3

From the definition of the syntactic congruence w

W'

-A

iff

A}

-A

in Z+ it follows that

Rw = Rw#

Consequently, the congruence class of w , i.e. the set w66-l is the set

(4.2)

n

u-l~v-l-

(u,I I ) E R ~

u

u-'Av-1

( u , v )#Rw

Since A is recognizable, only a finite number of the sets u-'Av-l are distinct. Therefore, the apparently infinite intersection and union in (4.2) are indeed finite. Repeated use of (3.3s) shows that the sets u-'Av-' are in Z + T . Since Z + Y i s closed under boolean operations, it follows that (4.2) is a set in Z + Y . Thus w6E-I E Z + T for all w E Z+ and consequently sE-1 E Z + T for all s E SA 5. Operations on Varieties

Let V be an S-variety (or an M-variety). If for each semigroup (or monoid) S in V we consider the reversed semigroup (or monoid) Se, we obtain again an S-variety (or an M-variety) which we shall denote by Ve. Observe that if V is closed, then Ve need not be closed; this is because of the non-symmetric nature of the semidirect product. Note, however, that if V is a variety of groups or if V is a closed variety containing U , then V = Ve.

VII. Varieties of Sets

200

Y be a

+-variety (or a #-variety) of sets. If we replace each set A E Z + Y (or A E Z * Y ) by its reversal AP, we obtain a +-variety (or a x-variety) denoted by T o . Let

PROPOSITION 5.1.

If V o Y, then VBo Y e .

In this statement V could be either an S-variety or an M-variety. The proof follows from the elementary observation that the syntactic semigroup S,, is simply SAe,Similarly for syntactic monoids I Let X be a set and 3- a class of subsets of X.The class $2- is said to be boolean closed if it satisfies the following conditions : (5.1) 0 E F. (5.2) If A E .F, then X - A E $2-. (5.3) If A, B E K,then A n B E 3-.

y

If is any family of subsets of X,then the least boolean closed family of subsets of X containing is denoted by @J4k and is called the boolean closure of 6$Y. We shall also say that Y i s a set of boolean generators for

yk PROPOSITION 5.2. Let V o Ywhere V is an S-wariety (or M-variety) i E I } of semigroups (or monoids). Then generated by the family {Si, Z + Y (or Z * Y ) is the boolean closure of the family of all sets of the form sp-' where p: Z++ Si (or p: Z* -+ Si) is any morphism with i E I and s is any element of Si. Proof. Clearly SF-' E Z + F . Conversely, assume A E Z + Y . Then SA < S, X . . . X S,, for some semigroups S,, . . . , S, in the family (Si,i E I ) . By Proposition 2.1s, there exists a morphism p: Z+-+ S , x . .. xS, such that A = By-l for some B c S,x . . . x S,. Let pi: Z+-+Si, 1 5 i 5 n be the coordinates of p, Then

A=

un

Sip;l

15i S n

the union extending over all elements s = (s,,

. . . , sn)

of B

I

We now turn our attention to lattice operations on varieties. Given any family (Vi,i E I} of S-varieties, the intersection nV, is again an S-variety. The union (JVi in general need not be a variety but it generates an S-variety that we shall denote by VV, and call the join of the family

5. Operations on Varieties

201

{Vi}. If the family {Vi} consists of the varieties V, , . . . ,V,, then we shall write V, v . . . v V, for the join. Clearly, if V,, . . . ,V, are nonempty, then S E V, v . . . v V, iff S S,X . . . x S , with Si E Vi for i = 1, . . . ,n. The same notions and notations apply also to M-varieties. i E I } of +-varieties, the +-variety is Given a family {Ti, defined by setting = n(Z+Yi).Next we define Z + T to be Boolean closure of the family U Z + F i . It is easy to check that T is a +-variety and is the least +-variety containing Tifor all i E I . Thus T is the join V T i . The same notions and notations apply also to rr-varieties.

<

nTi

Z+(nTi)

PROPOSITION 5.3.

If Vio Tifor all i E I , nvi n r i

then

0

VVi 0 V T 4 This is an immediate consequence of the fact that e is a bijection which preserves inclusion I PROPOSITION 5.4.

Let V be an M-variety and let

V-Y;

(v)s=dw-

be the corresponding #-variety and +-variety of sets. Then

Z + W = { A n Z+I A E Z # T } If, further, U, E V , then

Z + Y = {AI A

c Z+, A E

Z*Y}

We need two auxiliary propositions. PROPOSITION 5.5,

If A is a recognizable subset of Z", then sAnZ+

<

MA

Proof. Let p A : Z* -+M A be the syntactic morphism of A , and let B = ApA. Let v: Z+ M A be the restriction of pa. Then A n Z+ = Bp-' and thus the conclusion follows from Proposition 2.1s I ---f

PROPOSITION 5.6. Let A c Z+ and let M be a monoid such that SA < M . Then either MA < M 01 MA", < M .

202

VII. Varieties of Sets

Proof. By Proposition 2.1s there exists a morphism p: 2+--+ M such that A = Bp-I for some B c M . Extend p to a morphism y : 2*+ M be setting l y = 1. Then By-’ is either A u 1 or A depending on whether 1 E B or 1 4 B. Thus the conclusion follows from Proposition 2.1 1 PROOF OF PROPOSITION 5.4. Let A E 2*T.Then MA E V and thus SAnz+ < M A , by Proposition 5.5. Consequently SAnz+E (V)s and A n Z+ E ZfW: Assume B E Z + Y , i.e. SE E (V)s. I t follows that S, M for some M E V. By Proposition 5.6, we have MB M or MEvl M. Thus setting A = B or A = B u 1 as the case may be, we find A E 2*T and B = A n Zf, as required. If U, E V, then 1 E 2 ° F and thus B = A - 1 is in 2*T I

<

<

<

We recall that the assumption U, E V signifies that V is not a variety of groups. If V is a variety of groups, then 1 4 2*Tprovided Z # 0. Thus both B and B u 1 cannot simultaneously belong to 2*T. EXERCISE 5.1. Let V be an S-variety such that U , E V and let V o F. Establish the equivalence of the following conditions :

(i) (ii)

V is monoidal.

If f : r*+ Z* is a morphism of monoids and i f A E 2.W; A f -l E F + Y . (iii) If A E 2+Y,then A E f + F .

then

Hint: Use Proposition 5.4 and Exercise 3.5. 6. The Syntactic trn and t s of a Set

All automata considered here are deterministic and the word “deterministic” will systematically be omitted. Let

a‘= (9,i, T ) be a 2-automaton. In particular Q is a right Z*-module. Thus the pair (Q, 2.) and the action of Z*on Q represent a tm

TMi/

=

(Q, Mi/)

where Mi/ is the monoid of all partial functions f : Q

--f

Q of the form

203

6. The Syntactic t m and t s of a Set

qf = qs for some s E 2*.There results a surjective morphism

l of d is a subset of L'f, i.e. 1 4 I dl. If i @ T , then the behavior I d Thus only paths of strictly positive length in d a r e needed to compute 1 &I. Consequently it is natural to regard Q as a right Z+-module (rather than a Z*-module). The pair (Q, L'+)then represents a ts Tsd=

(8,SJ)

and we have the surjective morphism [&0:

Z+-+S,

Clearly

T M d = TSJ* Sd c M d = S ~ 1, U

Consider a recognizable subset A of Z* and let &A

= (QA,

i ~ T, A )

be its minimal automaton (see A,III,S). The tm associated with the automaton dA is then denoted by

TMA = ( Q A M A ) 9

and we have the surjective morphism Z*

-+

MA

The fact that MA and p A coincide with the syntactic monoid and the syntactic morphism of A, as described in Sections 1 and 2, is a fairly elementary consequence of the explicit description of the minimal automaton dk(see A,III,lO). Because of this we call TM, the syntactic tm of A. If, in the above, A is a subset of Z+, then iA 4 T A , and we may consider the ts of dA. It is denoted by

TSA= ( Q A , S A ) and we have the surjective morphism

204

VII. Varieties of Sets

which of course is nothing else than the syntactic morphism of A . The ts TS, is called the syntactic ts of A . Clearly

TMA = TSA' SA

~ M A = S A U1,

and the morphism p A agrees with tAon Z+. It would be erroneous to conclude from the above that MA = SA'. Indeed, it may happen that SA is a monoid in its own right, but that its unit element does not act as the identity transformation on QA. This situation is present in EXAMPLE 6.1.

Let Z # 0. The automaton , 0 2 + O c J E

is then the minimal automaton of the set Z+. The semigroup S+ , has a single element and this is a monoid. However, ME+= U , = S&+. PROPOSITION 6.1.

Let &be a Z-automaton with behaaior A

c Z*. Then

TM, TM,'

=

Id I

< TM, < TMdc

MA < M d

If A

c Z+, then also

TSA < T S d TSAc < TSdo SA

<S d

Proof. The passage from the automaton d = (&, i, T )to the minimal automaton d,of A consists of two steps. First we pass from d to the trim automaton d t = (Qt,it, T t ) in which all the states in Q which either are not accessible or not coaccessible are removed. The tm TM,t is then the restriction TMd I Qt and therefore TM,t TM,. The passage from d tto dA is obtained by merging equivalent states in Qt. This yields a surjective function fp: Qt-+ QA such that for each s E Z* the partial function QA ---t QA given by s is covered by the partial function Qt + Qt given by the same s. Thus T M . TMdt.

<

<

6. The Syntactic t m and t s of a Set

205

The second inequality is proven similarly using the complete minimal automaton dAc of A. The assertion MA Md follows from the second inequality. The inequalities for the case A c C+ are proved similarly I

<

PROPOSITION 6.2.

For all recognizable subsets A , A’ of Z* the follow-

ing hold: (i) (ii) (iii)

TMAnA,< TMA x T M A , . TM,,-1.4< TMA and TMA,-l TMAfor all u E 2. TM,,-, < TMAfor all monoid morphisms f : r*.--t Z*.

<

If A, A‘ are subsets of 2+,then also (i.s) TSAnAg (iiis) TSA,-l Proof.

< TSAX TSA,; < TSAfor all semigroup morphisms f: I’+

-

Z+.

Let

a’= (Q, i, T ) ,

a”

=

(Q’, i‘, T ’ )

be the minimal automata of A and A’. We define the 2-automaton =

( Q x Q’, (i, i’), T X T ’ )

with action

{

(40, q‘c) (4, 4’)u = 0

if qu # 0 and otherwise

Q‘U f

0

It is clear that

1 9 1= I d l n Id’[ = A n A’ and that

T M s c TMJx TMJt

=

TMAx TMAt

Consequently (i) is implied by Proposition 6.1. If A, A’ c 2+,then we also have

TS, c T S ~ TSdg X so that (i.s) follows from Proposition 6.1. To prove (ii) we construct the automata u - l d = (Q, iu, T ) ,

da-l

=

(Q, i, Tu-’)

VII. Varieties of Sets

206

where Ta-l

=

(q I qa E T ) . Then

Thus (ii) follows from Proposition 6.1. To prove (iii) we convert d into a r-automaton 9 5 'by setting qy = q(yf ). Then IS/= Idlf-'=Af-'

T M a c T M d = TMA Thus (iii) follows from Proposition 6.1.

To prove (iii.s) we extend f : r+ Z+ to a morphism f : and define 9 just as above. Then --+

r*

3

Z*

T S g c T S d = TSA Thus again (iii.s) follows from Proposition 6.1

I

Comparing Proposition 6.2 with the analogous Propositions 2.2 and 2.2s we notice the absence of the assertion concerning complementation. This is the main difference between the syntactic invariants MA, SA and TM, , TSA . In addition, the analogs of the assertions for a-lA and Aa-I for TSA are absent. EXERCISE 6.1.

Show that

CHAPTER

vlll Examples of Varieties of Sets

In this chapter we assemble a number of interesting examples for the correspondence F e V which was the main topic of Chapter VII. 1. General Comments

In Chapter VII we have defined the syntactic invariants TMA and MA for any recognizable subset A of Z". If, further, A c Z+, then we also have the invariants TSA and S,. The difference between Z* and Z+ being slight, we actually have four syntactic invariants to choose from. If we start out with a class &'of sets which we try to characterize by their syntactic invariants, it becomes important to know which of the four available invariants to choose. For this we do two acid tests. The first test is to see whether the class &'is closed under complementation (in Z+or in Z"). If it is not, then the invariants SA and MA cannot be used. The second test is to see how the class behaves under inverses of morphisms f: r*-+ Z". If it is not closed under f-' for such morphisms, then the invariants MA or TMA cannot be used. If it is closed under f-' for non-erasing morphisms only (i.e. morphisms f: r+-+Z+), then there is a possibility of using S, or TSA. There is always the possibility that the given class &'cannot be adequately described by the syntactic invariants of its members. For "naturally" defined classes d this is not likely to happen. In some cases we start with a variety V of semigroups and try to describe the +-variety T of sets given by V e Y. Here we first test V 207

VIII. Examples of Varieties of Sets

208

to see whether or not, it is monoidal. If V is monoidal, then we replace it by the M-variety V n M and try to find the corresponding *-variety y. There is also the question as to how one "describes" the +-variety (or the x-variety) Y. We recall that T defines a family C + Yof subsets of ,Z+ for each finite alphabet L'. We give preference to descriptions that give C + Y i n terms of a single alphabet Z (i.e,, descriptions that do not utilize morphisms f: r+ + Z+). Sometimes we can give a global description of Z + Y . Sometimes we have to be satisfied with indicating boolean generators for Z + Y . The examples in this chapter were selected so as to illustrate the various points made above. More examples will be given in the subsequent chapters as stronger tools become available. 2. Finite and Cofinite Sets

As our first example we consider the class of all finite subsets of Zx (or of Z+). Since this class is not closed under complementation, the syntactic invariants MA or SA cannot be used. Further, if f: r* ,Z* is a morphism of monoids and if A f 0 is a finite subset of C*,then Af -l is not finite unless f is non-erasing, i.e. f is the extension of a morphism f: ,Z+ r+. Thus by Proposition VII,6.2(iii), the syntactic invariant TM, cannot be used. Thus the search is narrowed down to the syntactic invariant TS,, and this implies that we should consider only finite subsets of Z+. Here the search succeeds and we have -+

-+

PROPOSITION 2.1. A subset A of

Z+is jinite if and only if the ts T S ,

is strongly nilpotent. We recall that a ts X = (Q, S) is said to be strongly nilpotent if 8 is the only idempotent of S. The class Nil of strongly nilpotent ts's was introduced in Example III,6.1 where we have shown that

Nil = L[O'] This class reappeared in IV,4 where it was shown that

Nil

= (1')

Thus the class Nil (as an exclusion of a prime) is closed. Actually more is true: Nil is a two-sided ideal in Ts,since 1' X o Y implies 1' X and 1'< Y.

<

<

209

2. Finite and Cofinite Sets

Let d =(Q, i , T ) be the minimal automaton of A and let TS, = (Q, S ) . Assume that A is finite and let n be an integer exceeding the length of all words in A. We assert that QZn = 0. Indeed, assume qw = q’ for q, q’ E Q and w E Zn. Since q is accessible and q’ is coaccessible, we have iu = q, q’w E T for some u, v E P.Consequently, iuwv E T so that uwv E A contrary to assumption. Having shown that QSn= 0 it follows that 8 is the only idempotent in S. Thus T S , is strongly nilpotent. Conversely, assume that TS, is strongly nilpotent. By Proposition III,9.2 we have QSn c QES for some n 2 1, where E is the set of all idempotents in S. Since E reduces to 8, we have QSn = 0 and thus QZn = 0. Consequently all successful paths in A have length < n . Thus A is finite I PROOF OF PROPOSITION 2.1.

Observe that we have actually shown that QSn = 0 hoIds iff all elements of A have length less than n. The picture changes if instead of considering the finite subsets of Z+, we consider the subsets A of Z+ that are either finite or are cofinite (i.e. are such that Z+- A are finite). We shall denote this class of sets by Z+X. We shall see below that JT is a +-variety. A semigroup S will be called nilpotent if it satisfies either one of the following three equivalent conditions : (2.1) Se = e = eS for each idempotent e in S. (2.2) Either S = 0 or S is a semigroup with a zero and no other idempotents. (2.3) Either S = 0 or S is a semigroup with a zero 0 such that S” = 0 for some integer n 2 1. The equivalence (2.1) e (2.2) is clear and so is the implication (2.3) + (2.2). The implication (2.2) s. (2.3) follows from Proposition I 11,9.2. PROPOSITION 2.2. The class N of nilpotent semigroups is an S-wariety. The class JT= {Z+JT}of subsets of C+that are either finite or cofinite forms a +-variety of sets. Further

N + X Prmfi The first statement is clear. If A c Z+is finite, then by Proposition 2.1, TSA = (QA, S A ) is a strongly nilpotent ts. Thus SA E N.

210

VIII. Examples of Varieties of Sets

If A c Z+ is cofinite, then Z+ - A is finite. Since S,+-, = S,, it follows that SA E N. We have thus shown that A E Z + X implies SA E N . Conversely, assume A c Z+ and S, E N. Let tA:Z+ -+ S, be the syntactic morphism, and let A = B&l. Let 0 be the zero of SA and let n be such that SAn= 0. It follows that O&I contains all the words w of length I w 1 2 n. Thus O t i l is cofinite. Consequently A is finite or is cofinite depending on whether 0 $ B or 0 E B. In either case A E Z + X I Let N , be the class of semigroups consisting of the empty semigroup and all semigroups S with a zero 0 such that Sk+l = 0. Show that N k is an S-variety. Define N k o x k . Show that Z + x k is the boolean closure of the singletons s E Z+ satisfying I s I 5 k. EXERCISE 2.1.

EXERCISE 2.2.

Show that the variety N k is defined by the equations

Derive from the above that N k is dejned by the single equation xg

. . . xk=yo

...Yk

3. Finitely Generated Varieties

Let

V be an M-variety with generators MI, . . . , M k . Thus

Replacing M I , . . . ,M k by Mo = M , x

. . . x M k we obtain

v = (Mob Thus V is generated by the single monoid M o . Let F be the +-variety corresponding to V. From Proposition VIIJ.2 we have

(3.1) Z * F i s the boolean closure of the sets my-' for all m E Mo and all morphisms y : Z#--* M , .

As a consequence we obtain

(3.2) Z * T is finite.

3. Finitely Generated Varieties

211

We also show

(3.3) Let y : Zo* M,, be a surjective morphism. For any finite alphabet 2, the family Z * T is the boolean closure of the sets --f

Am,? =

where m

E

(T-llrp-'

M , and rp: Z* + Zo* is any morphism.

Indeed since y is surjective, any morphism y : Z* - M 0 admits a factorization y = 'py. Thus my-,= Am,p. We shall now calculate the classes Z*w^where

v=

VOY,

(MO)M

for various specific monoids M , .

EXAMPLE 3.1. Let M , = 2, be the cyclic group of order p 7 1. The M-variety V = ( Z P his easily seen to be the M-variety defined by the pair of equations xy = y x , xp = 1 Let a be a generator of 2, and let y : N 2, be the surjective morphism defined by l y = a. The sets ky-1 with k E 2, then are --f

B k = {n I n

E

N , n = k mod p }

for k E p. We may use N in place of Zo*in (3.3) and thus consider the sets Bkrp-l for all possible morphisms 'p: ?,? -+ N . Since N is commutative, we may define the sum p = p1 p2 of two morphisms y l , 'p2: Z* N by setting srp = s'pl srpz. This implies that

-

+

+

+

the union extending over all pairs K, ,K, such that k , , k, E p and k, k, = k mod p . This shows that it suffices to consider morphisms p: Z*+ N that generate the monoid of all morphisms Z* N . Such a generating family of morphisms is obtained by defining for each letter u -+

1 lo: Z* -+ N

I.I~={:

if if

z=u t f a

Thus I s lu is the number of times the letter u appears in the word s E z*.

VIII. Examples of Varieties of Sets

212

As a consequence of the above discussion we obtain

(3.4) Z + T i s the boolean closure of the sets

($1

I s Iu

E

R mod p )

for all R E p, a E Z. EXAMPLE 3.2.

Let M , = Z(l,r).If a is the generator, then the ele-

ments of Z(,,') are

1, a,

...,

a2,

=

The M-variety V = (Z(l,r))M is easily seen to be the M-variety defined by the pair of equations x'+1 = x'

xy = yx,

A discussion quite analogous to that of the preceding example leads to the following conclusion: (3.5) Z * T is the boolean closure of the sets {s for all R E r, a

E

I Is

Iu

=R

)

Z.

Consider the special case of Example 3.2 with Y = 1. Then Z,,,,) = U , and V = (Ul)Mis defined by the equations xy = yx and xa = x. Thus V = J, is the variety of commutative and idempotent monoids. From (3.5) we deduce that Z * T i s the boolean closure of the sets EXAMPLE 3.3.

{s

I I s lu

=O} =

(Z - a)+

Replacing each of these sets by its complement we obtain

(3.6) Let

xl

o J,.

is the boolean closure of the sets

The class

Z*aP for a E Z. Consider M, = Z(p,r)with p > 1, Y > 0. We have seen in Example II,9.2 that Z(p,r)is isomorphic with a submonoid of Z p ~ Z ( l , rSince ) . further 2, Z(p,r)and Z(,,,) Z ( p , r )it, follows that EXAMPLE 3.4.

<

(z(p,r)xcI

<

= (2,sZ(1,r)xcI

3. Finitely Generated Varieties

Thus the variety

V

=

21 3

(Zcp,+))Mis the join

v=v,vv, of the varieties V, = (Zp)M, V, = (Z(l,7))M.Thus Proposition VII,5.3 (for M-varieties) applies. We obtain

(3.7) P Y i s the boolean closure of the sets k

{ s I I s l , = k modp},

~

p

kEr

{sIIsI,=K},

EXAMPLE 3.5. Let V = Corn be the M-variety of commutative monoids. We show that V is generated by the cyclic groups Z, where p is a prime and by the monoids Z(l,r,with r > 0. Indeed, by Proposition V,1.3, V is generated by the cyclic monoids Z(p,r)with p >_ 0, r 2 1. Further Z(p,r) 2, x 2(,,?) and 2, may be decomposed into a direct product of cyclic groups with prime period. The above argument shows that V is the join of the M-varieties ( Z p h with p a prime and (Z(,,?))Mwith r > 0. This proves

<

(3.8) Z * Y is the boolean closure of the sets {s

I I s,

I Ek

mod p } ,

{SII&I=k),

K

E

p, p prime

KEN

Note that P T i s no longer finite. EXAMPLE 3.6. Consider M,, = U , . Let U = (UZ),and M o U.Let 1, a, b be the elements of U , . For any morphism v: Z* --+ U , set

r, = .{

r, = {a I ag,= a>, r, = {u I up = q

ug, = 11,

From this we easily deduce

(3.9) Z*P is the boolean closure of the sets

z*o with 'I c ,Z and

CT

E Z (or equivalently with u E Z - T).

Vill. Examples of Varieties of Sets

214 EXERCISE 3.1.

In Example 3.2 show that C * T is the boolean closure

of the sets (C*U)kZ*

with u

E

Z, 1 5 k 5 Y .

EXERCISE 3.2.

Let A c Z*. Show that MAis commutative ifand only if uutv

for all u, v

E

Z*, u,

E

tE

A

implies

uzuv

E

A

Z.

+-

EXERCISE 3.3. Formulate the analog of (3.3) for semigroups, and varieties. Consider the examples where the generator So of the S-variety is or 2.

4. The Variety D

Next we consider the S-variety D of all semigroups S such that

(4.1) Se

= e for all idempotents

eES

This variety was explicitly introduced in V,3 but implicitly already appears in IV,4. We have shown that

D

= (2')

n S = [TI n S = [2] n S =

[z],,

Thus D is the closed S-variety generated by the semigroup 2 which is obtained from the monoid U , by the removal of the unit element. Let S E D and let E be the set of all idempotents in S. We claim

(44

ES=E=SE

The equality S E = E follows directly from (4.1). Let s E S, since se = e we have eses = ees = es so that es E E and thus ES = E. Since S" = SES for n 2 card S, we obtain

(4.31

Sn = E

for all n 2 card S

Then .Z+9 is the boolean closure in PROPOSITION 4.1. Let D e 9. Z+of the family of all sets of the form

(4.4)

22%

4. The Variety D

for v

E

215

Z+. Alternatively, Z + B consists of all sets of the form

A

(4.5)

=

Z*B v C

where B and C areJinite subsets of Zf Proof. We assume that the reader will verify that the boolean closure of the sets (4.4) coincides with the family given by (4.5). Consider A = 2% with ZI E Z+, I v I = n > 0. We claim

(4.6)

yx

-A

whenever y , x E Z+,

x

Ix I 2n

Indeed, uyxw E A signifies that uyxw has CJ as a terminal segment. Since I x I 2 I v I, this will take place iff u r n has v as a terminal segment. Thus (4.6) holds. From (4.6) we deduce

st = t

for s

E

SA,

t E SAn

Thus, in particular, se = e for all idempotents e in S, and thus SA E D. Conversely, assume A c Z+ and SA E D. Let n = card SA. Thus S A n = E is the set of idempotents of SA. Assume that

y x ~ A , IxI=n Then x t A E SAn so that xEL4= e is an idempotent. Since SAe follows that y x t A t z 1= e t i l = S A e f l 2 Z*x

= e,

it

and thus Z*x c A. Consequently, A has the form

A = Z*B u C where B is a subset of Znand C is a set of words of length t n . Thus B and C are both finite I Automata that recognize sets in the class 59 have been studied and are known as dejiinite automata. EXERCISE 4.1.

Let Dk be the S-variety dejned by the equation

VI.II. Examples of Varieties of Sets

216

boolean closure of the family of sets

I

{Z*v v

E

Z+, I v I 5 k}

Show that alternatively Z + g k consists of all sets of the form

A=Z*Bu C where B and C are subsets of Z+ containing only words of length Sk. 5. The Variety

Let

(5.1)

b

b be the class of all semigroups S such that eSe = e for all idempotents e E S.

This variety was explicitly introduced in V,3, but implicitly already appears in IV,4. We have shown that

D = (E')n s = ~ [ 2 ' ]n s Thus f)is a closed S-variety. It is also clear that

b = L(1) so that b is local ; it is the variety of all semigroups that are locally trivial. Let S E 0 and let E be the set of idempotents of S. Since eses = es and sese = se, we .obtain

ES=E=SE

(54

Thus E is a two-sided ideal in S. Since Sn= SES provided n 2 card S we obtain

(5.3)

Sn= E

for all n 2 card S

We next prove (5.4) cSd = cd for all idempotents c, d in S. Indeed,

CSd = CdCSd c CdSd = cd

5. The Variety

b

217

PROPOSITION 5.1. Let Zf of the family of sets

(5.5)

b o @.

Then Z+B is the boolean closure in Z"v

VZ",

with v E Z+. Alternatively, Z+Bconsists of all jinite unions of sets of the form

(5.6)

uZ:"v,

v

with u, v E Z+

Proof. We first observe that each singleton v E Z+ is in the Boolean closure of the sets (5.5). Indeed v = vZ" - vZZ*. If A = P v , then SA E D by Proposition 4.1.Consequently SA4 E b since D c b. If A = v Z w , then SA E DQand thus again S, E 0 since De c 0.Thus the boolean closure of the family (5.6) is in Z.9. Conversely assume A E Z+B, i.e. SA E D. Let n = card S A . Thus SAn= E is the set of idempotents of S,. Assume that

XEA,

IxI>2n

Then

x=yuz,

Iyl=lzl=n

Thus, by (5.4), xEA

= csd = cd = CSAd

with c = Y ~ AE

E,

d = XE;q E E

Consequently, since xEAEjl c A we have

yZ*z c A Thus A is the (finite) union of sets of the form yZ*z with I y I = I z I = n and of a finite set of words of length <2n. Since further yZ"z differs from yZ" n Zxz only by a finite set, all the conclusions follow I Since the sets (5.5) are boolean generators for Z+& while the sets 2% are boolean generators for Z + 9 and the set v Z * are boolean generators for Z+Bewe deduce that 9= 9v ge. This implies PROPOSITION 5.2.

b = D v De I

VIII. Examples of Varieties of Sets

218 COROLLARY 5.3.

A sem+roup S is in

b

i f and only i f

s <SIXS, with S, E D and S ,

E

De

I

EXERCISE 5.1. Show that S E (unique) minimal two-sided ideal.

;ff its set of idempotents E is the

6. Locally Testable Sets

Let k 2 1 be an integer. We define an equivalence relation as follows s %kt

-k

in Zt

provided : Any initial segment of s of length < k also is an initial segment of t and vice versa. (ii) Any terminal segment of s of length < k also is a terminal segment of t and vice versa. (iii) Any segment of s of length k also is a segment o f t and vice versa.

(i)

These conditions may be restated as follows: (i') If z, E Z+, I v I < k, then s E vZ# o t E vZ*. (ii') If z, E Z+, I v I < k, then s E Z#v o t E Z%. (iii') If v E Z+, 1 z, 1 = K, then s E Z%Z" o t E Z%P. We make a number of easy observations that the reader should verify.

(6.1)

w kis a congruence, i.e. s -k for all u E 2".

t implies us N k ut and su %k tu

(6.2) If s ' V k t and I s I < k , then s = t . ( 6 . 3 ) T h e quotient semigroup Z+lNk = ZTk is finite, i.e. there are only a finite number of congruence classes under -k. Let tk:

Z++ Z

T k

be the natural factorization morphism. A subset A of Z+ is called k-testable if A = B -zk1

6. Locally Testable Sets

219

for some B c Z T k . Equivalently A is k-testable iff A is saturated with respect to - k , i.e. iff s -k t and s E A imply t E A. Since ZTk is finite, every K-testable set is recognizable. The class of K-testable subsets of Z+ will be denoted by Z+Fk. A subset of Z+ is said to be locally testable if it is k-testable for some K 2 1. The class of locally testable sets in Z+ will be denoted Z + 9 F . Thus

u Z+Fk

Z+98-=

k21

The following intuitive description of the equivalence relation -k may be illuminating. For simplicity assume k = 3. Suppose that the word s E Z+ is written on a very long tape, but that this tape is covered except for a window which allows us to examine exactly three consecutive squares on the tape. We shall assume that the window can be moved so as to inspect any three consecutive squares, and that the results of the inspection are recorded, without reference however to their sequence or multiplicity. A complete scan of s will thus tell us exactly what are the segments of s of length 3 and also (by moving the window until blank spaces on the tape appear) what are the initial and terminal segments of length <3. Thus s w 3t means that s and t are indistinguishable under the circumstances described above. PROPOSITION 6.1. ing family of sets

The class z + % k is the boolean closure of the followv

with

wZw, Z*v

with

I v I
Z%Z*

with

I z, I = k

-

1

Proof. From the definitions it follows directly that Z + g k is closed under boolean operations. Let 99 be the family of sets tabulated above. It is clear that each of these sets is K-testable and thus 9 c z + % k - , . Let s E Z+ and let [s] be the congruence class of s under -k (i,e. [ s ] = S t k t i ' ) . We note that

Consequently Z+&k c 9 9 b . Thus [s] is in the boolean closure S'b of 9. Since 9c 2+sk and 2 + 8 - k is boolean closed it follows that Z+gk =99k I

VIII. Examples of Varieties of Sets

220

COROLLARY 6.3. The class C + 9 5 of locally testable subsets of C+ is the booZeun closure of the family of sets z*v,

with any v

E

.Z+

vz*,

Z*vZ#

I

We shall be interested in the +-variety

9&= {Z+9@-} of locally testable sets. The fact that this is a +-variety is far from obvious and, in fact, a direct proof would be fairly cumbersome. However, we shall prove THEOREM 6.4. LJ, o 9%.

This automatically implies that 9% is a +-variety. We recall that J1 is the variety of idempotent and commutative monoids. Thus LJ, consists of all semigroups S that are locally idempotent and commutative, i.e., of all semigroups S satisfying the conditions

esese = ese esete = etese for all idempotents e and all elements s, t . A key fact about LJ, is THEOREM 6.5. LJ, = J,

* D.

The inclusion J, w D c LJ, follows from V, (12.7). The proof of the opposite inclusion requires preparation and is postponed to Section 8. Taking Theorem 6.5 for granted we can now prove Theorem 6.4. Proof. Let A E z+5?6.Then A E C+6, for some integer k 2 1. I t follows that A = Bti' where t k : Z+ CTk is the natural factorization morphism for the congruence relation By Proposition VII,Z.ls, we have SA C T k . Thus to prove S, E LJ,, it suffices to show ZT, E LJ,. For this we inspect the equivalence relation N k and note ---f

<

6. Locally Testable Sets

221

the identities

for all x , s, t E Z+provided I x I 2 k. Consequently the identities (6.4) and (6.5) hold in ZTk for all idempotents e. Thus ZTk E LJ,. Conversely, assume that A c Z+and that S, E LJ, . Thus by Theorem 6.5, SA E J1 i* D and thus

SA<Ti*W with T E J1 and W exists a morphism

E

D. Consequently, by Proposition VII,2.ls there q: Z + - + T s W

such that A = Bq-l for some B c T r W . To prove that A is locally testable, it thus suffices to show that ( t , w)q-* is locally testable for all t E T , W E W. Since W E D, there exists an integer K 2 2 such that

for all w l , . . . ,wk-l E W (see Example V, 3.7). We shall show that for u, v E Z+ u -kv implies up = vq Let y : Z+-+W,

y : Z+-+T

be the coordinate functions of q. Then y is a morphism of semigroups, while y satisfies the identity

(6.7 1

(XY)Y = X Y

+ (XYJ)(YY)

Now assume u -k v and u # v . This implies I u I 2 k and I v I 2 k. From formula (6.6) it follows that uy depends only on the terminal segment of u of length k - 1. Thus since u and v have the same terminal segment of length k - 1, it follows that uy = v y . Assuming that u = o1 . . . o, with Y 2 k we deduce from (6.7)

VIII. Examples of Varieties of Sets

222

with (a, . . . a,-,)y interpreted as 1. From (6.6) it follows that the summation above breaks into two parts

c

k-1

(6.8)

(GI

i=l

* * '

ddw)

and

The first summation involves only the initial segments of length < k of u. These are the same for u and for v and thus (6.8) for u and for v agree. The summation (6.9) depends only on the segments of length exactly k in u. Because T is idempotent and commutative, repetitions and order are immaterial. Thus only the set of segments of length k of u determines (6.9). Thus (6.9) is the same for u and for v. Consequently uy = v y and thus u v = v9 I It is not too difficult to see that if A is any of the sets

v,

Z#V,

V P ,

PVZ*

then S, E J1 * D.For A = v or A = Z#v this is clear because S, E D. The remaining two cases A = vC" or A = 2YvZQ follow from the fact

SBz.


#

S,

for some

T E J1

that will be established later (Proposition IX,2.5). If we use the observation just made we obtain a proof of (6.10)

J, * D o 9%

without making use of the rather difficult Theorem 6.5. The relation (6.10) is not regarded as a satisfactory solution of the problem for the following reason. If a finite semigroup S is explicitly given (e.g. by its multiplication table), then it can be effectively determined whether or not S E LJ,. Membership in J, IC D is much more elusive. 7. A Theorem on Graphs

A theorem on directed graphs that is of independent interest is needed for the proof of Theorem 6.5.

7. A Theorem on Graphs

223

A (directed) graph G is given by two (possibly infinite) sets V and E and two functions a, w : E -+ V subject to no conditions whatsoever. The elements of V are called the vertices of G while the elements of E are called the edges of G. For each edge e E E the vertices ea and ew are called the start and the end vertices of e. We shall use the notation e : v, -+ v2 or e

v1-

v2

if v, = ea and v2 = em. The edges e l , e2 are called consecutive if elm = e2a, i.e. if

We must emphasize that vl, v 2 , and v3 need not be distinct. In the free semigroup E+ generated by E we consider the set

P = Ef - E+CE" where C c E2 is the set of words e1e2where e, and e2 are non-consecutive. The elements of P are called paths. Thus

p

= el

... en

is a path iff ei and e i f l are consecutive for all 1 5 i < n. The integer n is called the length of path and is denoted by 11 p 11. Clearly E c P and each edge is a path of length 1. The functions a and w are extended to functions a, w : P - + V by setting pw = enw pa = ela,

-

As in the case of edges we shall use the notation p: v1-+v2 or vl P vz if vl = pa and v2 = p w . If v = pa = PO,thenp is called a loop about v. If

p = e, . . . en,

q = el'

. . . em'

are consecutive paths (i.e. if en and el' are consecutive), then

pq = el

. . . enel' . . . em'

VIII. Examples of Varieties of Sets

224

+

is a path and I Pq I1 = I P I II q 11. Clearly ( P 4 b = Pa and ( P d w = qw. The triple product p q ~ is associative provided it is defined (i.e. pq and qr are defined). P could be regarded as a partial semigroup. It could be converted into a genuine semigroup by the adjunction of a zero 0. One would then define pq = 0 whenever p and q are not consecutive. This procedure is not recommended. One could adjoin to P the set V with each element v E V regarded as a path 1,: w v of length 0 and with lVp= p provided v = pa and pl,, = p provided v' = pw. When this is done we may obtain a category with V as the set of objects and the paths p: vl v, as the morphisms. This is the free category C represented by the graph G. An equivalence relation rv in P is called a congruence if it satisfies the following two conditions : --+

--f

(i) Ifp -p', thenp andp'are coterminal (i.e.pa = f a andpw = p'w). (ii) If p -p', q q' and p and q are consecutive, then pq -p'q'. N

Observe that as usual in (ii) it suffices to require pq We define the function t: P

-

p'q and pq

-

pq'.

+2Z

by the following conditions: for e E E

et =e

(P4b = Pt u q t Thus for each pathp, p t is the set of edges traversed b y p without regard to order and multiplicity; we may call p t the support of p. THEOREM 7.1. (Simon).

in P satisfying

Let

-

be the smallest congruence relation

(7.1)

xx "x

(7.2)

XY "YX

for any two loops x, y about the same vertex. Then for any two coterminal paths X

u=:w il

the conditions X"Y

7. A Theorem on Graphs

225

and X t =yt

are equivalent. Proof. Let = be the congruence relation in P defined by p = p ’ iff p and p’ are coterminal and p t = p ’ t , If x and y are loops about the same vertex v , then ( x x ) ~= x t u x t = x t and (xy)t = x t u yt = ( y x ) t . Thus the congruence = satisfies (7.1) and (7.2). Consequently p -p’ implies p E p‘ and thus also p t = p‘t.

Conversely assume that 2

u-.v Y

are coterminal paths such that x t = yt. We must prove that x - y . For the proof it will be convenient to enlarge P by adjoining the trivial paths 1, for each v E V. The function t is extended to V u P by setting let= 0. Each congruence in P is extended to V u P by agreeing that each trivial path 1, is congruent only to itself. LEMMA 7.2. Given paths

V A W - Yt such that y t c xt, there exists a factorization 21

v - t$0- w of x such that x

-

xyxl

Proof. The proof is by induction on the length 11 y 11 ofy. If 11 y 11 = 0, then y = 1, and x = xl, is the required factorization. Now assume that y = ze, with

WZ-UC-t By the inductive assumption there exists a factorization

v - uXO - w such that

(7.3)

x

-

Xl

XZXl

VIII. Examples of Varieties of Sets

226

We thus have the diagram xo

0-u-w

21

u 2

Since y t c xz, it follows that e E x t . Thus x admits a factorization x = x2ex3

(7.4)

so that we have the diagram XI

21

v-ud.t.w

W z

We assert that x = (x,e)x, is the required factorization of x. We thus must prove x

We have x

-

-

xyx, = xzex,

by (7.3)

XXXl

by (7.4) by (7.1)

= x,(ex,z)x,

-

--

x,(ex,z)(ex3~)xl

by (7.4) by (7.2) by (7.3)

= x(zex,)(zx,)

x(zxd(zex3) xzex,

I

LEMMA 7.3. Given paths

such that y t c xz, we have xy

-

x

Proof. By Lemma 7.2, x admits a factorization x = x,,x, such that

and x

-

xyx,

227

7. A Theorem on Graphs

I t follows that

We can now prove Theorem 7.1. Let then 2

U*W Y

be paths such that x t = y t . Replacing the set E by x t , we may assume that E = x t = y t . Let V l be the set of vertices accessible from w by a path, and let Vo= V - V,. The existence of trivial paths implies that w E V,. We now consider two cases.

Case 1. u E Vl . There is then a path z : w we deduce

-

xzy

(7.5)

x

(7-6)

Y -yzx

Therefore x

--

xzyzx

by (7.5) and (7.6)

YZXZX

by (7.2)

yzx

by (7.1)

“Y Thus x - y

u. From Lemma 7.3

by (7.6)

as required.

Case 2. u E V o . Define for i, j = 0, 1

Clearly El, = 0. Since u E Yo and w E V , and since x is a path leading from u to w, it follows that E,,, # 0. Let eo E Eel. This yields factorizations 20

co

Z1

UO

eo

Y1

u-s-t-w u-s-t-w

of x and y . From this it is clear that

xot

= Eoo,

X l t

= Ell,

Yot

= Eoo,

Ylt

= Ell

VIII. Examples of Varieties of Sets

228

and since x and y traverse all the edges, it follows that xOr = y o t = E00 I

x1r = Y l t = Ell I

eo

=

El0

-

Since Eoo and Ell have lower cardinality than El we may, arguing by induction, assume that xo -yo , x1 m y 1 . Thus x y as required I The reader should note the motive behind the enlargement of P by adjoining trivial path. Thanks to this device the induction in Lemma 7.1 could begin with 11 y 11 = 0. Without it, the induction would have to begin with 11 y 11 = 1, thus lengthening the proof. Also, the definition V , near the end of the proof would have to be slightly revised.

Let S be a semigroup. Define a graph G as follows. The vertices are the idempotents of S. The edges EXERCISE 7.1.

e: i - j where i and j are idempotents are all the triples

(i,s, j ) ,

s E

iSj

Call s the label of the edge. Thus to each path p : i -+ j corresponds a label p l E i S j obtained by multiplying out the labels of the edges. Show that S is in LJ,z pfor any paths . P

7

2--i

in G

pt

=

p't

implies

pl

=p ' l

8. Proof of Theorem 6.5

We shall say that a ts X is idempotent and commutative if the semigroup S, is idempotent and commutative. The reader should review the definition of the trace in III,8. THEOREM 8.1. Let X Q, Y be a relational covering of ts's, such that for some parametrization, the trace of q~ consists of idempotent and commutative ts's. Then X
some complete, idempotent, and commutative tm Z.

229

8. Proof of Theorem 6.5

Proof. Let X = (Q, S ) , Y = (R, T ) and let (52, a,8) be a parametrization of p for which the hypothesis of the theorem holds. We shall regard Q and R as right Q-modules by setting = Y(o/~)

YO

qo = q(oa),

Then, for o E Q we have YfPW

c

YWfP

and thus upw c rwp

for all Y E R and all w E 51". We now construct a graph G with R as set of vertices. T h e edges e: r

+ T I

are all the triples e = (I,w ,Y')

with

w E

52 and

( I , w , Y') with Y, I' then the path is

(Y, ~

Thus a path in G may be recorded as triple E R, w E Q* and YW = Y'. Indeed, if w = o1. . . m n ,

YO

= Y'.

1 ~o1)(~ol,oz , YW,~,)

-.

*

(yo1

* * *

we-1

1

a n 9 Y')

This convention includes the trivial paths (Y, 1, Y). Let E be the set of all edges of G and P the set of all paths in G (including the trivial paths). We recall that we have the function t : P+2"

which to each path assigns its support. We shall regard 23 as a semigroup with union as addition. Then 2x becomes an idempotent and commutative monoid with 0 as unit element. We propose to establish a covering

X
with We first define the relation y: QxRx~ExR-+Q

(9, y ,

4 Y')Y

=

{qw I

(I,

w, Y ' ) E

P, (Y,

w, ~

A}

' ) =t

if q E YV if q $ v

VIII. Examples of Varieties of Sets

230

For any q E Q we may choose Y E R so that q E Y Q ~ .Then q E (4, I , 0, Y)Y and thus y is surjective. Next we show that y is a relational covering. Given s E S choose u E Q+ such that s = UOI. Let t = up. Then 9 s c t p Define the function

f: R (I,

Then for some path E

(Y,

u, Y

w, i )E

-+2E

U ) ~

P we have

pr = qw,

YQ~,

if YU # 0 otherwise

(Y, W , Y ’ ) T

=A

Suppose further that qrs f 0. Then qrs = qru = qwu c YWU f 0. Since u, Y ’ U ) E P. Consequently

and therefore (Y’,

(Y,

wu, Y’U)T =

((Y,

Y’ = Y W ,

(YWU)Q1

it follows that

Y‘U

# 0 and thus

w , Y ’ ) ( Y ‘ , u, Y k ) ) T

= ( I , w, Y r ) t

u

(Yl,

u, Y

k ) t

=A

u Ylf

Thus qfs E (q, y , A

u Ylf, W

y = (4, y , A , y r ) (

f,t ) y

and this proves that s is covered by (f,t ) . T o conclude the proof we now show that y is a partial function. For this we must show that if

are paths satisfying (I,w , Y ! ) t = ( I , u, Y ’ ) t

(8.2) then

(8.3) for all q

E

~p.

8. Proof of Theorem 6.5

231

We define the equivalence relation

-

( I ,w , y o

( r , 21, r')

by the condition qw = qu for all

qE~ p ,

We assert that this is a congruence in P. Indeed, suppose also that (Y',

w',

-

Y'')

(Y',

u,, Y ' , )

Then for all q E r y

qww' = quw' = quu'

and thus ( Y , WW', Y " )

Next we prove

-

w, y)(y, w, I>

(8.4)

(y,

(8.5 )

(r, w, y ) ( y , u, r )

(Y,

uu', r")

( r ,w , y ) ( r , u, y ) ( r , w, y )

Indeed, the word w yields a transformation s E S and also a transformation t E T such that p,s c tv. Since ( Y , w , Y ) is a path, we have YW = I , i.e. rt = Y . Consequently YVS

c rtp, = rp,

and thus s defines a transformation of Tr,. However, Tr, is assumed to be idempotent and commutative. Thus for each q E rp, we have qss = qs. Equivalently qww = qw. Thus ( r , ww, Y ) ( r , w , I ) . Condition (8.5) is proved similarly using the commutativity of Tr,. Once conditions (8.4) and (8.5) are verified, Theorem 7.1 applies and thus (8.1) and (8.2) imply (8.3) as required I

-

From Theorem 8.1 we can now rapidly deduce the inclusion

using the delay covering of III,9. Indeed, let S E LJ,, S f 0, and let S dPY be the nth delay covering with n = card S . By Proposition 111,9.3, Y E = ( 2'). Since Y is complete, it follows from IV, (4.9), that SyE D. Next, let T = Tr, be the trace of y for some element

[z]

VIII. Examples of Varieties of Sets

232

p E Q y . We note that T is complete. Further, from Tilson's TraceDelay Theorem (111,9.5), we deduce that T S, for some idempotent e in S. Since T is complete, this implies S , <eSe. Thus S, is idem,potent and commutative. Consequently the conditions of Theorem 8.1 are satisfied and we have

<

s
2

z+

Let J be the M-variety treated in Example V,3.5, and let J be the corresponding #-variety of sets. Given u, v E Z# we shall write u c v if u is a subword (but not necessarily a segment) of o. Thus u c z1 iff u can be obtained from v by erasure of letters. We define for each w E ZlwI

[w = {v I v

E

Z", w c v}

Thus [w is the internal shuffle product (see A,II,3) w u Z". Clearly [l = Z*. If w = al . . . a,, then

[w = Pa,Z*a,

(9.1)

. . . a,Z#

Z * y is the boolean closure of the family of

THEOREM 9.1. (Simon). sets [w with w E Z+.

The proof requires some preliminaries. For each integer n 2 1 and each w E Z* we define

wan = {u I u E Z+, I u I 5 n, u c w } Further we set

w It is easy to check that monoid is denoted by

-,w'

iff wa,

= w'a,

-,is a finite congruence relation. Jz,, = Z*l-n

PROPOSITION 9.2. Jz,, E

Jn.

The quotient

9. The +-Variety

2

233

Proof. We observe that the conditions w E znanand wal c xal are

equivalent, i.e. that if I w I 5 n then w c zn holds iff each letter of w is also a letter of z. This immediately implies

X"f1 mnX", Since each set [w with der - n , we obtain

(xy)"

N n

(yx)"

I

I w I 5 n is the union of congruence classes un-

If w E Z+, I w I 5 n, then M I , E Jn

COROLLARY 9.3.

I

This proves the "easy" part of Theorem 9.1. For the other part we need PROPOSITION 9.4. For each Jinite alphabet 2 and each monoid S there exists an integer n such that for each morphism q ~ Z+ ; ---+ S

x

-,,

XI

E

J,

implies xrp = x'y

Another way of expressing the conclusion is to say that y admits a factorization

Z+-5 J Z , ,

S

where n is factorization morphism for the congruence ular, if rp is surjective, then so is y. This implies COROLLARY 9.5.

-n.

In partic-

The variety J is generated by the monoids

J2,n.

The remainder of this section is devoted to the proof of Proposition

9.4. Taking this proposition for granted, the proof of Theorem 9.1 can be rapidly concluded. Indeed, let S E J be the syntactic monoid of a subset A of Z*. Proposition 9.4 implies that A is a union of congruence classes of ference

T h e congruence class of any element x E Z*is the dif-

wn.

n[u - u [ v the intersection extending over all u E xan and the union extending over all v E Z+ such that I z( I 5 n, v p xan. Thus A is in the boolean closure of sets [w with 1 w I 5 n.

VI11. Examples of Varieties of Sets

234

S

T o prove Proposition 9.4, we begin with a detailed study of a monoid i.e., of a monoid satisfying

E Jn,

S" = snfl,

(St)" = ( t s ) n

for all s, t E S. Let E be the set of all idempotents in S. We obtain a function n: S - + E

sn

= Sn

We define a law of composition in E by setting el

(9.2)

* e2 = (ele,)n = (ele2)n

The composition law (9.2) converts E into a commutative and idempotent monoid and the function n: S + E is a surjective morphism of monoids. For each e E E, the set en-l is a nilpotent subsemigroup of S with e as zero. PROPOSITION 9.6.

Proof.

For each e E E define

N,

= en-'

= (s I s E

S , sn = e ]

We first prove

(9.31

s, t

E

N,

implies st

E

N,

Indeed, we have (st)" = ( s t y s t and since S E R we obtain (st)" = (st)"s. This implies (st)" = (st)%" = (st)"e. However, te = e and se = e and thus (st>"e= e. Consequently, (st)" = e, i.e., st E N , . Statement (9.3) asserts that N, is a subsemigroup of S. Since se = e = es for all s E N o , it follows that e is a zero for N,. Since e is the only idempotent in N , , it follows that N , is a nilpotent semigroup. Next we observe that st E N , implies t s E N , since (st)" = (ts)". From this we deduce that if st E N , and u E N , ,then sut E N,. Indeed, we have ts E N , and thus tsu E N , and consequently sut E N , . From this it follows by induction that st E N , implies siti E N , for all i 2 1. In particular, we have st E N , implies sntn E N , Using the function n and the multiplication in E given by (9.2), the

9. The *-Variety

2

235

formula above states (st)n = s7E

x

tn

This implies that E with the multiplication x is a monoid and that n is a surjective morphism. Since (st)n = (ts)n and (ss)n = sn, it follows that E is idempotent and commutative I COROLLARY 9.7. For each monoid S E J there exists an integer q such that N,Q= e

holds for each idempotent e in S

I

Next we need a number of properties of the congruences For each subset A of Z we define

TA =

{wI

wal

=

-%

in P.

1 uA}

Thus TA consists of all words in A* in which all the letters of A appear at least once. PROPOSITION 9.8.

If w a -, w Y

with

2 q card Z,

q >0

then there exists a subset A of Z such that

Proof. We first show that w admits a factorization w

= u1 . . . I(,

such that a E u,al c u,-,a, c

. . . c ulal = wa,

If r = 1 then w = u, is the required factorization. We proceed by induction and assume Y > 1. Let w = ulv be a factorization with wal = ulal and with u1 shortest possible. By the inductive hypothesis, it suffices to show that vu - r _ l z1. Let then x E (z1a)aT-,. Since u1 is shortest possible, we have u l= yt with t E Z, z @ ya,. It follows that TX c w = yzv and this implies x c z1 since z does not appear in y. Thus V b -,-I

v.

Since r 2 q card 2, it follows that the sequence of sets u,al c u,-lal . . . c ulal contains a block of length q of equal terms, i.e., there exists

VIII. Examples of Varieties of Sets

236

+

an index i such that 1 5 i < i q 5 r and such that uia, = A for i 5 j 5 i q. Then a E A and ui . . . u, E TA'Jas required I

+

PROPOSITION 9.9.

If uav

then either ua -,. u or av

-2r-1

uv

-, v.

Proof. Assume that xu E (ua)a, - ua, and let ay E (av)a,. Then xay c uav and I xay I = 2r - 1. Therefore by assumption, xay c uv. Since xu u it follows that ay c v. This implies av -, v I

+

For each integer p 2 1 there exists an integer n 2 p such that x w nx' implies the existence of y such that PROPOSITION 9.10.

Proof. An element s E C" is called p-minimal if s' c s and s' w Ps implies s' = s. Equivalently s is p-minimal if s = uazt implies s +p uv. We assert that if s is p-minimal, then

I s I 5 card(sa,) Indeed, for each 0 5 i 5 I s I consider the factorization s = uiviwith I ui I = i. If uiap = ui+lolpthen (uivi+l)ap = (ui+lvi+l)ap = sap contrary to the assumption that s is p-minimal. Thus the sequence

0 # uoap c ulap . . .

c ulslap = sap

is properly ascending showing that I s I 5 sap. The argument above shows that there exists an integer n L p such that I s I 5 n for all s that are p-minimal. Now given x w nx' choose y c x such that y w Px and such that y is p-minimal. Then I y 1 5 n and it follows that y c x' I We can now prove Proposition 9.4. First given S E J we determine q > 0 so that N,q = e for all idempotents e in S (Corollary 9.7). Then we set p = 2r - 1 r = q card 2,

9. The *-Variety

3

237

Finally n is determined using p in Proposition 9.10. We must show (9.4)

x

w,,

x'

implies x q = x'p

By Proposition 9.10, there exists z E Z" such that x c x, x c x', z - p x - p x'. Thus it suffices to prove

x

and

cx

z -px

implies

ZP,= XP,

+

For this it suffices to consider the case when I x I = 1 I z 1, i.e., x = uuv and x = uv. From Proposition 9.9 we deduce that either uu -, u or uv -, v . Thus it suffices to establish the following two implications : (9.5)

uu rvr u

implies

(uu)p = up

(9.5')

uv

-, v

implies

( o v )= ~ vv

Assume ua -, u. From Proposition 9.8 we deduce that there exists a subset A of 2 such that u E A and

Consider the composite morphism cpn: C" E. Since E is idempotent and commutative, the image of Trl is a single element e of E, i.e., TAP, c N , . Thus T,qp c N,Q= e and therefore up = (xq)e. However, 5 E A implies ta E TAP and thus by the same argument (uu)p = (zq)e = up1. This proves (9.5). Statement (9.5') follows by reversal I --f

Simon's original proof of Proposition 9.4 was long, but subsequently he published a simpler version. The proof presented here was developed by M. P. Schutzenberger jointly with the author. T h e essential difference between the two proofs is that Simon replaces Proposition 9.10 by the following (deeper and harder to prove) fact: x m Px' implies the existence of z such that x c z , c x, and x w Px. I n particular, this permits us to show that (9.4) holds already with n replaced by the much smaller integer p . XI

EXERCISE 9.1. Given w = al . . . a, E Z*write explicitly the minimal automata of the set [w. Deduce from that that M I , E R and thus, by reversal, also that M , , E J.

VIII. Examples of Varieties of Sets

238

Show that a monoid S is in J t$ there exists an idempotent and commutative monoid E (i.e., a monoid in J 1 ) and a surjective morphism n: S + E such that ep-' is a nilpotent subsemigroup of S for each e E E. Show that up to an isomorphism of E, the monoid E and the morphism n are unique. EXERCISE 9.2.

EXERCISE 9.3. Llse the binomial coejicients (;) section and show that

dejined in the next

Show that for each integer n and each w E W, the set

EXERCISE 9.4. Show that ([w,)n ( [ w z )is the union of sets of the form

[wwith I w I I I "I

I

+I

w2

I.

10. p-Groups

Let p be a prime, p > 1. We shall denote by G, the G-variety of all (finite) p-groups, i.e. the variety of all groups of order p' for some integer r. This variety is ultimately defined by the sequence of equations

Thus

where Gp,k,the variety of p-groups of exponent p k , is defined by the single equation xpk = 1. We shall now describe the +-variety Fpof sets defined by the G-variety

GP.

Let u, w E Z* with u = ul . . . C T ~ . T h e binomial coeficient defined as the number of factorizations

(E)

is

239

10. p-Groups

with v o , . . . , v,, E 2%. Thus the binomial coefficient counts the number of ways in which u is a subword of w . T h e following formal rules are easily verified

(10.1) if u = l , a otherwise

(3 {

(10.2)

=

1 0

(10.3)

if u = l otherwise

These rules may be used to define the binomial coefficient by induction on I w I. If w = a p and u = 8 with q S p , then

(3 (3

P!

= q ! ( p - q)!

=

Given u E Z* and 0 5 i < p we define

i

[u, i ] = w I w E Z*,

THEOREM 10.1.

):(

= i mod p }

Z*Vp is the boolean closure of the family of sets [u,i], u

E

0 5i < p

Z*,

T h e theorem will be proved at the end of the section after suitable preparation. Given u E Z* we define the equivalence relation -,, in Z* by setting

):(

w1 wYw, o

=

(T)

mod p

whenever

u E Z*v2'*

The condition u E Z%Z* expresses the fact that v is a segment of u. PROPOSITION 10.2.

For any u

E

C*,the equivazence

ence. Further

G,, = Z*/N,, is a p-group of exponent PI"'.

is a congru-

VIII. Examples of Varieties of Sets

240

Proof. Assume w -,w' and t -,t' and let v be a segment of u. We

have

and similarly with w and t replaced by w' and t'. Since v1 and v2 also are segments of u we have

w't' showing that wUis a congruence. Since this obviously Thus wt IV, is a finite congruence, it follows that G, is a finite monoid. T o show that G, is a p-group of exponent plu' it suffices to show that wp'"' -,1 or equivalently that

for all w E Z*. For K = 1 the result holds trivially. We proceed by induction and assume 0 < I v I 5 k 1. Then

+

...

(W;+l)=c(<)

(t)

the summation extending over all factorizations v = zll . . . up of v. If for some 1 5 i 5 p we have 0 < I vi I < k 1, then by the inductive assumption ($) 5 0 and the summand may be omitted. There remain summands with vi = v, vj = 1 for J f i. Each such summand yields ($) and there are exactly p such summands. Thus (w':') = 0 as required I

+

Given Z and an integer k 2 0 we define the congruence -k=

n

-U

lul-k

Thus w1 -k

wZ *

(7)= (T)mod p

whenever

I I 5k

The quotient Z * / w k will be denoted by GT,kwhere r = card Z. It follows from the above that GT,kis a subgroup of the product n G ,

10. p-Groups

241

extended over all u E Z#, 1 u 1 = k. Thus G,,kis a p-group of exponent pk. The reader will easily verify the following special cases

G,,* = 1,

Gr,l M 2,.

We shall now derive a more algebraic and less combinatorial description of the groups G,.k. Given a monoid S and a field F an algebra F [ S ] over F is defined as follows. As a vector space over F, F [ S ] has the elements of S as a base. The multiplication in S is then extended to yield the multiplication in F [ S ] . The unit element of S becomes the unit element of F [ S ] and the inclusion S -+ F [ S ] is multiplicative. Each element x of F [ S ] has an expansion

with only a finite number of the coefficients x, f 0. The elements x satisfying Z x , = 0 form an ideal denoted by I. This is easily seen to be the ideal generated by elements of F [ S ] of the form 1 - s with s E S. If S = Z*where Z = { a l , . . . , a,} is an alphabet with r letters, then F [ S ] is the polynomial ring F[a, , . . . , a,] with non-commuting variables al,. . . , a, and with coefficients in F. In our application F will be the prime field 2, with p elements. PROPOSITION 10.3.

The condition

(10.4)

w1 - k

w2

in Z# is equivalent with the condition (10.5)

w1

- m, E

P + 1

Proof. Let Z = {al, . . . , a,}. Since Z,[Z*] is the free 2,-algebra generated by the elements a l , . . . , a,, there is a unique endomorphism R:

ZP[Z*] + Z,[C"]

such that ain = 1 - ai

for i = 1, . . . ,Y. Clearly

3t

is an involution, i.e. mz is the identity. It

242

VIII. Examples of Varieties of Sets

follows that n is an automorphism. Thus condition (10.5) is equivalent with

(10.6)

wln - w2n E Jk+"

where J = In. We now calculate the ideal J . T h e ideal I is generated by the elements 1 - w with w E Z+. The identity

1 - WlW2 = (1

- w1)

+ w,(l - w,)

implies that I is the ideal generated by the elements 1 - u with u E Z. Thus J is the ideal generated by the elements (1 - u)n = u. Consequently J k + l is the ideal generated by the elements z, of Z" with I o I > k. Next we establish the identity

(10.7) u EE*

Observe that the summation is finite since )(; = 0 whenever I w I < I u I. The identity clearly holds if w = 1 or if w = u. Assume that (10.7) holds for w , , w, E Z". Then

as required. From (10.7) we deduce

This makes it clear that (10.4) and (10.6) are equivalent

I

PROPOSITION 10.4. Let G be a p-group of order q and let I , be the

ideal in &[GI generated by the elements 1

-g,

g E G. Then

=0

Proof. The assertion clearly holds if q = 1, i.e., if G = 1. If G # 1, then by a well-known theorem (one of the Sylow theorems) G has a non-

10. p-Groups

243

trivial center. Let then C be a subgroup of order p in the center of G, let H = G/C,and let n: G H be the factorization morphism. This morphism extends to a morphism of Zp-algebras n: ZP[C;l

-

ZP[ffl

and its kernel K is an ideal in Zp[G]. We assert that

(10.8)

K

(1

=

- c)ZP[G]

where c is a generator of C. The inclusion (1 - c ) Z p [ q c K is clear since (1 - c)n = 0. To prove the opposite inclusion we choose a function v: H G such that vn = identity and extend this function to a linear transformation y : Z p [ q+ Z p [ H ] . Since each element g E G can be written as g = ci(hv) for some 1 5 i 5 p , h E H , it follows that each x E Z , [ q has the form

-

c P

x=

Ci(X#)

i-1

for some elements xl, . . . , xp E Z p [ H ] .Since xn

2 xiyn = c xi P

P

i=l

i=1

=

it follows that x E K iff Cxi = 0. Thus for x E K , we have

c (ci P

x =

-

l)(X#)

i=l

Since ci - 1 E (1 - c)Zp[Gl for all integers 1 5 i 5 n it follows that x E (1 - c ) Z P [ q .This proves (10.8). Arguing by induction, we may assume that IH7= 0 where r = q/p = card H . This implies

I$Z = (Icn)I = I H I

=0

and thus

10' c K

=

(1

-

c)Zp[q

It follows that

I$

= (I$)P

c (1

-c)pzp[q

It thus suffices to show that (1 - c ) p = 0. This, however, is clear,

VIII. Examples of Varieties of Sets

244

since in the binomial expansion of (1 - c)* all the binomial coefficients (7) with 0 < i < p are divisible by the prime p. Thus (1 - c)* = 1 - c p =1-1=0 I The least integer n 2 1 for which 1;" = 0 holds will be called the nilpotency index of the p-group G. PROPOSITION 10.5. Let G be a p-group of nilpotency index n and let pl:

Z*+G

be any morphism with r = card Z. Then

pl

admits a factorization

where t is the natural factorization morphism corresponding to the congruence w nin Z*. Proof. The morphism pl induces a morphism pl: Z,[Z*] + Z p [ G ] of 2,-algebras and we have Ipl c IG. This implies In+1pl

cg+1 =0

Now assume w , , w, E Z* and WI N n

WZ

By Proposition 10.3 this implies

w, - w2

€ In+'

This in turn implies w,pl

Consequently w,pl

= w,pl.

- w2pl E I"+lpl = 0

It follows that y may be defined as

t-lpl

I

PROPOSITION 10.6. Let G be a p-group of rank 5 r and nilpotency index n. Then G is a quotient group of G,,n.

r, G is generated by elements gl , . . . ,g, Proof. Since G has rank I and thus there exists a surjective morphism pl: Z* -+ G where Z is an alphabet of r letters. Then the morphism y : Gr,n-+ G of Proposition 10.5 is surjective I

245

References

COROLLARY 10.7. The G-variety G, is generated by the groups Gr,k for all integers r 2 1, k 2 1 I COROLLARY 10.8. The G-variety G, is generated by the groups G, for all elements u E C" and all jinite alphabets Z.

This follows from Proposition 10.2 and the fact that Gr,kis a subgroup of a finite direct product of groups G, I We are now ready to prove Theorem 10.1. The set [u, i ] for u E Zs and 0 5 i < p is a union of congruence classes of the equivalence relation -,. Thus Mr,,i, G, and therefore Mr,,i, E G,. Conversely let A be a subset of Z" such that M A E G, and let

<

pz

C" +MA

be the syntactic morphism of A. Let r = card Z and let n be the nilpotency index of M A . By Proposition 10.5 ,u admits a factorization Z* 1 ,GT,n -LMA

Since A = App-' we have A = Bt-I with B = Apy-'. Thus A is the union of congruence classes of w n . Since each such congruence class is the intersection of sets of the form [u, i ] , it follows that A is in the boolean closure of the sets [u, i ] I References

T h e variety D discussed in Section 4 is closely related to the definite automata of Perles-Rabin-Shamir (1963). The variety 0 (of Section 5 ) was considered by Perrin (unpublished). T h e hardest example treated in full, namely the variety LJ, and the corresponding +-variety of locally testable sets (Sections 6-8), is a reworking of results of Brzozowski-Simon (1971) (see also Simon (1972)). In particular, Theorem 7.1 has been extracted from the arguments of Brzozowski-Simon. Results in the same direction but somewhat less complete have been obtained by McNaughton (1971) and by Zalcstein (1972). I n particular, Exercise 7.1 is from McNaughton. The notion of a locally testable set is due to McNaughton and Pappert (1971). Section 9 is modeled after Simon (1972).

246

VIII. Examples of Varieties of Sets

The study of p-groups that occupies Section 10 has been suggested by Schutzenberger. Conversations with Patrick Gallagher proved to be helpful. J. A. Brzozowski and Imre Simon, Characterizations of locally testable events, IEEE 12th Annual Symposium on Switching and Automata Theory (1971). Robert McNaughton, Algebraic decision procedures for local testability. Math. Systems Theory 8 (1974), 60-76. Robert McNaughton and Seymour Pappert, “Counter-Free Automata,” M.I.T. Press, Cambridge, Massachusetts, 1971. M. Perks, M. 0. Rabin, and E. Shamir, The theory of definite automata, IEEE Trans. Electronic Computers, EC-12 (1963), 233-243. Imre Simon, Hierarchies of events with dot-depth one, Dissertation, University of Waterloo (1972). Imre Simon, Piecewise testable events, to appear in Lecture Notes in Computer Science, Springer-Verlag, Berlin (1975). Yechezkel Zalcstein, Locally testable languages,j . Comput. System Sci. 6 (1972), 151-167.

CHAPTER

Ix Aperiodicity

The key result of this chapter is Schutzenberger's Aperiodicity Theorem 3.1. It gives a description of the recognizable subsets A of C* having aperiodic syntactic invariants. The concatenation product AB of sets plays an important role and the proof utilizes sequential functions. The theorem leads to the Brzozowski hierarchy for aperiodic sets, which is also treated in this chapter. 1. Recognizable Sets and Sequential Functions

Let f : P -+ 2 3 be a partial sequential function and let A be a recognizable subset of C*.Then B = Aj-' is a recognizable subset of r* and PROPOSITION 1.1.

If, further, f is a function, then also

If A c Zf, then B c l'+ and

and i f f is a function, then also

247

248

IX. Aperiodicity

Proof. Let d =(Q, i, T ) be a (deterministic) Z-automaton with

behavior A and let

A = ( ~ , A): j,

r z --+

be a sequential machine with result f . Define the r-automaton

9 =( Q x P , ( i , j ) , T x P ) (4,P)Y

=

(q(P, Y ) 4

Pr)

where, as usual, it is understood that ( q , p ) y = 0 if either ( p , y ) L = 0 or q(p, y ) A = 0. A simple calculation shows that

(Q,Pk! = (!?(Ag ) l , Pg) for all g E I'". Taking (q, p ) = ( i ,j ) and noting that gf = ( j ,g)A, we obtain (i,& = M g f ) , j g > Consequently

I .%? I = { g I ( i , j ) g E T x PI = { g I i(gf 1 E TI =

{g I gf E A = A f

-1

=

B

The definition of ( q , p ) y shows that y viewed as a transformation of Q X P is in the wreath product TM, o TfVd and thus

Taking d to be the minimal automaton of A and A to be the minimal sequential machine o f f we obtain the inequality (1.1). If A c Z+, then the inequality (1.3) also follows. Iff is a function, then k i s complete. If we take &to be the complete minimal automaton of A, then, as constructed above, 9also is complete. This implies the inequalities (1.2) and (1.4) PROPOSITION 1.2. Let A be a recognizable subset of Z+. There exists a sequential function f : Z* + 2* such that

A

=

Cf-'

TM, < TM,"

<

o

TM,

2. The Concatenation Product

where

249

c= 2"l

is the set of all binary words terminating with 1. Proof. Let &'= (Q, i, T ) be the complete minimal automaton of A. We convert d into a complete sequential machine

A = (Q, i, A ) : 21-2 by defining the output function

A: Q x Z - 2 as follows

Let f be the result of

A = Cf-' is then clear. The relation

TMJ < TM,g = TM,

1

TMA

is immediate. Since f is a function, we have, by Proposition 1.1,

TMA" < TIM,"

o

TMj

The minimal automaton of C is

EXERCISE 1.1. Prove the conclusion of Proposition 1.2 with TM replaced by T S . 2. The Concatenation Product

T h e concatenation product AB of two recognizable subsets A and B of Z*, turns out to be a rather complicated operation when looked at from the point of view of the syntactic invariants. It requires a new operation on semigroups due to Schutzenberger.

IX. Aperiodicity

250

Given semigroups S and T we consider the set V = 2S’xT’of all subsets of the product S’xT’. T h e set V is converted into a monoid using the union of subsets as multiplication. Thus V is an idempotent and commutative monoid, i.e. Y E Jl = (Ul)M. Next we define a left action of S on V by setting sv

for all v

E

=

{(su, w ) I

(24, w ) E

v}

V. Similarly a right action of T on V is defined by vt

=

{ ( u , w t ) I (u, w ) E v }

Clearly the two actions commute, i.e. (sv)t = $ ( a t )

The Schiitxenberger product

TO s = ( T , V , S ) is now defined using the triple product introduced in V,9. We recall that the elements of T 0S are triples ( t , v, s)

with the product ( t , v, s)(t’,ZJ’,s’)

=

(tt’, vt‘

+ sv’,ss’)

Note that S O = 0 (where 0 = 0 is the unit element of V ) so that the action of S on V is right unitary. If further S is a monoid, then the action of S on V is unitary. Similarly for the action of T on V. Thus if both S and T are monoids, then T 0S is a monoid with (1, 0, 1) as unit element. THEOREM 2.1.

If A and B are recognizable subsets M A B< M B O M A

If, further, A

If B

c Z+, then sAB

<M B 0

SAB

<S B 0

c Zf, then

sA

MA

of Z*, then

2. The Concatenation Product

251

Finally, ;f A c Zf and B c Z+, then

Proof. We shall consider only the case A c Z+, B c Zx as the proof in the other cases is similar. Consider the syntactic morphisms E.4

z Zf

---f

s.4,

)(lg:

zx* MI]

We extend l Ato a morphism of monoids

and define the function

by setting for s E Z+

The following formula is then clear by inspection (47=(

for s, t

E

S Y N W

+

(S(A')(tY)

Z+. It follows that the function

is a morphism of semigroups. T h e definition of y implies that s E A B iff

SY n ( A ~xAB P B )f 0 It follows that

A B = Cp-' with C

= {(x,

v y ) I u n ( A ~ A x B Pf B )01

Thus, by Proposition VII,Z.ls,

252

IX. Aperiodicity

v

be the variety THEOREM 2.2. Let V be a non-empty G-variety, let &fined by V in V,10, and let be the corresponding +-variety. Then the class of sets Z + pis closed under concatenation.

ve v

v

v.

We have Proof. It suffices to show that S, T E implies T 0S E T 0S = ( T , W, S ) with W = ZSxT. Since W is aperiodic, we have W E (1).Thus, by Proposition V,10.7, (T, W, S) E (1) + V = Thus T O S E I ~ COROLLARY 2.3.

V

o

- v.

Let V be a closed M-variety (or S-variety) and let

F. If U , E V, then the class Z + Y (or Z + T )is closed under con-

catenation.

-I

Indeed, Corollary V,10.11 implies V = V n G

Let V be a closed M-variety (or S-variety) and let V T.If U , E V and V = Ve, then the class Z + T (or Z f T ) is closed unakr concatenation. COROLLARY 2.4.

Indeed, we have Uze= L, PROPOSITION 2.5.

< U , + U , and therefore U, E V I

For each recognizable set A c Z+

MA,* < V + M A

< *

sAZ+

MA

for some V E J,. I f , further, A c Z+, then

Proof. We have M,, MA,*

=

1 and S,+

< =

=

1 (if Z f 0). Therefore

0MA= 1 0M A (1,

v,MA) = v # MA

The other case follows similarly

I

3. Schutzenberger’s Theorem

253

EXAMPLE 2.7. Let G be the variety of groups and let G o 29. Let Z = a and consider the sets A = (c2)+,B = (u3)+.Then M,, = Z , , M s = Z , , so that A, B E 2”9.However, AB = Z+ - a and thus M A , = M , = Z,,,,,. Since Z0,,, I G it follows that AB 4 Z* 9. Thus the hypothesis U , E V made in Corollary 2.4 is not redundant.

Let R = [U1IM. Then U , @ R. However, UZe [U1lMand thus R f Re. Let R o 9, 2 = {a, T } and consider the sets Z+ and a. Then ME*= 1 E R and M , = Z 0 , , , . From Exercise III,2.2 it then follows that M, E R. However, Mz,, = U , @ R. EXAMPLE 2.2.

E

( F ) nM

=

It follows that Z* and a are in Z * 9 , but P a is not. This shows that the hypothesis V = Ve made in Corollary 2.4 is not redundant. EXERCISE 2.1. (Difficult).

(i) a E Z+Y” for all a E (ii) Aa E Z + Tfor all A

Let T be a +-variety such that

C. G

Z + T and all a E Z.

Let Z + Y be the boolean closure of all sets of the form

A,

AZ*

or A E Z * T . Show that W - is a +-variety and that if Y o V , then

Y e J L * V 3. Schutzenberger’s Theorem

T h e M-variety A of aperiodic monoids was discussed in Example V,3.3. We now introduce the corresponding +-variety doA of aperiodic sets. Schutzenberger’s Theorem describes the classes Z + d and, next to Kleene’s Theorem, is probably the most important result dealing with recognizable sets. THEOREM 3.1. (Schiitzenberger). The class Z + d is the least class of subsets of Z+ satisfying the following conditions:

(i) Z w dis boolean closed. A , B E Pd implies AB E Z * d . (ii) C T E Z e d for each u E Z. (iii)

254

IX. Aperiodicity

Let 2*Y be the smallest class of subsets of Z* satisfying conditions (i)-(iii) of Theorem 3.1. We first show that 2°F c Pd. For this it suffices to show that 2"dsatisfies conditions (i)-(iii) of Theorem 3.1. Condition (i) clearly is satisfied since Z * d i s boolean closed. T o prove (iii) note that M , (the syntactic monoid of the singleton o) is aperiodic. Thus Mo E A and CT E Z * d . T o prove (ii) consider A, B E Pd, so that M A ,MB E A. By Theorem 2.1, it suffices to prove M B 0M A E A. This follows from Theorem 2.2 since A = ( 1)where (1) is the G-variety consisting of 1 alone. The proof of the opposite inclusion Pd c 2"T is more elaborate and utilizes sequential functions. Let A E Z * d . Without loss we may assume A c 2+.By Proposition 1.2 there exists a sequential function f: Z* + 2" such that TM, is aperiodic and A = Cf-I where C = 2*1. Since C E 2 ° F it suffices to prove Proof.

-

r" be a sequential function such that T M , is (3.1) Let f: Z" aperiodic. Then B E PT implies Bf -I E Z * T . Since TM, is aperiodic, it follows from Corollary VI,3.4 that f is the 2.. composition of sequential functions f i , . . . ,fk such that TM,, Consequently it suffices to verify (3.1) under the additional assumption

<

Let 8 be the class of all subsets B of P such that Bf -l E P T f o r all sequential functions f: Z* +T" satisfying (3.2). Since f commutes with boolean operations and Z * T is boolean closed, it follows that 8 is boolean closed. For each y E F the set yf is a subset of Z and thus yf-' E Z * T so that y E 8. It remains to show that 8 is closed under concatenation. We now interrupt the argument to prove a general property of output modules (i.e. sequential machines without a specificed initial state) from which the fact that 8 is closed under concatenation will follow rather quickly. PROPOSITION 3.2. Let A = (Q, A): Z 'l be an output module. For each state q E Q let f,: Z* -+ P be the result of the sequential machine (8,q, I ) . For each pair p , q E Q let A,, be the behavior of the automaton ( Q , p , 4). For each subset B of P let

Bpp= Bf;' n A,,

3. Schiitzenberger's Theorem

255

Then

for any subsets B , C of P.If, further, A? is complete, then

(3.3 1

(BC)fi'=

U

BprCrq

7.9 E Q

Proof.

T h e condition sE

(BC),,

(BC)f;' n A,,

=

is equivalent with

PS = 4,

(p,s)A

=

b E B, c

bc,

E

C

This is equivalent with the existence of a factorization s PU = r,

YZ, =

( p , u)A = b,

9,

(Y,

z,)A

= UZ,such

that

=c

i.e. with E

Bpr,

v

E crq

This proves the first assertion. If, further, is complete, then 2* = UgcQA,, and therefore (BC)f;;' = (),EQ(BC)P,.Thus the second assertion follows from the first I Now assume that B, C E 8 and let f: Z* + P be a sequential 2.. Let then = ( Q , i, A): Z -+ T be a function such that TM, complete minimal sequential machine with result f =fi. Then (3.3) applies. Thus to show that BC E 8 it suffices to show that B,, E Z * Y for all p , q E Q . Since B,, = Bf;' n A,, and since T M f p TM, 2. it suffices to show that A,, E Z * T for all p, q E Q. We recall that A,, is the behavior of the automaton ( Q , p , 4 ) . Since TM, Z', Q has at most two states. If Q has only one state, then A,, = Z* and the result is obvious. Thus we may assume that Q = (0, 1 ) . Since A,, = L'* - A,, and A,, = 2# - A,, it suffices to consider the case p f q. Arguing by symmetry, it suffices to prove A,, E 2°F. Define

<

< <

2,= { o l o E L ' , l a = 0 , O a = 0 } Q=

{,lcTEZ,

I,=

1, Oo=O}

With this notation we have

A,,

=

Z*Z,sZ*

<

IX. Aperiodicity

256

and since

52" it follows that A,,

E

= Z*- Z*(Z - Q)Z#

Z#T

I

All that was said above about the M-variety A applies also to the S-variety AS.I n particular, (3.1)-(3.3) hold with obvious changes in the notation. We have the +-variety d p of aperiodic sets defined by

As 0 Jc9 Since U, E A, Proposition VII,5.4 implies Z + d y = { A n Z+1A =

{AI A

E

Z*d)

c Z+, A E

2Fd)

From this and Theorem 3.1 we easily deduce THEOREM 3.1s. The class Z + d y is the least class of subsets of Z+ satisfring the following conditions :

(i) Z + d is boolean closed. (ii) A, B E Z + d implies AB E Z + d . (iii) cr E Z + dfor all cr E Z I In this statement we use the simplified notation Z + d instead of Z+dp. 4. The Brzozowski Hierarchy

The Brzozowski hierarchy is a doubly indexed hierarchy for the aperiodic subsets of Z+, i.e., for the class Z + d . It grows out in a natural manner from Theorem 3.1s by systematizing the applications of concatenation and boolean operations. We proceed inductively starting with

Assuming that Z+9,+, is already defined for some n > 0, the class Z + S n is defined as the boolean closure of all sets of the form

(4.1)

A, . . . Ai

4. The Brzozowski Hierarchy

with A , ,

. . . , Ai E

257

Z + 9 n - l . Clearly

...

93, c g1t . . . c Snc

and the union is the class of all aperiodic sets d9. The last assertion is a reformulation of Theorem 3.1s. Within each SVn (n > 0) one can again establish a hierarchy by defining Z + 9 n , to z be the boolean closure of the sets (4.1) with i 5 1 and with A,, . . . , Ai E Z+Sn-las before. Then

Bn., c

c

...

c

c

.. .

and the union is SVn. We note that Bn,o =

9 R - I

for n > 1. The case n = 1 is an exception and 2+9,,, is the boolean closure of Z + 9 , which is not boolean closed. We first consider the classes S1,zFrom . the definition it is clear that Z+91,zis the boolean closure of all singletons s E L'+ with 1 s 1 5 1. Thus coincides with the +-variety X i of Exercise VIII,2.1 while 9, coincides with the +-variety Xconsisting of all sets that are finite or cofinite. The corresponding S-variety has been shown in Proposition VII1,Z.Z to be the variety of nilpotent semigroups. We now move on to the next level of the hierarchy and consider the families 92,1 with 1 5 1. The family 92,z is the boolean closure of the family of sets

(4.2)

A,

... A d ,

1 sill

where each of the sets A,, . . . , Ai is in 9,, i.e. is either finite or cofinite. Each finite set is a union of singletons, while each cofinite set is the union of a set of the