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. Define v < w if and only if q} < r3 (as rational numbers) where j is the greatest i G IQ such that qz ^ r*. Then V is an ordered group.
4
CHAPTER 1. DEFINITIONS
AND
EXAMPLES
*Example 1.3.8 Every torsion-free Abelian group G can be embedded (as a group) in a rational vector space; call it V Totally order a basis for V By the previous example, V is an ordered group. With the inherited order, so is G. Thus every torsion-free Abelian group can be totally ordered so as to be an ordered group. A group G is said to be divisible if for every positive integer n and g £ G, there is h £ G such that hn = g. It can be shown that every group can be embedded in a divisible group [Lyndon k. Schupp, page 189]. We have just shown that every orderable Abelian group can be embedded in a divisible such. One of the biggest unsolved problems in the study of ordered groups is: Question: Can every orderable group be embedded in a divisible orderable group ? We will prove later (Corollary 8.6.3) that every lattice-ordered group (and so every ordered group) can be embedded (as a group and a lattice) in a divisible lattice-ordered group. *Example 1.3.9 Let X be an arbitrary topological space and C(X) be the additive group of all continuous functions from X into R (where R is equipped with the usual topology); i.e., (/ + g)(x) = f(x) + g{x) for all a; £ X. Then C(X) is a lattice-ordered group under the pointwise ordering (i.e., / < g if and only if f(x) < g(x) for all x £ X). A related example is: Example 1.3.10 Let X be an extremally disconnected topological space; that is, a Hausdorff space in which the closure of every open set is open. Consider the set D(X) of all continuous functions / from X into RU {±00} such that {x £ X : f(x) <£ R\ is nowhere dense. Then D(X) is a lattice-ordered group under addition of functions and the pointwise ordering. For more details, please see Sections 5.5 and 5.6.
5
1.3. EXAMPLES
Example 1.3.11 Let D be the group of all differentiable functions from 1 into R. Then D is a subgroup of C(K) (Example 1.3.9). It is a di rected group under the inherited pointwise ordering. However, it is not a sublattice of C(R) because the supremum of 0 (the 0 function) and the differentiable function g : x i-» 2x agrees with 0 if x is negative and with g if x is positive and thus is not differentiable at x = 0. So a subgroup of a lattice-ordered group is a partially ordered group but need not be a lattice. In general the subgroup need not even be directed (see Example 1.3.19 below). Example 1.3.12 Another classical example is provided by the "divisi bility" condition: Let D be a domain, U(D) be its group of units, and K be its field of quotients. Let K* — K\{0}. For a, b € K*, define a\b if and only if b/a € D. Note that | is reflexive and transitive but not necessarily antisymmetric: a\b and b\a are simultaneously possible, but only if b — au for some u G U{D). Thus, if we define aU(D) < bU(D) if and only if a\b, then we obtain a well-defined partial ordering on G{D) = K*/U(D). The group G(D) is called the group of divisibility and is an Abelian par tially ordered group. If v : K* —> G(D) is the natural map given by v : a H-> aU{D), then x\a, b if and only if v(x) < v(a),v(b). Thus: G(D) is a lattice-ordered group if and only if D is an hcf (gcd) domain; G(D) is an ordered group if and only if D is valuation ring (with respect to v); G(D) = Z if and only if D is a discrete valuation ring (with respect to v). The directed group Ci ® Z of Example 1.3.4 provides a uesful coun terexample to several conjectures in this subject [Lattice-ordered Groups: advances and techniques, Chapter 4]. *Example 1.3.13 Let {Gt : i e 1} be a collection of right partially ordered groups. Then their Cartesian product
G = Y[Gl is a right partially ordered group where the ordering on G is given by: g < h if and only if g% < h% for all i G /. Similarly, for families of par tially ordered groups, directed groups and lattice-ordered groups. This particular ordering is known as the cardinal ordering.
6
CHAPTER 1. DEFINITIONS
AND
EXAMPLES
Let {Gi : i € / } be a collection of right partially ordered groups. Then their direct sum iei
is a right partially ordered group under the ordering inherited from the cardinal ordering. Similarly, for families of partially ordered groups, di rected groups and lattice-ordered groups. Another set of interesting subgroups arises in the special case that / is the set of strictly positive integers and each G1 is the ordered Abelian group of real numbers (under addition and the usual ordering). The product is then just the set of real sequences under the pointwise order ing and is a lattice-ordered group. Under the inherited order, so are the subgroups of all bounded sequences of real numbers, all convergent se quences, all sequences of real numbers convergent to 0, and all sequences of real numbers that are eventually 0 (the above direct sum). *Example 1.3.14 Let / be a well-ordered set; that is, every non-empty subset of / has a least element. Let {Gi : i € / } be a collection of right partially ordered groups. Then so is their Cartesian product
G = Y[Gt where the ordering on G is given by: g < h if and only if gj < hj where j is the least i (E I such that g+ ^ hi- Similarly, for families of partially ordered groups, directed groups and ordered groups, and for inversely well-ordered index sets (every non-empty subset has a greatest element). *Example 1.3.15 Let / be a totally ordered set and {Gi : i € 1} be a collection of right partially ordered groups. Then so is their direct sum iei
where the ordering on G is given by: g < h if and only if g3 < hj where j is the greatest i G / such that gi ^ hi. Similarly, for families of partially ordered groups, directed groups and ordered groups. We could instead use the ordering on G: g < h if and only if g* < hj where j is the least i £ I such that gt ^ hi. In the special case that I is well ordered, this is just the order inherited from the full Cartesian product of the previous example.
1.3. EXAMPLES
7
E x a m p l e 1.3.16 Let F be an inversely well-ordered set and V = V(T, E) be the set of all functions from F into E. For each g G V, let supp(g), the support of g, be the set {7 G T : 5(7) ^ 0}. Then V is an ordered Abelian group where the group operation is addition of functions, and / < g if and only if /(/?) < g(fi) where 0 is the greatest element of supp((? - / ) . More generally, if F is any totally ordered set, let V = V(F, E) denote the set of all functions from F into E whose support is either empty or inversely well-ordered. Then V = V(F, E) is an ordered group under addition and the ordering just given; it is known as the full Hahn group on F. For more details, please see Section 4.5. E x a m p l e 1.3.17 Let T be a partially ordered set such that every non empty totally ordered subset of F has a greatest element. Let V = V(r, E) be the set of all functions from F into M, and supp(<7) be defined as above for all g € V Then V is a partially ordered Abelian group where the group operation is addition of functions, and / < g if and only if /(/?) < g(/3) for all /3 € F that are maximal elements of supp(g — / ) . More generally, for F an arbitrary partially ordered set, let V = V(r,R) be the set of all functions g from F into E such that supp() is either empty or every totally ordered subset of supp(g) has a maximal element. Then V is a group under addition and is partially ordered by: g > 0 if g(6) > 0 for every maximal element 6 of supp(g). This too is called the full Hahn group on F. E x a m p l e 1.3.18 Let F be a root system; i.e., a partially ordered set such that 7 and 5 have a common lower bound only if 7 < 6 or 5 < 7. Suppose further that every non-empty totally ordered subset of F has a greatest element. Let V — V(F, E) be the set of all functions from T into E, and supp(g) be defined as above for all g £ V Then V is a lattice-ordered Abelian group where the group operation is addition of functions, and / < g if and only if /(/?) < g(/3) for all j3 <E F that are maximal elements of supp((? - / ) . More generally, if F is an arbitrary root system, let V = V(F, E) be the set of all functions g from F into E such that supp(g) is either empty or every totally ordered subset of supp(g) has a maximal element. Then
8
CHAPTER 1. DEFINITIONS
AND
EXAMPLES
V is an Abelian lattice-ordered group under addition where g > 0 if g(S) > 0 for every maximal element 5 of supp(g). This too is called the full Hahn group on F. For more details, please see Section 4.6. Each of the previous three examples is known as a Hahn group with spine T. *Example 1.3.19 Let (fl, <) be a partially ordered set and A(Q) — Aut(Q, <) be the group of all permutations of Q that preserve the order < ; i.e., a permutation g of fl belongs to A(fl) if ag < (3g whenever a < P in Q. Let /, g £ A(£l). Define / < g if and only if af < ag for all a £ fl. Then A(Q) is a partially ordered group. In analogy with Example 1.3.16, we define supp(^), the support of an element g £ A(£l), to be the set {a £ fl : ag ^ a}. If (Q, <) is a totally ordered set, then A(fl) is a lattice-ordered group. In particular, with this ordering A(Q) is a lattice-ordered group. Let H be the subgroup of all maps i 4 s i , where a ranges through the entire set of positive rational numbers. Then no two distinct elements of H have an upper or lower bound (to be an upper bound of both the identity map and x i-» ax, the value of c in x t~¥ ex must be at least max{a, 1} for positive values of x, and at most min{a, 1} for negative values of x). Hence H is not directed under the inherited order (and therefore not a lattice). *Example 1.3.20 Let (Q, <) be any totally ordered set, G = A(Q), and -< be any well-ordering of 0.; i.e., every non-empty subset of Cl has a least element. Let / , g G G be distinct. Define / < g if and only if /3f < fig where (i is the least a G 0, (under -<) such that af / ag. Then G, with respect to this partial ordering, is a right ordered group. In general it is not an ordered group. For example, let fi be the totally ordered set of rational numbers and -< be any well-ordering of Q with 0 -< a for all a e fi\{0}. Let / be the map x H-> x+1 and g be the map x i-4 2x. Then g < f, since Og = 0 and 0 / = 1. However, if h is the map x H* X + 2, then hf < hg, since Ohf = 3 and Ohg = 4. Indeed, we now show that A(Q) cannot be made into an ordered group in any way at all. Define g £ A(q) by: _ I ap
~
{a- 2n) 2 + 2n
{ 2n + l +
if a e [2n,2n + l)
y/(a-2n-l) if a £ [2n+ l,2n + 2),
1.3. EXAMPLES
9
(n 6 2). If t € A(Q) is translation by 1 (i.e., at = a + 1), then t~lgt = g~l But if G is any partially ordered group, then G + is closed under conjugation by an arbitrary element of G; and we cannot have g ft 1 and g i> 1. Therefore A(Q) cannot be made into an ordered group. Example 1.3.21 Let G be a right partially ordered group and H a subgroup of G. Suppose that the relation Hf < Hg if and only if h\f < h2g for all h\, h2 € H gives a well-defined total order on TZQ(H), the set of right cosets of H in G. We define the G-core of H as coreG(H) — [\{g~xHg : g £ G}. It is the largest normal subgroup of G contained in H. If G is clear from context we will abbreviate this to core(H) and call it the core of H. Let C = core(H). Then G/C acts in the natural way as a group of orderpreserving permutations of the totally ordered set 7Zc{H). Moreover, the only coset Cg that acts as the identity is C. It is easily checked that G/C is a partially ordered group where Ca < Cb if and only if Hga < Hgb for all g € G. If instead, as in the previous example, we use a well-ordering -< on TZG{H), then we obtain a right (total) ordering on G/C via: Ca > C if and only if H fa > Hf where Hf is the least element (under -<) in the support of Ca; that is, G/C is right orderable. Example 1.3.22 Let G be the subgroup of GL(2,Q) comprising all 2 x 2 invertible upper-triangular matrices over the field of rational numbers Q, with positive entries on the diagonal. Let A and B be distinct elements of G. Define A < B if either a22 < b22, or a22 = b22 and both an < bu and ai2 < &i2- Then G is a partially ordered group. The group G can also be partially ordered as follows: define A < B if either (a22 < b22) or (a22 = b22 and an < bn) or (a22 — b22 and an — bn and a12 < 612)- Then G is an ordered group. The subgroup H of G comprising those elements with a22 = 1 can be right ordered analagously to Example 1.3.2: A G H+ if au + a12£ > 1. Example 1.3.23 We recall the definition of the upper central series of a group G. Let ((G) denote the centre of G, that is, the set of elements of G that commute with every element of G. Let (0(G) = {1} and Cn+i(G)/Cn(G) = C(G/(„(G)). A group G is said to be nilpotent of class c if Cc(G) = G. So an Abelian group is nilpotent of class 1. Since
10
CHAPTER
1. DEFINITIONS
AND
EXAMPLES
(n+i(G)/Cn(G) is a n Abelian group that is torsion-free if G is [P. Hall, Lemma 4.8 page 435], it follows that in this case it can be totally ordered (please see Example 1.3.8). If G is nilpotent and g £ G, there is a unique n with g £ Cn+i(G)\Cn(G). We define g > 1 if g(n(G) > lCn(G). With respect to this total order, G is an ordered group. Thus every torsion-free nilpotent group can be totally ordered so as to be an ordered group. We utilise this idea but instead take advantage of the lower central series of a group: Example 1.3.24 We recall the definition of a commutator in a group G: let g,h £ G; then [g, ft] = g~1h~lgh. If H and K are subgroups of G, then we define [H, K] to be the subgroup of G generated by the set {[h,k] : h £ H,k £ K). Define jm{G) inductively as follows: 70(G) = G and 7 m +i(G) = [jm(G),G]. The sequence 70(G), 71(G),... is known as the lower central series. A group G is nilpotent class c if and only if 7 c+ i(G) = {1} (please see, e.g., [M. Hall, page 151]). Let F\ be the free group on the set X of free generators. By a theorem of Magnus [Lyndon & Schupp, page 68], f]{jm{Fx) : m = 1,2,3,...} = {1}. Since Jm(Fx)/jm+i(Fx) is a torsion-free Abelian group, it can be made into an ordered group (see Example 1.3.8). If g £ Fx, then 5 = 1 or there is a unique m with g £ rym(Fx)\^m+i(Fx)In this latter case, define g > 1 ifgjm+l(Fx) > ljm+\{Fx). With respect to this total order, Fx is an ordered group. Actually, with the aid of two easy results from the next two chapters, it follows that any lattice ordering of Fx must be a total ordering: In any free group, [g, h] / 1 unless g and h belong to the same cyclic subgroup. But if g A h = 1 in a lattice-ordered group, then [g,ft]= 1 as is shown in Lemma 2.3.4. So, by Lemma 3.3.3, the lattice ordering on Fx must be total. It is possible to totally order the free group on continuum many gen erators to make it an ordered group, so that any other group structure that can be put on this ordering resulting in an ordered group necessarily generates the variety of all groups. This very deep theorem can be found in [Holland, Mekler, & Shelah].
1.3. EXAMPLES
11
Caution: In general, the lower central series for nilpotent groups can not be used for ordering. Let G be the group generated by a, b, c where c commutes with both a and b and [a, b] = c2. Then G is nilpotent class 2 with 71(G) = (c2); but G/ji(G) is not torsion-free. *Example 1.3.25 Let C be a right partially ordered group, and
G = l[C where G is given the cardinal ordering. Let <j) be a permutation of / and A be the group of permutations of / generated by tj>. Let H = G a A, the semi-direct product of G by A ordered by: (g, <j>n) < (/, (f>m) if either n < m, or n = m and g < f. Then H is a right partially ordered group. Similarly, for right ordered groups, partially ordered groups, or dered groups, directed groups and lattice-ordered groups. We will use G * A to denote H — G » A equipped with this partial ordering, and say that H is a lexicographic extension of G by A. Let Go = Y1 in A or (a = 1 and g £ G + ). We say that H is a lexicographic extension of G by A and write H = G Vf A. If G and A are partially ordered groups, then so is G x A with the above ordering if conjugation by the elements of A map G + to itself. *Example 1.3.27 Let A be an ordered group and G be any partially ordered group. For each a £ A, let Ga = G and B = UaeA^a- We let W = B » A with multiplication: (6,ai)(c,a 2 ) = (d,aia 2 ), where d(a) = b{a)c{aai) (a G A). We call W the (unrestricted) Wreath product of G and A, and write W - G Wr A. It is a partially ordered group
12
CHAPTER 1. DEFINITIONS
AND
EXAMPLES
under: (6, a) > 1 if and only if either a > 1 or (a = 1 and 6(i) > 1 for all i & A). Indeed, W is a lattice-ordered group if G is. In the next chapter we will see that W cannot be made into an ordered group even when G is (if A , G ^ { 1 } ) . Consider the subgroup of all elements (b, a) where b(i) = 1 for all but a finite number of i € A. We call this subgroup the (restricted) wreath product of G and A, and write it as G wr A. It is a partially ordered group under the inherited order from the Wreath product: (b, a) > 1 if and only if either a > 1 or (a = 1 and b(i) > 1 for all i G A). Example 1.3.28 Let G be any partially ordered group. Let B = ]Cnez ^ and W = B x Z; again (b,n)(c,m) — (d,nm), where d(i) = b(i)c(i + n) (i € Z). We call W the right wreath product of G and Z, and write H/ = G wr Z if W is partially ordered by: (b, n) > 1 if and only if either n > 0 or (n = 0 and 6(j) > 1) where j is the greatest integer i such that 6(i) 7^1. If G is an ordered group, then so is W We call the above group W the left wreath product of G and Z, and write W = G wr Z if H^ is partially ordered by: (6, n) > 1 if and only if either n > 0 or (n = 0 and b(j) > 1) where j is the least integer i such that 6(?) / 1. If G is an ordered group, then so is W Example 1.3.29 Let / be a set and V(I) be the set of all subsets of /. A non-empty subset T of V(I) is said to be a filter on I if for all
J,K&V(J\ (i)(JCA-andJef)^i(6f
(ii)
J,KeJr=^JnK€Jr
The set V{I) is a filter as is {/}. Hie I, then {J e V{I) : i e J } is also a filter. All other filters on / are said to be non-trivial. For example, if / is any infinite set, then the set of cofinite subsets of / (those whose complements are finite) is a non-trivial filter. If T is a non-trivial filter that is not contained properly in any other such, then T is called an ultrafilter. Equivalently, a non-trivial filter T is an ultrafilter if and only if for each J C I, either J 6 f o r I\J G T. [Chang & Keisler, page 166]
1.3. EXAMPLES
13
By Zorn's Lemma, every filter on I is contained in an ultrafilter on /. Suppose that 7 is a set and U be an ultrafilter on I. For each i € / , let Gi be a right partially ordered group. Define an equivalence relation on Uiei Gi by: / ~ g if and only if {i £ I : f(i) = g(i)} e U. The group operation is well-defined on the set of equivalence classes. A well-defined partial ordering on the set of equivalence classes is pro vided from Wi€lGl by: f < g if and only if {i £ I : f(i) < g(i)} € U. With this operation and relation, the set of equivalence classes becomes a right partially ordered group known as an ultraproduct of Y\ieI Gt. It is denoted by {Utei Gt)/U. If all the d are partially ordered groups, the same is true of any ultraproduct. Also with directed, lattice-ordered, ordered and right ordered in place of right partially ordered. If all the Gi are equal to G, we will call (YlieiG)/l(
an ultrapower of
G. We can embed G in (l\iei G)/U via the natural map; this maps each g € G to the equivalence class of the constant function g; i.e., g(i) = g for all i £ I. We will identify G with its image and so regard it as a subgroup of ([[G)/U. i€l
In the course of the book, we will use three well known results from model theory [Chang k Keisler, pp. 172 & 314]:
Theorem l . A (i) For any first-order structure G and ultrafilter U, the structures G and (JJ G)/U satisfy the same first order sentences. iei (ii) If G and H satisfy the same first-order sentences of some lan guage, then there is an ultrafilter U on some set I such that, with respect to this ultrafilter, the ultrapowers of G and H are isomorphic. (Hi) For any algebraic structure G with a one element subalgebra {I}, one can embed G in an ultraproduct of its finitely generated subalgebras.
14
CHAPTER 1. DEFINITIONS
AND
EXAMPLES
The proof of (iii) is easily achieved as follows: Let / be the set of finitely generated subalgebras of G. For each g G G, let Ig be the set of all elements of / to which g belongs. Then for any finite subset X of G, : C\{IX : x 6 X} is non-empty. So there is an ultrafilter on / containing each Ig (g e G). The natural map g i-> g' (given by: g*(J) = g if g e J and g*{J) = 1 if g $ J) of G into this ultraproduct is an embedding. /
Chapter 2 Basic Properties The purpose of this chapter is to obtain restrictions on the groups that can occur as various types of partially ordered groups, and the partial orders that can occur as the same. We will show that our answers can, at best, be partial; in many cases, there can be no classification. We also discuss the closure of certain classes under various algebraic construc tions.
2.1
Basic group-theoretic properties
In this section, we give easy group-theoretic consequences of the defini tions. Every group is a partially ordered group under the trivial order (1.3.3), so there are no restrictions on a group that can occur as a (right) partially ordered group. Since a total order is the most restrictive, we expect constraints and consider this case first. A group G is said to be torsion-free if gn ^ 1 for every g £ G\{1} and non-zero integer n. L e m m a 2.1.1 Every right ordered group is torsion-free. Proof: Let G be a right ordered group and g £ G. If g ^ 1, then g > 1 or g~l > 1. In the former case it follows by induction on n that gn > 1 for all positive integers n; hence gn ^ 1 for all non-zero integers n. Similarly, this holds in the latter case, too. / 15
16
CHAPTER 2. BASIC
PROPERTIES
The same conclusion holds for any lattice-ordered group. This follows since every lattice-ordered group is a subgroup of a right ordered group (Corollary 7.1.7). However, we give a direct self-contained proof now: L e m m a 2.1.2 In any lattice-ordered group, if gn > 1 for some positive integer n, then g > 1. Proof: If G is a lattice-ordered group and g e G, then, (g A l ) n = g Agn~l A...AgAl for all positive integers n (by induction on n). Hence if gn > 1, then (g A l)n = gn'1 A ... A g A I = (g A l)"" 1 . Therefore gAl = l,*>g>l. II n
Corollary 2.1.3 Every lattice-ordered group is torsion-free. In the special case that the group is an ordered group, we can strength en the conclusion of the first lemma. A group G enjoys unique of extraction of roots if for all positive inte gers n and g,h € G, gn = hn implies g — h. Lemma 2.1.4 IfG is an ordered group, then G enjoys unique extraction of roots. Proof: If a < b in G, then an easy induction on the positive integer m shows that am < bm Hence, in any ordered group, gn = hn implies 9 = h. /I The result fails for right ordered groups and lattice-ordered groups: consider A(R) under the right ordering given by Example 1.3.20 or the lattice ordering given by Example 1.3.19. Let h0 be the map x i-> 2x2 + | on the interval [0, | ] ; so h0 : [0, |] -> [|, 1]. Define hi on [|,1] by: ah\ = Q/IQ1 + 1. Let h £ A(R) be defined by the following rule: if a E [m, m + ~] for some integer m, then let ah = (a — m)hQ + m; and if a £ [m + | , m + 1) for some integer m, then let ah = (a - m)hi + m. Let g be the map x i-» X + | . Then ^ 2 = ^i2 but g ^ h. The converse of Lemma 2.1.4 also fails: There exist non-orderable groups with unique root extraction ([G. Baumslag, I960]). Recall that a subset X of a group G is said to be isolated if 5" e A' for some non-zero integer n implies that g e X. We now show that centralisers in ordered groups are isolated:
2.1. BASIC GROUP-THEORETIC
PROPERTIES
17
Corollary 2.1.5 Let G be an ordered group. If [g,hn] = 1 for some non-zero integer n, then [g,h] = 1. Proof: If [g, hn) = 1, then [g-lhg)n Lemma 2.1.4./
= g~lhng = hn\ so g~lhg = h by
We write [a, b, c] for [[a,i],c]. Corollary 2.1.6 Let G be an ordered group. If[gm,h,gn] positive integers m and n, then [g,h, g] = 1.
= 1 for some
Proof: For any positive integer m, [gm,h,gn] can be written as a product of conjugates of [g~l[g, h)g\gn] (i = 0, ...,m—1), and so a product of conjugates of [g, h,gn}. Thus [gm,h,gn] - 1 implies that [g, h,gn] - 1. By the previous corollary, this implies that [g, h,g] = I.// The analogue of Lemma 2.1.4 for lattice-ordered groups is still sur prisingly strong: Lemma 2.1.7 In any lattice-ordered group, g and h are conjugate if gn = hn for some positive integer n. Proof: Assume that gn = hn and let / = gn~l V gn'2h V ... V ghn~2 V h ~ Then it is easily verified that gf — fh (replace gn by hn) whence n l
f-'gf
= h. H
As we will see immediately after the proof of Theorem 2.B, Lemma 2.1.7 can fail for right ordered groups. The situation for directed groups is not well behaved; every group can occur as a subgroup of a directed group. For let G be an arbitrary group equipped with the trivial order as in Example 1.3.3 and Z be the additive group of integers. Then GZ ordered as in Example 1.3.4 is a directed group with G as a subgroup. However, even in this case, some information can be gleaned. Recall from Chapter 1 that if G is a partially ordered group, then G+ is a normal subset of G (i.e., g~lag € G+ for all g € G and a e G+). If A is a subset of a group G, we denote by (A) the subgroup of G generated by A. If A is a normal subset of G, every element of (A) can be written in the form ab~l for some a,b G A.
18
CHAPTER 2. BASIC
PROPERTIES
Lemma 2.1.8 A partially ordered group G is directed if and only if (G+) = G. Proof: Suppose that G is directed. Let g £ G. There is ft G G such that h > 0,1. Then h, hg~l £ G+ and g = {hg-l)-lh. Therefore (G+) = G. Conversely, if (G+) = G, let g £ G. There are f,h £ G+ such that g = }h~l Now / = gh > g and / > 1. Thus / is an upper bound for g and 1. Given a,b £ G, we have shown that there is an upper bound for 1 and ab~~l, call it c. Then cb is an upper bound for both a and b. Similarly, every pair of elements of G has a lower bound./
2.2
Orderability
In this short section we give conditions on a group for it to be (right) orderable. If G is a group and X is an arbitrary subset of G, we let Sa(X) be the subsemigroup of G generated by X and NG(X) be the normal subsemigroup of G generated by X. Recall that if G is a partially ordered group, then / > 1 if and only if g-lfg> l(g EG). Lemma 2.2.1 [Ohnishi] Let G be a group. Then G can be made into an ordered group if and only if for every finite set {gi,..., gn} of elements of G\{1}, there are e1, ...,en £ {±1} such that 1 £ Na(gl1, ■••,^'v). Proof: Clearly the condition holds in any ordered group. Conversely, if the condition is satisfied, let g £ G\{1} be such that for some fu...,fm,9i,..,9n e G\{1} we have 1 £ NG(g, /*', „.,/*») and 1 G NG{g~l,g\\--;9n) f° r a l l choices of 5U... ,8 £ {±1}. Then
l£NG{g\fl\...Jt,9\\-,9en) for all choices ot e,8i, ...,8m,ei, ...,en £ {±1}. This contradicts the con dition of the Lemma. Hence Zorn's Lemma applies and there is a normal subsemigroup P of G such that G is the disjoint union of P, P~l and {1}. If we define a < b if and only if ba~l £ P (a,b £ G), then G becomes an ordered group. // The analogous fact for right orders is:
2.2. ORDERABILITY
19
Lemma 2.2.2 Let G be a group. Then G can be made into a right ordered group if and only if for every finite set {gx,..., gn} of elements of G\{1}, there areex,...,en G {±1} such that 1 0 SG(9i, -,9n)The proof is identical except we drop the normality throughout. Finally, we provide a useful variant of this Lemma. Let A be a set and L be a group that acts on A via conjugation. An ordering of A is said to be L-invariant if g~laig < g~la2g whenever aY < a2 in A and ge L. Recall from Example 1.3.21 that if H is a subgroup of G, we write for the set {Hg : g G G} of right cosets of H and core(H) = f]{g~1Hg : g G G} for the largest normal subgroup of G contained in H.
TLG{H)
Lemma 2.2.3 Let H be a subgroup of a group G. Suppose that for any finite set of elements gi, ...,gn G G\H, there exist t\, ...,en G {±1} such that the subsemigroup generated by g{1,..., g^n has empty intersection with H. Then there is a G-invariant ordering on 1ZQ(H) and G/core(H) is right orderable. Proof: For each finite set D C G\H, let T{D) be the set of all functions e : G\H —► {±1} such that the subsemigroup Sodd^ : d e D}) has empty intersection with H. By hypothesis the sets T(D) are non-empty; they have the finite intersection property since J-{D\) n F{D2) 2 ?{Di U D2). By Konig's Lemma, there is e G C\{T{D) : D C G\H, D finite}. If we let P = {g £ G\H : e(g) = 1}, then it is easily checked that P is a subsemigroup of G and that G — P U H U P~l. Moreover, P = HPH:- for P C HPH since 1 € if, and if fy G H (j — 1, 2) and p G P , then /ixp G P . (If not, e(/iip) = —1, so /if1 belongs to the intersection of H and the subsemigroup generated by {hip)'1 and p, which is empty.) Similarly (hip)h2 G P so P P P C P We define a right partial ordering on 7lG(H) by: Hf < Hg if and only if gf'1 G P; it is transitive because P is a subsemigroup. Since P = HPH the partial ordering is well-defined; and total since G = P U H U P _ 1 . Thus G/core(H) is right orderable since it acts faithfully as a group of order-preserving permutations of the totally ordered set 1lG{H) (see Example 1.3.21). /
20
CHAPTER 2, BASIC
2.3
PROPERTIES
Basic order-theoretic properties
The group structure imposes a certain uniformity on the partial order. We now wish to study this more carefully. Rather than give an exhaus tive (and exhausting) list, we have provided only the most important immediate consequences in order to minimise the symbol gymnastics. Lemma 2.3.1 If G is a partially ordered group and g,h 6 G, then g < h if and only if h~l < g~l Proof: Ug
then h^gg'1
<
h^hg-1.//
Lemma 2.3.2 Let G be a partially ordered group and g,h € G. (i) If g\/h exists in G, then so does g l Ah (ffV/z)- 1
1
(ii) If gAh exists in G, then so does g 1Vh (9 A h)-1
l
. Moreover, g l Ah
l
=
. Moreover, g l\/h
1
=
Proof: (i) Since g V h > g, it follows that (g V h) l < g * Similarly, {g V h)~l < h~l If / < g~\ h~\ then Z" 1 > g, h. Thus f'1 > g\/ h. Hence / < (g\/h)~l. Consequently, g~l Ah~x exists and equals (gWh)'1 The proof of (ii) is similar./ Lemma 2.3.3 Let G be a partially ordered group and g,h G G. Suppose that g and h commute. (i) IfgVh exists, then so doesgn\/gn the latter equals (g V h)n
:
/iV ...V ghn 1Vhn;
(i) If gAh exists, then so does gu Agn lhA...Aghn the latter equals (g A h)n.
1
moreover,
Ahn; moreover,
Proof: By an easy induction on n (c.f., the proof of Lemma 2.1.2)./ Lemma 2.3.4 If G is a partially ordered group and g,h £ G, then g A h = 1 implies that g and h commute and g\/ h = gh. In this case, if G is a lattice-ordered group, then g Ahf = g A f for all f 6 G+.
2.3. BASIC ORDER-THEORETIC
PROPERTIES
21
Proof: If gAh = 1, then gh = g(gAh)~lh = gig'1 V h ' l ) h = hVg = g V h which, similarly equals hg. So g and h commute. In this case, / < hf so g A / < g Ahf < g2 Ahg Agf Ahf = (g Ah)(g A / ) = g A /.// A lattice L is said to be distributive if for all a, b, c e L, a V (b A c) — (a V 6) A (a V c) and a A (6 V c) ~ (a A b) V (o A c). Lemma 2.3.5 If G is a lattice-ordered group, then the lattice of G is distributive. Proof: Any lattice that satisfies the implication (aAb — aAcka\/b
= a\/c^b
= c)
is distributive [Gratzer, page 60]. So assume that aAb = a Ac and aVb = aVc. Then, by Lemma 2.3.2, b = (aA6)a _1 a(aA6) _1 6 = (aA6)a _ 1 a(a _ 1 V 6 _1 )6 = (aA6)a _1 (6Va) = (aAc)a _1 (cVa) = (aAc)a~la(a-lVc~l)c = c.jj Actually, an even stronger result holds. We will denote the supremum of {gi : i € / } by V{
= (V{ft : i £ / } ) " \
(b) V{/ft :iel}
= /(V{ft : i e / } ) , and
(c)\/{fAgl:i£l}
=
fA\/{gl:i£l};
also dually. Proof: Let g = V{ft : « € / } . Clearly, g'1 < g~\ fg > fgl and / A g > f A ^ for all igt for all % G / ; thus h~l > g. Consequently, h < g~l, so g~l = [\{g~l :i€ /}■
22
CHAPTER 2. BASIC
PROPERTIES
(b) If ft. > fgt for all i e 7, then f~lh > gt for all i € I. Hence f~ h > 5, so ft > fg. Thus fg is the supremum of {fgt : i 6 / } . l
(c) By (a) and (b), [\{gg~l : i € 7} exists and equals 1. But 1 < (/A 9 H / A & ) - 1 < 5r\ since g~\f/\g) = g~lf Al < ft X / A l = $f H / A f t ) . -1 Hence A{(/ A #)(/ A & ) : i e 7} exists and equals 1. The result now follows from (a) and (b). / Surprisingly, the above property can be captured by a first order sentence in the language of lattices. Lemma 2.3.7 [R.N. Ball, 1984] Let <j)\,<j)z be the respective sentences VxVyVz{x < y < z -> (3t)[x < t < z & Vw(w V y > z ^ w V x > t)]), VzVyVzfx < y < z ->• (3t)\x
Vw(w A t / < i - n u A z < t)}).
Then in any distributive lattice D, (1) D |= cj>i <& c V /\ ■ c(j = Aj(c V dj) whenever Aj dj exists in D (2) D \= fa <=> c A Vj rfj = Vj(c A dj) whenever Vj dj exists in 7). Proof: (1) Assume that d = f\dj exists in D but cVd a, we get dj — dj V d > t for all j because D |= ^ . Thus d = Ajdj > t, a contradiction. Conversely, if D [= ->^>i, let d < c < a witness it. If W — {w : wVc > a}, let {dj : j £ J } be an enumeration of {u; V d : u> £ W } . Then d = Aj dj since D (= -10! . But c V Aj dj = c V d = c < a < c V dj for all j , since if dj = wV d and i u V c > f l , then c\Zdj = cVwVd — w\/c>a. (2) is proved similarly. / By Lemma 2.1.8, in any lattice-ordered group G, we have (G+) = G. For each g e G, let g+ = 5 V 1, #" = ff_1 V 1 and \g\ = gV g~l = |p _ 1 |. L e m m a 2.3.8 7h any lattice-ordered group, (*) \9\ = 9+g~ > l; fuj 5+ A 5 = 1; fnij g = g+{g ) l ;
2.3. BASIC ORDER-THEORETIC
PROPERTIES
23
H Ug)+ < f+g+; (v) (+)" = (gn)+, (g~)n = ( 5 »)-, and \g\n = \gn\; M\fVg\<\f\V'g\<\f\\g\; (vH)\fg\<\f\\g\\f\. Proof: (i) If / > g,g~l, then f2 > 1. By Lemma 2.1.2, / > 1. Thus \§\ = 9 V 5 " 1 > 1 and g+g- = ( ^ 1 ) ^ - ^ 1 ) = l V p V f f ^ V l = gV g~l = (li) g+ Ag~ = (g V 1) A (g" 1 V 1) = ( 5 As" 1 ) V 1 = {g~l V g)~l V I = 1, using Lemma 2.3.5, Lemma 2.3.2, and (i). (iii) gg~ = g(g-1 V I ) = I V g = g+. (iv) fg V 1 < fg V / V g V 1 = ( / V l)(g V 1). (v) If 0 < fc < n, then (3"-* V g-*)71 = \j {g(n~k)J g~k^-J) ■ j = 0,..., n) > g(n-k)kg-k(n-k) _ L B y L e m m a 2.1.2, (5n~* v g-*) > 1. Thus gnWl > gfc Hence (g+)n = gnVgn~1y...gVl = (gn)+ Similarly, (g") n = ( S n ) - By (i), (ii) and Lemma 2.3.4, \g\n = (5+5-)" = {g+)n{g-)n = (gn)+(gn)- = \9n\(vi) Z , / " 1 < l/l and y . s " 1 < l
24
CHAPTER 2. BASIC
2.4
PROPERTIES
Characterisations of classes
In the proofs in this section, the key idea is that any subgroup of a (right) ordered group is also a (right) ordered group (Example 1.3.5) ; however it is quite possible for a subgroup of a lattice-ordered group to fail to be a lattice (Example 1.3.11) and for a subgroup of a directed group to be trivially ordered (Example 1.3.4). Suppose that G and H are groups that satisfy the same sentences of group theory. Then they have isomorphic ultrapowers (see Theorem LA). The class of groups that are carriers of ordered groups is closed under ultraproducts (Example 1.3.29); so if G is the carrier of an ordered group, then so is H (restrict the order on the ultrapower of H to the subgroup order-isomorphic to H). Thus: Lemma 2.4.1 There is a set of sentences of group theory that is satisfied by a group if and only if it can be made the carrier of an ordered group. By the same proof: Lemma 2.4.2 There is a set of sentences of group theory that is satisfied by a group if and only if it can be made the carrier of a right ordered group. Clearly, any element that is strictly positive has infinite order; there fore, a non-singleton group that is directed must contain such an element. Lemma 2.1.8 can be used to classify those groups that can be made into directed groups: Lemma 2.4.3 A group G can be directed if and only if there is a subsemigroup S of G such that S n S~l — {1} and (S) = G. The above results and the results of the previous section might lead one to believe that the collection of groups that are carriers of latticeordered groups can also be characterised. The following example shows this hope to be quite unfounded. Theorem 2.A [Vinogradov] There are metabelian groups G\ and G2 satisfying the same sentences of group theory such that only Gi can be made into a lattice-ordered group.
2.4. CHARACTERISATIONS
OF CLASSES
25
Proof: Let H = Q»Z, where (0, l)~l{q, 0)(0,1) = (-g,0). NowQ®Z and Z satisfy the same sentences of group theory [Szmielew], so the same is true of Gx = Q ® Z ® iJ and G2 = Z ® # . Define Gj" = {(a, m, 6, n) : m > 0 or (ra = 0 and n > 0) or (m = n = 0 and a > |6|)}. Clearly, with respect to this order, (G+) = G\\ so to shew that Gi is a lattice-ordered group, by Lemma 2.3.2 it is enough to show that g V 1 exists for all j e G j ; this can easily be verified. Let c = (0,1,0) € G 2 . If it were possible for G 2 to be the carrier of a lattice-ordered group, then \c\ = (0,1,0) V (0, - 1 , 0 ) would exist. Conjugation by d = (0, 0,1) gives (0, - 1 , 0 ) V (0,1,0) = |c|, so |c| is fixed. But the only elements of G2 that are fixed by conjugation by d have the form (m, 0, n) for m, n e Z. Now c 1 ^ G G 2 and (c 1 /^ = \c\l'k = (0, l/jfc,0) V (0, -l/fc,0) for any positive integer k (by Lemma 2.3.8(v)). Thus (m/k,0,n/k) 6 G 2 for all positive integers k. Therefore m = n = 0 and |c| = (0,0,0). By Corollary 2.3.9, (0, 0, 0) = (0,1,0), the desired contradiction./ The homomorphic image of a torsion-free Abelian group can have torsion elements: eg, Z/nZ for any positive integer n. By this example (for n > 2) the classes of (Abelian) groups that are carriers of an or dered, lattice-ordered or directed (Abelian) group are not closed under group homomorphisms. Although, as we will see in Corollary 7.1.7, every lattice-ordered group is a subgroup of a right ordered group (and so is right orderable), not every right orderable group can be lattice-ordered by Theorem 2.A. Here is a different proof: E x a m p l e 2.4.4 Let L = Z Wr Z. Then L can be right ordered as fol lows: Well order Z and define (/, n) > (0,0) if and only if either n > 0 or (n = 0 & f(j) > 0 where j is the least element of {i £ Z : f(i) / 0} under the well ordering). Let G be the subgroup of L generated by c = (1,1) and a — (a, 1) where a(m) = ( - l ) m ; so c^1ac = a - 1 . Then G inherits a right ordering from L. However, it cannot be lattice-ordered: Other wise a V 1 exists in G and so has the form ancm for some integers n, m. But then an~lcm = a~l{a V 1) = 1 V a~l = 1 V c~xac = c~x{\ V a)c = c~1ancmc = a~ncm Hence a 2 " - 1 = 1 which is impossible. A group G is said to be a subdirect product of {G, : i £ 1} if there is an embedding <j> of G into Y\ieI Gi such that the homomorphisms 07r; map G onto Gi for all i £ I, where 7T; is the natural projection of Yliei Gi
26
CHAPTER 2. BASIC
PROPERTIES
onto Gi. If we denote the kernel of 4>ni by Ni} then (~){Ni : i £ 1} = {1} and G is a subdirect product of {G/Ni : i £ / } . In the case that each Gt has property X, we say that a subdirect product of {Gt : i £ 1} is residually X. As we saw in Example 1.3.14, by well-ordering the index set /: Lemma 2.4.5 Being the carrier of an ordered group is a residual prop erty of groups. Similarly for right ordered groups. In contrast, using the Lemma concerning conjugacy of elements (with equal powers) in a lattice-ordered group (Lemma 2.1.7): Theorem 2.B [Holland, 1985] Being the carrier of a lattice-ordered group is not a residual property of groups. Proof: Let A — (Z © Z) V Z where , w. . > f (m + i,n+ i.r + s) if r is even [m, n, r)[i, i, s) = < ; ; . . . . . , v ' ' n '•" ' \ (m + j,n + i,r + s) if r is odd and (m,n,0) > (0,0,0) if m,n > 0 in Z. For each non-negative integer n, let An be an order-isomorphic copy of A. Let b = (bn) and c = (cn) where bn = (1, —1,1) and c„ = (—1,1,1) for all n. Let G be the subgroup of rin€z+ An generated by 6, c and £nez+ Ai- Note that b2 = c2 = (0,0,2) but ft and c are not conjugate in G. Hence, by Lemma 2.1.7, G cannot be made into a lattice-ordered group. However, for each i = 1,2,..., if Ni = {geG:gi = (0, 0, 0)}, then Nt < G and n,=i W« = {0}. Moreover, G/Ni = Ai is a lattice-ordered group./ The group A in the proof becomes a right ordered group if we use the ordering: (m,n,r) > (0,0,0) if r > 0 or (r = 0 < n) or (r = n = 0 < m). By Examples 1.3.14 and 1.3.25, G is right orderable. The proof above shows that b and c are not conjugate in G although b2 = c2 So Lemma 2.1.7 fails for right ordered groups. A group G is said to be locally X if every finitely generated subgroup of G has the property X. Theorem 2.C Being the carrier of an ordered group is a local property of groups. Similarly for right ordered groups.
2.4. CHARACTERISATIONS
OF CLASSES
27
Proof: The proof is similar to that of Lemmata 2.4.1 and 2.4.2. Let G be a group. Then G can be embedded in an ultraproduct of its finitely generated subgroups (Theorem l.A). By Example 1.3.29, this ultraprod uct is (right) ordered. Thus G itself inherits the (right) ordering./ When one tries to copy this proof for lattice-ordered groups, it breaks down on the last step — a subgroup of a lattice-ordered group need not be a lattice (the supremum of two elements of the subgroup may not belong to the subgroup). Indeed: T h e o r e m 2.D [Vinogradov & Vinogradov] Being the carrier of a latticeordered group is not a local property of groups. Proof: We construct a sequence Go,Gi,G2,... of finitely generated groups that are lattice-orderable, but whose direct limit is not. Let Go = (go)■ Assume that Gn, a finitely generated group, has al ready been constructed and that Gn is an isomorphic copy of it. Let Hn = Gn ® Gn and G n + 1 — Hn » (gn+i), where gn+i is a new symbol and generates an infinite cyclic group, and if (a, b) € Hn, then (a, b) conju gated by gn+i is (b,a). So Gn+1 is generated by {g0,g0, -,9n,9n>Sn+i}Clearly, even powers of gn+\ commute with all elements of Hn. More over, by induction, Gn+i is a lattice-ordered group under the ordering: (a, b, g™+1) 6 G*+1 if and only if m > 0 or (m = 0 and a > 1 in Gn and b > 1 in Gn with the order inherited from Gn by the isomorphism). We embed the group Gn in the group Gn+\ via the map: g, i-> (<7i, <7i~\ 1) (1 < i < n). To see that this is indeed a group embed ding we show by induction on n that if w(go,...,gn,g0, ...,gn) = 1, then w(gQl,...,g~l,g~Ql,...,g~l) = 1. For n = 0 this is obvious. Since all gt commute with all <7,-, the same is true of the inverses; and since conju gation by gn+i interchanges gt and g{ for 1 < i < n, the same is true of g^li- By induction on n, any word that holds in the gi and gt also holds in the inverses of9n+iNow suppose, to the contrary, that L can be made into a latticeordered group. Let c € L+; then c £ Gn for some positive integer n and
28
CHAPTER 2. BASIC
PROPERTIES
for some positive integer k, c2 has the form just described. Hence, in Gn+i, c2 = (d,d ,1). Since gn+i conjugates this element to its inverse, it cannot be positive. This contradicts c £ L+.// In sharp contrast to all the above results, the class of groups that are carriers of ordered groups is not closed under Wreath products (Example 1.3.27), but the class of groups that are carriers of lattice-ordered groups is closed under Wreath products (for the definition of the latter and the proof of the result for lattice-ordered groups, please see Chapter 7). L e m m a 2.4.6 The class of groups that are earners of non-singleton or dered groups is never closed under Wreath products. Proof: Let A and G be ordered groups, 1 < c G A and 1 < g 6 G.
Define b , d g f l = JJ G by ut \ — I 9 if a = c 2m f° r some integer m \ 1 otherwise and
, , , _ f g if a = c 2 m + 1 for some integer m \ 1 otherwise.
Then (b,c),(d,c) € W = A Wr G and (b,c) 2 = (d, c) 2 but (b,c) ^ (d,c). By Lemma 2.1.4, W cannot be made into an ordered group./ We close with a further limitation on the class of groups that are carriers for various classes of ordered and orderable groups. A class C of groups is said to satisfy the amalgamation property if for all A, Ci, C2 £ C and embeddings a3■ : A ->■ C3 (j = 1, 2) there is G G C and maps r, : C3 —> G (J = 1,2) such that <j\T\ = CT2T2The class of all groups satisfies this property [Lyndon & Schupp, Sec tion IV.2]. This can be used to prove that every group can be embedded in a divisible group [Lyndon k Schupp, page 189]. It is therefore natural to ask if the class of all ordered groups satisfies the property. T h e o r e m 2.E [Glass, Saracino k Wood] The class of all ordered groups fails the amalgamation property.
2.4. CHARACTERISATIONS
OF CLASSES
29
Proof: Let V be the rational subspace of E generated by 1 and \/2. Let U — V © V, a 4-dimensional ordered vector space. Let -0 = 2/4, an order-preserving automorphism of U, and A = U x (V1) be the resulting ordered group. Let [ 2 0 ' Then M is the order-preserving automorphism of V multiplication by V2 and M2 = 2I2. Let
L - \ °2 and
So LjM + MLj = 0
1_
I- °
L - f -1 ° ' (;' = 1,2). We may view i
[M [ Lj
0 ' M
as a non-singular order-preserving linear transformation of U (j = 1, 2). Let £ j = (i*) and C^ = [ / V Bj (j = 1,2). Now (L*)2 = 7/) (j = 1,2) so there is a natural order-preserving embedding of A into Cj (j = 1,2). Let a = (0,0,1,1) e f / C A Then (0, Lt)- 1 (a,0)(0,L*) = (a£|,0) = ( - 2 , 1 , 2,1, 0) ? ( - 1 , 1 , 2 , 1 , 0) - (aL5,0) = (0, Z^J-^a, 0)(0, L^). But in any amalgam in the class of ordered groups, (0, LJ) must equal (0, L*2) since their squares are equal (Lemma 2.1.4). Hence no such amalgam exists./ Note that the amalgam does not even exist in the class of all orderable groups. Hence Corollary 2.4.7 [Kokorin & Kopytov] The class of all orderable groups does not satisfy the amalgamation property. Things are no better when we turn to the class of right orderable groups. T h e o r e m 2.F [Kopytov & Medvedev] The class of right orderable groups fails the amalgamation property.
30
CHAPTER 2. BASIC PROPERTIES
Proof: Let C\ be the group on generators ai,b\ subject to the single defining relation a{lb\ai = frj-1; i.e., isomorphic to the quotient of the free group on two generators x and y by the normal subgroup generated by x~lyxy. Note that every element of C\ can be written uniquely in the form a™b" using the defining relation a^&iOi = frjf1. Hence C\ is a right ordered group under the ordering: a™6" > 1 if and only if (m > 0, or m = 0 and n > 0). Let C2 be the group on generators a2,b2 subject to the single defining relation b2la2b2 = a2l Similarly C2 is a right ordered group where 62n<22 > 1 if and only if (m > 0, or m — 0 and n > 0). So Cj and C2 are right orderable groups. Let A be the free Abelian group on two generators c and d, <j\ : A —> Ci be given by: c i-> af, d i-¥ bi and a2 ■ A ^ C2 be given by: c — i > a2, d H->■ 6j- If a right orderable amalgam G existed with maps r, : C3■ —> G (j' = 1, 2), then ajTi = co\T\ = ca"2r2 = a2T2 and &jTi = cfcriTi = da2r2 — b\r2. By replacing a,\ by its inverse if necessary, we may assume that a\T\ > 1. Let e £ {±1} be such that b2r2 > 1. Hence (using the above relations) we obtain 1 < (a-in ■ b2T2)2(b2r2
■ a X Ti) 2 = aiTib2r2aiTi
= a,\T\b2T2 ■ O\T\ ■ ax lT\b\r\a\T\
■ b2r2a\Ti
— a,\TXb2T2 ■ QjTx • b2 tT2b\r2alTl
= a,iTia2
1
T2O,ITI
■ (b2)2r2 ■ air 1 & 2 r 2 aiTi
— a1ri627"2ffliT"i&i
CT
\b2T2aiTi
= aXTy ■ b2T2a2T2b2 V2 • aiTj
— 1, the desired contradiction. /
Note that, in the proof, U\ and a2 cannot both be order-preserving for the same order on A; therefore the proof does not address the amalga mation property for right ordered groups. In Theorem 7.C we will prove that the class of all lattice-ordered groups also fails the amalgamation property. Since every lattice-ordered group is ^-embeddable in a right ordered group (Corollary 7.1.7) these results may lead us to conjecture that the class of right ordered groups fails the amalgamation property but this has yet to be proved (the proof of Theorem 7.C does not establish lt). Does t h e class of right ordered groups fail t h e amal gamation property?
Chapter 3 Values, Primes and Polars In this chapter we concentrate on certain kinds of subgroups that are appropriate when a partial ordering is present. These will play a central role in obtaining structure theorems. At the end of the chapter we obtain some consequences of our analysis: a classification of residually ordered groups, and of lattice-ordered groups containing no infinite pairwise or thogonal subset.
3.1
Values
Let G be a right partially ordered group and C be a subset of G. We say that C is convex if for all g € G and C\, c2 G C, ci < g < c2 = > g G C. Note that G itself is a convex subgroup of G as is {1}. All other convex subgroups of G will be called non-trivial or proper. For example, let G = Z® Z under addition; i.e., ordered by: (0,0) < (m, n) if 0 < n or (0 = n and 0 < m). Then {(m, 0) : m £ 1} is the only proper convex subgroup of G. A set of subsets that forms a totally ordered set under inclusion will be called a chain of subsets. L e m m a 3.1.1 If G is a right partially ordered group and {Ct : i £ 1} is a family of convex subgroups of G, then f]{Ci : i £ 1} is a convex subgroup of G. If {Ci : i € 1} is a chain of convex subgroups of G, then (J{Ci : i £ 1} is a convex subgroup of G. 31
32
CHAPTER 3. VALUES, PRIMES AND
POLARS
Proof: Obvious./
Lemma 3.1.2 If G is a right ordered group, the set of all convex sub groups of G is a chain. Proof: Let C and D be convex subgroups with C <2 D. Since G is totally ordered and every element of C is in G + or its inverse is, there is c e C+\D. For all d £ D + , we have 1 < d < c (otherwise 1 < c < d0 for some d0 £ D, whence c € D by the convexity of I?). Hence d 6 C for all d 6 D + But every element of Z) is in D + or its inverse is. Consequently, D C C.I Let G be a right partially ordered group and g £ G. Let C be the set of all convex subgroups of G that contain g. Then f|{G : G € C} € C by Lemma 3.1.1 and is contained in every element of C. It is the smallest convex subgroup of G containing g and will be denoted by C{g). Let G be a right partially ordered group and g € G\{1}. Let V be the set of all convex subgroups V of G such that g £ V By Lemma 3.1.1, the union of any chain of subsets in V is itself in V. Since {1} G V, we have maximal elements in V by Zorn's Lemma. Such a convex subgroup is said to be a value of g in G. We denote the set of all values of g by T(g), and by T{G) the set of all values of all g e G\{1}. Certain elements in a right partially ordered group will be especially interesting in this setting. An element g of a right partially ordered group G is said to be special if it has a unique value. We will denote this unique value Vg. By Lemma 3.1.2, every element of a right ordered group is special. Let A and B be convex subgroups of a right partially ordered group G. We will say that a convex subgroup A covers B if B C A and for any convex subgroup C of G, B C C C A implies that C — A or C — B. Lemma 3.1.3 If G is a right partially ordered group, then every value in G has a unique cover in G. Proof: Let g 6 G\{1} and V be a value of g in G. Let V(g) be the intersection of all convex subgroups of G that contain V and g. Then
3.1.
VALUES
33
V(g) covers V: for if D is a convex subgroup of G and D D V, then g G D by the maximality of V; hence V(g) C D.// In the context of lattice-ordered groups (^-groups) where V and A are algebraic operations, we restrict to sublattice subgroups; these are called ^-subgroups: so L is an ^-subgroup of G if xVy, x Ay e L for all x, y € L. Note that, by Lemma 2.3.2, a subgroup L of an £-group G is an £subgroup of G if and only if x V 1 G L for all x G L. Also, if LUL2 are ^-subgroups of G, then L\ C L2 if and only if L+ C L j . It is straightforward to use Lemma 2.3.8 to show that: Lemma 3.1.4 Let C be a subgroup of a lattice-ordered group G. Then C is a convex (.-subgroup of G if and only if C = {g G G : \g\ < \c\ for some c G C } . We will use C(G) to denote the set of all convex ^-subgroups of G\ it is partially ordered by inclusion. Theorem 3.A For any lattice-ordered group G, C(G) is a distributive lattice. It is a complete sublattice of the lattice of all subgroups of G. Moreover, for all A, Cx G C(G) (i E I), A D \J{C, : i G / } = \J{A H Ct : i Gl}. Proof: Let {Cj : i G / } C C(G) and C be the subgroup of G generated by {Cs : i G / } . Let |^| < \c\ for some c G C. Then c = ]Xl=lck with Ck G Cifc. By Lemmata 2.3.8(vii) and 2.3.10 (the Riesz Interpolation Property), g+ and g~ can be written as a product of elements from U{C+ : i € I}. Hence g G C; so C G C(G) by Lemma 3.1.4. Clearly V-U D Q : i € 2"} C >1 n V{Ct : i G 7} and if g G ,4+ n V{C, : i € / } , then ^ < rifc=i cfc with c t e (7^. By the Riesz Interpolation Property, g — njj = i 5A with 1 < gk < ck (1 < k < n). Since each 1 < gk < g G A, each 0fe G AnCik.// In analogy with the above definitions, let G be a lattice-ordered group and g G G. Let C9 be the set of all convex ^-subgroups of G that contain g. Then f]{C : C G Cg} contains g and is contained in every element of Cg. It is the smallest convex ^-subgroup of G containing g, and will be denoted by G(g). By Lemma 3.1.4:
34
CHAPTER 3. VALUES, PRIMES AND
POLARS
L e m m a 3.1.5 If G is a lattice-ordered group and g € G, then G(g) = {f € G : | / | < \g\nfor some n e Z+}. Corollary 3.1.6 If G is a lattice-ordered group and f,g 6 G+, then G(fAg)
=
G(f)nG(g)
and
G(fWg)
=
G(f)vG(g).
For any lattice-ordered group G and g £ G, we say that a convex ^-subgroup V of G is a value of g if g # V yet g & H for all convex ^-subgroups H which properly contain V The existence of values for all g ^ 1 is as before. Moreover, by Zorn's Lemma, for any convex tsubgroup C such that g 0 C, there is a value of g containing C. Again, for covers, etc., we restrict our attention to convex ^-subgroups. Since each g G G\{1} has a value, it follows that f]{V : V is a value of some g € G\{1}} = {1}. We again denote the set of all values of g by r() and write r(G) for the set of all values of all g e G\{1}. As above: L e m m a 3.1.7 If G is a lattice-ordered group, then every value in G has a unique cover in G.
3.2
Homomorphisms
Since we are dealing with groups equipped with a partial ordering, we will be primarily interested in group homomorphisms that preserve order; i.e., group homomorphisms cf> such that f <9
=> f4><9-
We will call such maps order-preserving homomorphisms. The kernels of such maps are clearly convex normal subgroups. A one-to-one order-preserving homomorphism will be called an order embedding. If both a map and its inverse are order-embeddings between (right) partially ordered groups G and H then we will say that G and H are isomorphic as (right) partially ordered groups. E x a m p l e 3.2.1 Let G be the additive group of integers equipped with the trivial ordering (Example 1.3.3) and H be the same group under the usual ordering. Then G and H are isomorphic groups under the identity map. The identity map from G to H is order-preserving but that from
3.2.
35
HOMOMORPHISMS
H to G is not. Hence there is an order embedding of G in H, but the two partially ordered groups are not isomorphic as partially ordered groups. Nor should they be! In the special case that we are dealing with lattice-ordered groups, there are other algebraic operations present, namely V and A. However, by Lemma 2.3.2, f Kg — ( / _ 1 V<7 -1 ) -1 ; hence any group homomorphism that preserves V also preserves A (and vice versa). Such a homomorphism will be called an (.-homomorphism. We have the trivial fact: Lemma 3.2.2 Let G and H be lattice-ordered groups andis an (-homomorphism; (ii) (g V l) = g4> V 1 for all g € G; (Hi) if f,g € G and f A g — I, then f<j> A g4> = 1. Proof: We have already noted the equivalence of (i) and (ii). Clearly (i) implies (iii). Conversely, since / ( / A )"' A g(f A g)^1 = 1, we have fd>((f
A g)<j>)~1 A g{{f A g)^-1
= 1. H e n c e f> A gcf> = ( / A g). B y
Lemma 2.3.2, <\> is an ^-homomorphism./ The kernels of ^-homomorphisms are precisely the normal convex £subgroups. These will be called (.-ideals. By Universal Algebra, all the Isomorphism Theorems hold for lattice-ordered groups and ^-homomorphisms (see [Burris & Sankappanovar, pp 42-50] or [Cohn, pp 57-62]). By Lemma 2.3.10, if K and L are £-ideals of a lattice-ordered group G, then so is KL: convexity is immediate from the lemma cited, and if a e K and b e L, then 1 < 1 V ah < (1 V a)(l V b) e KL. It is clearly the least ^-ideal of G that contains both K and L. Since K n L is also an ^-ideal of G, we obtain (in analogy with Theorem 3. A) Lemma 3.2.3 The set of (-ideals of a lattice-ordered group forms a dis tributive lattice. Corollary 3.2.4 IfG is a cardinal sum of (-ideals G\,..., Gn and C is an (-ideal of G, then C is a cardinal sum o/CflGi,..., CoGn. Consequently, any two direct cardinal decompositions of a lattice-ordered group have a common refinement.
36
CHAPTER 3. VALUES, PRIMES AND
POLARS
A group G is said to be subdirectly irreducible or monolithic if G can never be isomorphic to a subdirect product of groups {G; : i € 1} unless G is isomorphic to some G;, i £ I. Thus G ^ {1} is subdirectly irreducible if and only if f]{N : N < G and N ± {1}} ^ {1}. If G is subdirectly irreducible this intersection is called the monolith of G and is the unique minimal (non-trivial) normal subgroup of G. Every group G is a subdirect product of subdirectly irreducible groups: For each g € G\{1}, let Ng be a normal subgroup of G maximal with respect to not containing g (existence by Zorn's Lemma). Then f\{Ng : g € G\{1}} = {1} and each G/Ng is subdirectly irreducible (its monolith is the normal subgroup generated by gNg). Analagous considerations apply to lattice-ordered groups where "nor mal subgroup" is replaced by "£-ideal" That is: Lemma 3.2.5 Every lattice-ordered group is a subdirect product of sub directly irreducible lattice-ordered groups. In analogy with the group-theoretic case, we will call the unique nontrivial minimal £-ideal of a subdirectly irreducible lattice-ordered group the t-monolith.
3.3
Prime subgroups
In analogy with rings, we define a convex ^-subgroup C of a latticeordered group G to be prime if for all convex ^-subgroups A and B of G, HAnBCC then ACC or BCC. We will denote by 11(G) the set of all prime subgroups of a latticeordered group G. Note that, by Theorem 3.A and the definition: Lemma 3.3.1 IfG is a lattice-ordered group, then every value is a prime subgroup ofG. Corollary 3.3.2 Every convex ^-subgroup of a lattice-ordered group G is an intersection of values of G; hence it is an intersection of prime subgroups of G.
3.3. PRIME
37
SUBGROUPS
Proof: If C £ C(G), let 5 £ G\C. There is a value Vg of 5 containing C. Hence C = n{V 9 : 9 £ G \ C } . // We give some alternative statements that are equivalent to primeness: L e m m a 3.3.3 Let G be a lattice-ordered group and C be a convex £subgroup of G. Then the following are equivalent: (i) C is a prime subgroup of G; (ii) if f A g = 1, then f £ C or g £ C; (Hi) if f A g £ C, then f E C or g £ C; (iv) if / , g £ G, then there is c £ C such that f < eg or g < cf; (v) if Di and D2 &re convex ^.-subgroups of G containing C, then D1 C D2 or D2 C Dx. Proof: (i) => (ii). By Corollary 3.1.6, if fAg = 1, then G(f)nG(g) = G(f A g) = {1} C C. Since C is prime, G(f) C C or G{g) C C. Thus / e C or g £ C. (ii) => (iii). If / A g £ C, then as / ( / A g)-1 A (/ A g)~l = 1, we have, without loss of generality, / ( / A g)~l £ C. Therefore, f £ C. (iii)=> (iv). S i n c e / ( / A 5 ) - 1 A g ( / A 9 ) - 1 = 1 for all / , g £ G, without loss of generality, c = / ( / A g)'1 £ C. Hence / = c(f A g) < eg. (iv) => (v). Suppose that dx £ D?\D2 and d2 £ D$\DX. Without loss of generality, for some c £ C we have d\ < cd2. Thus d± £ D2 by convexity, a contradiction. (v) =* (i). If A,B € C(G) a n d A n B C C , then C = CV{AnB) = (CvA)D(CV B) by Theorem 3.A. By (v), C V A C C V B without loss of generality. Hence C = C V A, so A C C.// By (ii), if V is a value of g £ G\{1}, then V is a value of g+ or g~; thus values are always values of strictly positive elements. We will repeatedly use this important but trivial fact. We will write / _L g for / A g = 1. Corollary 3.3.4 If G is a lattice-ordered group, f,g£ g, then f and g have incomparable values.
G + \{1} and f _L
38
CHAPTER 3. VALUES, PRIMES AND
POLARS
A partially ordered set X is said to be a root system if for all x,y, z £ X, we have (x < y, z) => (y < z or z < y). By the lemma: Corollary 3.3.5 If G is a lattice-ordered group, then Tl(G) is a root system, and therefore so is T(G). Corollary 3.3.6 If G is a lattice-ordered group, then C(G) is a chain if and only if G is an ordered group. Let G be a lattice-ordered group and C e C(G). We can define Cf < Cg if and only if there is c £ C such that f < eg. It is easily verified that this is indeed a well-defined partial order on 11(C), the set of right cosets of C (c.f., Example 1.3.21). With respect to this partial order, 1Z(C) is a lattice. By Lemma 3.3.3, Corollary 3.3.7 Let G be a lattice-ordered group and C £ C(G). Then C is prime if and only if 1Z(C) is a totally ordered set with respect to the above order. We now wish to extend the definition of lexicographic extension from that given in Example 1.3.25. Let G be a lattice-ordered group and C 6 C(G). We say that G is a lexicographic extension of C (or G is a lex extension of C for short) if g £ G\{1} and / J_ g implies that / £ C. Note that such a C must be prime by Lemma 3.3.3. We now connect this concept with the original term given in Chapter 1. Lemma 3.3.8 Let G be a lattice-ordered group and C £ C(G). If G is a lex extension of C then C < d for every d £ G+\C; i.e., c < d for every ceC andde G+\C. Proof: Let c £ C+ and d £ G+ with d £ C. Then c ^ d implies that c / c A d . But d(c A d)~x 1 c(c A d)'1 Therefore d(c A d)~l G C. Hence d e C (since K c A d < c e C ) , a contradiction. // One importance of values can be seen from the following lemma: Lemma 3.3.9 If G is a lattice-ordered group, then f < g if and only if Vf £ ^9 for tt^ values V in G.
3.4. SPECIAL
VALUES
39
Proof: If / ^ g then fg ' V I / 1 and so has a value, say V Now 1 < fg"1 V l ^ V ; hence V < V{fg~l V 1) = V(f V g)g~l Thus V# < ' ' ' ( / v ff) = max{V/, Vg} since V is prime. Consequently, Vf ^ Vg. The converse is obvious./
3.4
Special values
Another important aspect of values is given by the next lemma. Note that, in any lattice-ordered group G, the set of values of an element g € G is the union of the sets of values of g+ and g~ (since g = g+{g~)~l, values are prime, and g+ J_ g~). Hence if g is special (i.e., has a unique value) then g~ = 1 or g+ = 1; so g > 1 or g < 1. In any lattice-ordered group, an element clearly has the same values as its inverse. We will therefore assume throughout that all special elements of a lattice-ordered group are strictly positive; and if g is special, then Vg will denote its unique value. As noted in the first section of this chapter, the converse (all strictly positive elements are special) holds in all ordered groups. L e m m a 3.4.1 Let G be a lattice-ordered group and f,g,h be special el ements of G. (1) Ifflg, then Vf C Vg or Vg C Vf. (2) If f Lg but f l h and g jLh, then Vf c Vh and Vg c Vh. (3) IfVfcVg, thenf^ig. (4) If Vf — V„, then f < gn for some positive integer n. Proof: (1) Since / A g i= 1, let M be a value of / A g. Then / , g £ M so VJt Vg D M. By Corollary 3.3.5, V} C Vg or Vg C Vf. (2) Since / A g = 1 and / 0 Vf it follows that g € Vj\Vg\ similarly / e Vg\Vf. By Corollary 3.3.5 and (1), we deduce that Vf C Vh and Vg C Vh since Vf and Vg are incomparable. (3) If / it. g, then for some positive integer n and value V we have Vfn > Vg > V by Lemma 3.3.9. Clearly, / " 0 V, so / £ V Thus V C Vf C Vg. Hence V, < Vgg < Vgfn = Vg (since / € Vf* C Vg), a contradiction. (4) If / ^
40
CHAPTER 3. VALUES, PRIMES AND POLARS
The unique value of a special element is called a special value. Special values are especially nice! L e m m a 3.4.2 Let V be a special value in a lattice-ordered group G. Then VV without loss of generality. Since V is prime, so is g~1Vg and both are maximal convex ^-subgroups of V* Hence fn € g~lVg for all positive integers n (otherwise V, the value of / " , would contain g~1Vg). Thus gfn G Vg and so V < Vfn < Vgfn = Vg for all positive integers n. But the convex ^-subgroup of G generated by V and / is V* (it properly contains V and is contained in V*). Therefore there are h\,...,hm+i € V+ such m that g < c = h\f ■ ■ ■ hmfhm+\. Now c = hf for some h € K4" since m Vf = fV Consequently, Vf < Vg < Vhfm = Vfm, the desired contradiction. / We close this section by tying special values to a natural class of lattice-ordered groups. A lattice-ordered group G is said to be finite valued if every element of G has finitely many values. So any ordered group is finite valued as is any cardinal sum of such groups. It is not hard to see that Z wr Z is also finite valued. However, neither the cardinally ordered full Cartesian product of an infinite family of ordered groups nor Z Wr Z is finite valued. L e m m a 3.4.3 A lattice-ordered group G is finite valued if and only if every value of G is special. Proof: Suppose that G is finite valued and let g € G+. If g has a unique value then it is clearly special, so assume that g has values Vi,...14+i- Since these are values in G(g) which are special in G(g) if and only if they are special in G, we may assume that G = G(g). Thus V\,..., Vn+i are the maximal convex ^-subgroups of G and every proper convex f-subgroup of G = G(g) is contained in at least one of these
3.5.
POLARS
41
(by Zorn's Lemma). Let ft G V+ +1 \V and ht £ V+\Vn+l (1 < i < n). Let / - Vr=i fi and h = ALi K Then / G V„ +1 \(U" = i K) and h G (fir=i K ) \ K + i - By replacing / by / ( / A h)'1 and h by h ( / A h)~\ we may further suppose that f ± h. Now Vi,..., Vn are values of / and Vn+l is a value of h. Further, V n+ i is the unique value of h and so is special: if M ^ Vn+i is a value of h, then as M C Vi for some i — 1, ...,n, we have f £ M C Vi (since f ± h), a. contradiction. Moreover, if M is a value of / other than V 1; ..., Vn, then it contains h and is contained in Ki+i- This is clearly false. Therefore, by induction on n, Vi,..., V„ (the values of / ) must all be special. Hence all the values of g are special, as required. Conversely, suppose that every value of G is special. If g G G+ has set of values {V; : i G / } , choose ft G G+ with unique value Vj (i G I). By Lemma 3.4.1 (1), /, J_ / , if i,j € I with i / j . Let C be the convex ^-subgroup of G generated by { / * : * € / } ; then C is £-isomorphic to the cardinal sum of {(/*) : i G / } . Since /j G C\Vi for all i e I, g must belong to C (otherwise there would be some i0 G / such that Vj0 2 C). Consequently, / is finite and each Ci is special. / Note that the proof yields (by induction on n) that every positive ele ment of a finite-valued lattice-ordered group can be written as a product of pairwise orthogonal special elements.
3.5
Polars
If G is a lattice-ordered group and g G G, let g1- — {/ G G : |/|A|g| = 1}, the polar of #. By Lemmata 2.3.4 and 2.3.8(vi),(vii), it follows that gL G C{G). EX CG let XL = f]{xL : x G X), the po/or of X; so A' 1 G C(G). Moreover, since G{g)L = gL, it follows that XL = CL where C is the convex ^-subgroup of G generated by X. We will also write f±g if / G gL and say that / and g are orthogonal or perpendicular or disjoint. We will denote by P(G) the set of all polars in G. We call a subgroup C of a partially ordered group G a cardinal summand if there is a subgroup D of G such that G is the cardinal sum of C and Z?. Note that any cardinal summand of a lattice-ordered group G is a polar; indeed, if G is the cardinal sum of its (necessarily convex
42
CHAPTER 3. VALUES, PRIMES AND
POLARS
£-)subgroups C and D, then D = C 1 and C = D"1 A lattice-ordered group G is called projectable if each g 1 (5 G G) is a cardinal summand; and strongly projectable if every polar X 1 is a cardinal summand. Even very nice Abelian lattice-ordered groups can fail to be projectable (please see Example 3.6.2 below). It is easy to see (c.f., Corollary 3.2.4) Lemma 3.5.1 If G is a lattice-ordered group and G = ^Al i£l
=
^Bj, j£J
then G=
£
(AiHBj).
By Zorn's Lemma, each prime subgroup contains a minimal prime subgroup. For each g £ G\{1}, let Pg be a minimal prime subgroup of G contained in some value of g. Since g £ Pg, we have f\{Pg : g G G\{1}} = {1}. Hence the intersection of all minimal prime subgroups of G is {!}. Minimal prime subgroups are closely related to polars. Lemma 3.5.2 Let G be a lattice-ordered group and P E C(G). The following are equivalent: (i) P is a minimal prime subgroup of G; (u) P = {j{9L ■ 9 t P}; (in) P e 11(G) and hL g P whenever h£ P. Proof: (i) => (ii). If / , g £ G+\P, then / A j ^ l since P is prime. Hence there is an ultrafilter F on G+ containing G+\P Let Q = [){gL : g € F}. Observe that if g g F, then gx C ( G + \ F ) C P; thus Q C P We prove that Q 6 C(G): If a,6 e Q, then a £ fL and 6 G g1 for some f,g e F Now 1 < (|a6|v|aV6|)A/A5 < (|o6|V|a||6|)AfAg < |a||&||a|A/Ag by Lemma 2.3.8(vi), (vii). Moreover, \a\\b\\a\ A f Ag < (\a\ Af)(\b\ Ag)(\a\ A / ) = 1. Hence ab,aWb £ Q. Since Q is clearly convex and closed under inverses, it follows that Q eC(G). If c 0 Q and c A d — 1, then as c A g > 1 for all 5 G P , we have that c G P since P is an ultrafilter. Thus d G c x C Q proving that Q
3.5.
POLARS
is prime. By the minimality of P, we have Q = P and F = G+\P c g P, then c e d 1 for some d G F ; hence c 0 F.)
43
(If
(ii) => (iii). Clear. (iii) => (i). Let Q 6 11(G) be minimal contained in P If Q =£ P , then there is g G P \ Q . By (iii), there is ft G S ^ V , whence h & Q. But g $. Q, so ft G gL C Q, a contradiction. / Corollary 3.5.3 If G is a lattice-ordered group and P G C(G)\{{1}}, £ften P 1 is a minimal prime subgroup of G if and only if P is totally ordered. Proof: If P x were a minimal prime subgroup of G, let f,g € P with /.Lfl and / ^ 1. Since P n P x = {1}, we have g $ P1 ii g ^ 1. By Lemma 3.5.2, / G gL C F- 1 ; so 5 = 1. Thus F is totally ordered. Conversely, if P is totally ordered, let fLg. If / , 5 £ F 1 , there are cuc2 G F with / A ci,g A c2 > 1. But (/ A Ci) A (5 A c2) = 1. Since F is an ordered group, / A C\ = 1 or g A c% — 1. This contradiction shows that F- 1 G n(G). If ft G F 1 , then F C ft1; thus ft1 £ PL By Lemma 3.5.2, F x is a minimal prime subgroup of G.jj It is straightforward to verify that: Theorem 3.B For any lattice-ordered group G, V{G) is a complete Boolean algebra under inclusion, where A{F : ! 6 / } = f1{F : i G / } and V { F : % G / } = ( n j F 1 : t G / } ) x = (U{P : i G 7 } ) 1 1 Finally, we give an easy application of polars. Lemma 3.5.4 Any subdirectly irreducible Abelian lattice-ordered group is ordered. Proof: Let G be an Abelian lattice-ordered group and g be a strictly positive element of G. Then g1 D g 1 1 = {1} and g G g 1 - 1 Hence gL = {1} if G is subdirectly irreducible. Let f,heG. Then / ( / A f t ) - 1 A 1 1 1 ft(/ A ft)" = 1. Thus / ( / A ft)" = 1 or h(f A ft)" = 1 since gL = {1} if g > 1. Therefore / < ft or ft < / ; i.e., G is totally ordered. // If G is a lattice-ordered group, then the complete sublattice of C(G) generated by V{G) has been studied in [Zheng Xiqiang k Qi Zhinan]; it is a surprisingly rich sublattice and provides almost the same information about G as C(G) does.
CHAPTER 3. VALUES, PRIMES AND POLARS
44
3.6
Closed subgroups
Another class of convex ^-subgroups plays a role in later work. Let G be a lattice-ordered group and C be a subgroup of G. We say that C is closed if for any subset {ci : i £ 1} of C, \/l€l ct £ C whenever it exists in G. Note that, by Lemma 2.3.6, if { c , : i e / } is a subset of C and f\i€l ct exists in G, then this infimum belongs to C if C is closed. So all closed subgroups of G are ^-subgroups of G. Clearly the intersection of any family of closed (convex) subgroups of G is a closed (convex) subgroup of G. Since G itself is closed, given any subgroup C of G, there is a unique smallest closed (convex) ^-subgroup of G containing C. This is called the closure of C. We can construct the closure of a (covex) ^-subgroup by transfinite induction: Let C be a (convex) £-subgroup of a lattice-ordered group L. Let C* be the ^-subgroup of L generated by all V{c; ■ i £ 1} for which {ci : i € 7} C C+ has a supremum in L. By Lemmata 2.3.10 and 2.3.6, C* is convex if C is; and if a set of elements of C has a supremum in L, that supremum must belong to C* We set C(0) = C, and for a any ordinal with a < \L\, define „ , . _ J IJ{C(/?) : /? < a}
C(Qfj
[
C{P)*
if a is a limit ordinal
if a = fi + l.
Clearly C(|L|) is the smallest closed (convex) ^-subgroup of L that con tains C. Again, by Lemma 2.3.6: Lemma 3.6.1 Let G be a lattice-ordered group. (i) gL is a closed convex (.-subgroup of G for all g € G. (ii) Every summand of G is a closed subgroup of G. (Hi) If C is a convex subgroup of G, then the closure of C is a convex (.-subgroup of G. (iv) If C is a subgroup of G and C <\ G, then the closure of C is a normal (-subgroup ofG. (v) If C,D G C(G), C C D and C is a closed prime subgroup of G, then so is D.
3.6. CLOSED
SUBGROUPS
45
(vi) All special values are closed. (vii) Polars that are prime are maximal, and maximal polars are prime subgroups. (viii) If K is a closed (.-ideal of G and A G C(G) with A D K, then A is closed in G if A/K is closed in G/K. We need only prove (v), (vi) and (vii): Proof: (v) Let d = Vie/ ^» w * t n e a c n ^ e D+ ^ e a c n ^ e C> t n e n d G C C D; so assume that there is dk G D+\C. Since Cdkld ^ Cd, we must have Cd^ld < Cd. Hence Cd^ld < Cdj for some j £ I since l TZQ(C) is a totally ordered set. Thus 1 < dk d < cdj for some c € C. Therefore d £ D since c,dj,dk £ D and D is convex. (vi) Suppose that V is special and that g = \li^igi with all <7i 6 Vr+ If c? ^ V, then V / {1} and we may assume that each gt > 1. Let / G G be special with unique value V Thus f A g £ V since V is prime, and f Ag= Vi S /(/Aft) with each / A & € V + Also / A 5 G V* and V is the unique value of / A g: if M is a value of / A g, then / 0 M, so M CV. We may therefore assume that g itself is special with unique value V. Hence g~lg is special for all j € I with unique value V {gjlg 0 V and 1 < 9jl9 < 3> s o if M is a value of gjxg, then g £ M; hence M C V). Since # = Vig/5i, there is /c G I such that gk <£ g7lg. By Lemma 3.3.9, there is a value M such that Mgk > Mg~lg. Now g £ M (otherwise gk,9j G M and so Mgk — M = Mg~lg). Therefore M C V since is special with unique value V Thus V — Vgk > Vg~lg = Vg > V, the desired contradiction. (vii) Let C1 be prime and suppose that CL C D1- Clearly, CL n DL is the polar of the convex ^-subgroup of G generated by C and D. We may therefore assume that D C C. Then D C CL since D n Z) 1 = {1} and C1- is prime. Thus D C C n C x = {1}; so D 1 = { l } 1 = G*. Conversely, if C1" is a maximal polar. Then as C 1 C c x C G for all c G C + , it follows that c x = CL for all 1 < c G C (c £ cL). Let f,g £G+ with / -L 5. H g & CL, then 5 A c > 1 for some c G C + Since g Ac £ C and 5 A c !_ / , we get j G (5 A c)L — C 1 Hence CL is prime. / The converse of (viii) is false. E x a m p l e 3.6.2 Conrad (see, e.g., [Darnel 1995, page 95]) Let C — Y[n£Z+Z and D = Za © Zp where Za and Zp are copies of Z. Let
46
CHAPTER 3. VALUES, PRIMES AND POLARS
L — C ® D with the cardinal ordering, an ^-group, and G be the lsubgroup of all g G L such that g restricted to Z + is eventually constant and g(/3) = Zimn_>00c;(n). Let K — {g 6 G : (Vn G Z+)g(n) = 0} and 5 = (E„ez+Z)©{0}. Then K is closed since K = S1 It A = ( £ n g z + Z ) 0 i f , then A/K is not closed in G/K since A/K = £ n S Z + z a n d G/K is £isomorphic to the eventually constant sequences. However, A is closed in G since if {s; : i G 7} C £ n g z + Z n a s a supremum £ > 0 in rin€z+ z such that ^m n _ >0O t(n) exists and exceeds 0, then {(sj;rj,0) : i G 7} G .4 fails to have a supremum in G no matter what the values of rt G Z may be (as (i;r, Zimn-joo^n)) are upper bounds of {(s;;ri,0) : i G 7} for all r G Z).
Note that if g = (0; 1,0), then h & gL ® gix where h = (1;0,1). Hence G is neither projectable nor strongly projectable. There is another source of closed values that results from generalising the definition of special values: a value V in a lattice-ordered group G is said to be essential if it contains all the values of some g G G\{1}. L e m m a 3.6.3 Every essential value of a lattice-ordered group is closed. Proof: Let G be an £-group, g G G + \{1} and V a value in G containing every value of g. By Lemma 3.6.1 (vi), we may assume that V is not special (whence g € V), Suppose that {/j : i G 1} C V+ with / = Mia fi & v- L e t Mi b e a n v v a l u e °f 9~lffi~l If 5 G Mu then Mi ^ M^fff1 = Mlff~1 > M ; . If g $ Mu then there is a value of g containing Mf, hence M» C V by hypothesis. Therefore, Mig-1ffr1 > Mx since Vg~lff^ = Vff~l > Vf~l = V By Lemma -1 3.3.9, o / > / l . Since this holds for all i G I, we get a contradiction to / = V i 6 / / i . // In the next chapter (Lemma 4.2.9) we will prove that every closed value is essential under appropriate hypotheses. For any lattice-ordered group G, let D(G) be the intersection of all closed prime subgroups of G\ so D(G) is a closed convex ^-subgroup of G. Since each closed prime subgroup is an intersection of the values that contain it (all of which are closed by Lemma 3.6.1(v)), D(G) is the intersection of all closed values of G. Hence:
3.6. CLOSED
SUBGROUPS
47
Corollary 3.6.4 (i) D{C) = CnD(G) for all C G C(G); so D{D(G)) = D(G). (ii) Every non-identity element of a lattice-ordered group G has a closed value if and only if D(G) = {1}. We consider a generalisation of the distributive property to include possibly infinite suprema and infima. A lattice-ordered group G is said to be completely distributive if and only if A l€ / V j e j 9ij = Ma.i-^j A ie / 9i,a(') for all {g%j ■ i G / , j G J } C G for which the indicated suprema and infima exist in G. Note that any ordered group is completely distributive. We now tie this concept to D(G). Lemma 3.6.5 [Byrd & Lloyd] A lattice-ordered group G is completely distributive if and only if D(G) = {1}. Proof: Suppose that G is completely distributive and that D{G) ^ {1}. Let g G D(G)+ with g ^ 1 and {P{ : i G 1} be the set of all prime subgroups of G. So g G cl(Pi) for all * G I where we write cl(P) for the closure of P Then g = V, e j 9i,j for some {gtj : j £ J } C P+ (where we have used repetitions if necessary to get the same index set J for all i £ I). Hence g = A i e / V ; e j ^ j = Ma-.i^j Aie/5i,./ A,e/ft,a(i) = \frj-+j 1 = L the desired contradiction. Suppose that D(G) = {1} but G is not completely distributive. Then there is {gitJ : i G i", j G J} C G such that 3 = A l6 / V j e j ft,; > \Jaj^jl\iei9iAi) = /■ B y Lemma 2.3.6, gf~l = AieI\fj6j 9i,jf~l > Vai^j f\iei9i, gtj > 1 for all i € I, j € J and that g = A; e /Vjej 5*,j > V J : h j A , e ; f t , 4 ) = 1- So 1 < 5 = yjeJglij for all i £ / and AjeJ 9z,a(i) = 1 for all cr : / -» J. If £>(G) = {1}, then g has a closed value M by Lemma 3.6.1(v). Since M is closed we get Mg = M{\f3eJgttJ) = \fjeJMgl:] for all ? G /. If there is h G G + such that M < Mh < Mg, then % , T ( t ) > M/i for some r{i) G J; and if no such h exists, then Mgi
48
CHAPTER 3. VALUES, PRIMES AND POLARS
that Mgitj = M for all j G J; so g = \Jj€J gitj e M since M is closed — a contradiction). By replacing h by 5 in the latter case, we obtain the existence of h £ G+ and r(i) € ./ such that Mgi:T^) > Mh > M for all i G / . Hence M(Ai € J 9i,r(i)) = AieJ M9i,r(i) > Mh > M. Therefore 1 = Aiej 9',r{i) & M, the desired contradiction. / Although it may seem more natural to call D(G) the closed radical of G, in view of this lemma D(G) is actually called the distributive radical
of a Since special values are closed (Lemma 3.6.1(vi)), we get Corollary 3.6.6 If every non-identity element of a lattice-ordered group G has a special value, then G is completely distributive. We close this section with a strange result concerning complete distributivity. Lemma 3.6.7 [R.N. Ball 1984] Let fa and fa be the sentences given in Lemma 2.3.7, and d> be the sentence Vx\/y(x < y -4 3z13z23ti3t2[kk=h2(zk
xAtk < zk k yVzk > tk) &
& {Vw)(zi V w > ti or t2 A w < z2)\). In any distributive lattice, D is completely distributive if and only if D \=k fa k fa. Proof: Assume first that D \= 4> k fa k fa. If D is not completely distributive, let b = A, Vj dij > VCT A, dhcr^ = a. Since D (= fa & fa, we may assume that Vj <2y = b for all i G I and Ai diiCr^ = a for all functions a : / —> J. Let c < d with a A d < c and b V c > d. If for all z G / there is j € J such that rf*j V c > d, then let r(z) be such a j e J. Hence d < Ai{diiT(i) V c) = (A; ^i,T(i)) V c = a V c by Lemma 2.3.7; so (o A d) V c = (a V c) A (d V c) > d > c, contradicting a Ad < c. Thus for some to € J, ^10 j V c ^ d for all j (E J. Consequently, if c^ < d^ with a Adk < ck and 6 V e& > dj. (fc = 1, 2), then cx A dioj ^ dx for all j € J. If d2 A di0tj < c2 for all j & J, then we would have d2 A b < c2; so c?2 = d2 A (b V c2) = d2 A c2 < c2, a contradiction. Thus D |= -i0. Conversely, if D \= -'(fa k fa k <^>), then by Lemma 2.3.7 we may assume that D \= fa k fa k -^fa Let a < b witness the failure of 4>
3.7. A LIMITING
49
EXAMPLE
and X = {(c, d) : c < d & a Ad < c k, b\/ c > d). Let / index X; so -^ = {(ci) di) '■ * ^ 0 " For e a c n ' ^ ^i l e t {^ij : i e J} D e a n enumeration of X{ — {e : a < e < 6 & eVc, ^ d;} (with possible repetitions). To prove that D is not completely distributive, it suffices to shew that Vj dl)3 = b for all i e I and At dit<7^ — a for all a : I -> J. If Vj ^ij 7^ 6 for some i 6 7, let r 6 L> be such that d y < r < 6 for all j e J. Now (r, 6) € X and if s G D satisfies s V c* ^ d* and 6As ^ r (D |= -i0), then an easy computation gives (sAb) Vo G X t ; say, (s A 6) V a = diyj0. Then s Ab < (s A 6) V a = dy,, < r, a contradiction. Hence Vj d y = b for all i G 7. Finally, if A* djjff(j) ^ a for some a : I -> J, let r G D be such that a < r < dj,CT(j) for all i 6 i . Now r\Jc{< dl
3.7
A limiting example
In this section we give an important example that shows that very dif ferent lattice-ordered groups can have the same lattices of convex £subgroups and the same partially ordered set of values. Example 3.7.1 [Kenoyer] Recall Examples 1.3.9 and 1.3.19. Let C = {g G C(K) : (3o0 < ... < an)(supp{g) C \a0,an] & g\[aj,aj+i] is linear {j = 0, . . . , n - 1))}. We turn the picture through TT/4 to obtain G = {g G .4(E) : (3a 0 < ... < an)(supp(g) C [a0,an] k g\[aj,aj+i] is linear (j — 0, ...,n — 1))}. Thus C is an Abelian lattice-ordered that C(C) =C{G), etc. We first show that the maximal Ca = {g G C : g{a) = 0} (a G R). the closure of the support of g in R,
group but G is not. We will show convex ^-subgroups of C are just If g G C, then A = cl(supp(g)), is compact. Hence {g(x) : x G R}
50
CHAPTER 3. VALUES, PRIMES AND
POLARS
is bounded for each g £ C. If L E C(C) and L %Ca for all a G R, let g G C+ and m = sup{g(x) : x G R}. By hypothesis, for each a G A there is fa G L+\Ca. So fa(a) > 0 and as fa is continuous, Ua — {% G R : /o(^) > fa{a)/2} is an open neighbourhood of a. Thus {£/0 : a G A} is an open cover of A. Hence there is a finite subcover {Uaj : j = 1,..., s} by the compactness of A. Let n = min{faj (o,-)/2 : j = 1,..., s} and p G Z + be such that pn > m. Then / = Vj=i faj 6 L + and 0 < g < pf- So g £ L and therefore L — C. Consequently, the maximal convex ^-subgroups of C are C a (a G R). Similarly, the maximal convex ^-subgroups of G are Ga = {g €G :ag = a} (ae R). Let C a , r = { 9 e C : ( 3 e > 0)(V6 G [a, a + e])5(6) = 0} Ca,i = {geC:(3e>
0)(V6 G [a - 6, a])5(6) = 0}.
Then Ca:r,Ca,i Q Ca. Again, using the compactness of the closures of supports of elements of C, it is easy to prove that any prime subgroup of C properly contained in Ca contains either Ca>r or Ctt,j • For if P C Ca is a prime subgroup of C contained in neither Caj nor C0><, let h\ G P + \ C a > r and /12 G P+\Caj. Let g e C£ By hypothesis, there are e > 0 and mj > 0 (j = 1,2) such that /ii(x) = m i ( i - a) if x G [a, a + e] and /i2(a:) = —mi(x -a) if x G [a - e, a]. By decreasing e if necessary, we may further assume that g is linear on each of these intervals with slopes rii and —n2 respectively (where ni,n2 > 0). Let n G Z + be such that 771,-n > rij (j = 1,2). Then 0 < g(x) < n(hi V h2)(x) for all a; G [a - e,a + e}. Let A = cl(supp(g))\(a — e,a + e), a compact set not containing a. Since P is prime and P %Gb for all 6 G A, we can use the compactness argument of the previous paragraph to obtain / G P+ with f(x) > g{x) for all i S A . Then 0 < 5 < / V n(/ii V h2), so 5 G P Hence P = Ca establishing the claim. The corresponding result holds for the analogues in G by the same method. Note that GaiT <\ Ga and Gaii < Ga. We complete the proof by showing that every proper prime subgroup of C is either Ca for some a G R, or C a r or „,/ for some a G R. For if P is a proper prime subgroup of C, then P C Ca for some a G R. If P^Ca, then P C Go>r or P C Ca_(. Without loss of generality, P C C 0ir .
3.8. RESIDUALLY
ORDERED
GROUPS
51
Let g e C+r\P and e > 0 with g(x) = 0 for all x <= [a, a + e]. Let
/(*) =
if x G [a, a + e/2] x—a a + t — x if x 6 [a + e/2, a + e] 0 if x 0 [a, a + ej.
Then / e C\Ca,T and f A g = 0. Since P is prime and g 0 P , we get / G P C C ajr , a contradiction. Thus P = Ca,r- The same method establishes the corresponding result for G. Consequently, C(C) and C(G) are isomorphic lattices by Corollary 3.3.2. Also every prime subgroup of C is a value (and similarly for G). Since C is Abelian, these values are all normal in their covers (indeed in C itself). However, the values Ga are not normal in their covers (G) since f~lGaf = Gaf / Ga if af i= a — but every non-identity g € G has a value (Gaj) that is normal in its cover (Ga), where a — inf(supp(g)). Actually, the only closed prime subgroup of C is C (Lemma 4.4.3) but each Ga is closed ([McCleary 1969] or [Glass 1981, Theorems 8D and 8C]); so the lattices of closed prime subgroups of C and G are not isomorphic.
3.8
Residually ordered groups
In Section 2.4, we defined the notion of a residual property for groups. We extend the definition to the algebra of lattice-ordered groups. A lattice-ordered group is said to be residually ordered if it is a subdirect product of ordered groups. The Americans and French call this property "representable", whereas the Russians refer to it by the more evocative name "o-approximable" Though less standard, the name used here seems more descriptive and in keeping with its group-theoretic roots (and universal algebraic generali sation). In many — but not all — ways, residually ordered groups play a role similar to that played by residually finite groups in group theory. We now characterise residually ordered lattice-ordered groups. T h e o r e m 3.C [Lorenzen] Let G be a lattice-ordered group. Then the following conditions are equivalent: (i) G is residually ordered;
52
CHAPTER 3. VALUES, PRIMES AND (ii) g+ A f-'g-f
= 1 for all
POLARS
f,g£G.
L
(iii) g(ii). Let G be residually ordered; say G is an ^-subgroup °f ILe/G;, cardinally ordered, where each G{ is an ordered group. It is enough to show that (<77Tj)+ A {f'^i)~1{g^i)~{f^i) — 1 for each i £ I\ that is, we can reduce to the case that G itself is ordered. But in any ordered group, a+ — 1 or a~ — 1, so the conclusion is immediate. (ii)=> (iii). Let g £ G. Then \g\ = g+ V g~ Since g± = |#| x = (g ) n {g~)L and g~ = (g~l)+, it is enough to show that (g+)± < G for all g € G. If a £ G+ n (9+)-1-, let c = g+a" 1 Then c+ = g+ and c - = a by Lemma 2.3.8. By (ii), f~laf £ {g+)-L for all f £ G. Since + x + L ( 5 ) £ C(G), it follows that (g ) < G, as required. + L
(iii)=> (i). Let V be a value in G, say V is a value of g £ G. Let M be a minimal prime subgroup of G contained in V. Since every minimal prime subgroup of G is a union of polars (by Lemma 3.5.2), M is normal in G. Since the convex ^-subgroups of G that contain M form a chain (Lemma 3.3.3), so does the set of all conjugates of V Let Ng — f]{f~lVf :f £ G}. Then Ng is an £-ideal of G and g £ Ng. Since M C Ng, we have that ./V9 is prime by Lemma 3.3.3. Hence G is a subdirect product of the ordered groups {G/Ng : g £ G+\{1}}. // Corollary 3.8.1 Every Abelian lattice-ordered group is residually or dered. Corollary 3.8.2 A lattice-ordered group G is not residually ordered if and only if there are f,g £ G with g > 1 and f~lgf Ag = 1. Proof: The condition clearly implies that G is not residually ordered by (iii). Conversely, by (ii), there are a,b £ G with g = a+ Ab~1a~b > 1. Then 1 < gAbgb'1 < a+Aa~ = 1 by Lemma 2.3.8(H). Hence bgb~lAg = 1 and g > I.// Corollary 3.8.3 A lattice-ordered group G is residually ordered if and only if f2 A g2 = (/ A g)2 for all f,g£G.
3.9. FINITE PAIRWISE ORTHOGONAL
SETS
53
Proof: Clearly the condition holds in any ordered group, and hence in any residually ordered lattice-ordered group. Conversely, if the condition holds then (fg)2 A f2 = (fg A / ) 2 < fgf. Hence fg A g~lf < / , and so g A f-lg~lf < 1. Thus, g+ A f~lg~f = l l (9 A f~ g~ f) V 1 = 1 by Lemma 2.3.5.// In Section 6.1, we will use the commutator calculus to show that every locally nilpotent lattice-ordered group is residually ordered. We will also study its role in the set of all varieties of lattice-ordered groups in Chapter 10.
3.9
Finite pairwise orthogonal sets
Let G be a lattice-ordered group. If G\{1} contains no two element subset of pairwise orthogonal elements, then by Lemma 3.3.3, {1} is a prime subgroup of G. By Corollary 3.3.6, G is an ordered group. Let C = 1 © Z with the cardinal ordering (Example 1.3.13), and G = C v T z , where (a, 6,0) conjugated by (0,0,1) is (6, a, 0). Then G, a lexicographic extension of C by Z, is a lattice-ordered group (Example 1.3.26). Every pairwise orthogonal subset of G\{1} contains at most two elements, yet G is not residually ordered (since (1,0,0) x is not a normal subgroup of G). However, G = (1,0,0) L © (0,0,1) L < G We now show that this example is, in some sense, the most complicated for a latticeordered group having no three element pairwise orthogonal subset. Suppose that G is a lattice-ordered group such that G\{1} has no three element pairwise orthogonal subset, but that {/, g} C G\{1} is pairwise orthogonal. We may clearly assume that f,g > 1. Since every element of A = fL is orthogonal to / , A must be a totally ordered subgroup. Similarly, so is B — gL. By Lemma 2.3.4, the subgroup generated by fL U g1- is just A © B. Call this cardinal sum G. Note that G G C(G) and f,g € G. We wish to shew that C <\G and G is a lexicographic extension of G. To achieve this we need a lemma: L e m m a 3.9.1 Let G be a lattice-ordered group, 1 < / , g G G with f _L g. If G\{1} contains no three element pairwise orthogonal subset and C,D € C(G) are totally ordered, then CC\D — {1} or {C, D} is a chain.
54
CHAPTER 3. VALUES, PRIMES AND
POLARS
Proof: Assume the hypotheses of the lemma and that {C, D} is not a chain. Then there are c G C+\D and d G D+\C. So c(c A d)" 1 JL d(cAd)'1 and c(cAd)" 1 € C+\D and d(c A d)" 1 e D + \ C . Hence, by convexity, without loss of generality we may assume that c _L d. Let a G C + and 6 € £>+; s o a A i e C n D . Since G is totally ordered, either a < c or c < a. Similarly, d < b or b < d. If a A b > c, then c e D, & contradiction. Thus a A b < c. Similarly, aAbl = {1} or {h~lAh, A} is a chain. In the first case, h~lAh C <7X since g G ^4; so h~lAh C 5 . In the second case, since {^4, / I - 1 J 4 / I } is a chain and h~xAh is totally ordered, {£?, /i - M/i} is not a chain. By the lemma again, Bf\h~lAh = {1}. Hence h~xAh C fL = A. Thus we obtain, in both cases, that h~lAh C C. Similarly, h~lBh C C, and so C < G. Let a;i,X2 € G + \ C . Then {x, A / , a;; A g} is an orthogonal subset of G\{1} for j = 1,2. Hence { x i , ^ } is not an orthogonal set (otherwise G\{1} contains a four element pairwise disjoint subset). By Lemma 3.3.3, the ^-ideal C of G is prime and G/C is an ordered group. Thus: Theorem 3.D [Clifford & Conrad] Let G be a lattice-ordered group and f,g £ G\{1} be such that f J. g. If no three element subset o / G \ { l } is pairwise orthogonal, then G = C ¥ H for some ordered group H and lattice-ordered group C that is the cardinal sum of two ordered groups. The structure of a lattice-ordered group having a three element pairwise orthogonal subset but none such of four elements is now easy to describe (as is its extension to lattice-ordered groups containing no infi nite pairwise orthogonal set of elements that is left to the reader). Theorem 3.E [Conrad, 1960] Let G be such a lattice-ordered group. Then there are totally ordered A, B G C(G) such that the subgroup gen erated by A and B is l-isomorphic to their cardinal sum. There is an ordered group K such that the subgroup H generated by A U B U K is £isomorphic to (A © B)V K. There is a further totally ordered C G C(G) such that the subgroup generated by C and H is (.-isomorphic to their cardinal sum. Finally, there is an ordered group L such that G is (.isomorphic to (H © C)^L = {({A © B)4* K) © G ) t T L .
Chapter 4 Abelian and Normal-valued Lattice-ordered Groups 4.1
Simple Abelian lattice-ordered groups
In any right partially ordered group G, we write a
f,geG+\{l},91tf. Let G be a lattice-ordered group such that C(G) = {G, {1}}. By Corollary 3.3.6, G is an ordered group. Lemma 4.1.1 If G is a lattice-ordered group having no non-trivial con vex (.-subgroups, then G is an Archimedean ordered group. Proof: Let g 6 G+\{1}. Then G(g) = G, by hypothesis. Hence if / € G + , it follows that / < gn < gn+1 for some positive integer n. Therefore, g ^ / . / Lemma 4.1.2 Every Archimedean ordered group is Abelian. Proof: Let G be an Archimedean ordered group. If G' + \{1} has a least element, say o, let g € G+ Then for some positive integer n, a" < g < an+1. Thus 1 < ga~n < a, so g = an Hence G is a cyclic group and so is Abelian. 55
56
CHAPTER 4. ABELIAN & NORMAL VALUED
^-GROUPS
If G + \{1} has no least element, then for every strictly positive element o, there is a strictly positive c with c2 < a: Let 1 < b < a with b2 •£ a. Then 6"1 < b'xab~l < 1, whence 1 < {ah'1)2 < a. Now suppose that / , g G G do not commute. We may clearly assume that / , g G G+ By interchanging / and g if necessary, we may suppose that a = [f,g] > 1. Let c be such that 1 < c < c2 < a, / , g. Since G is Archimedean, there are positive integers m and n such that cm < f < c m + 1 and cn < g < cn+1. Now a = [f,g] < c-mc-ncm+1cn+l = c2, a contradiction. Therefore G is Abelian./ We are now ready to obtain our first far-reaching result: T h e o r e m 4.A [Holder] Every Archimedean ordered group is isomorphic (as an ordered group) to a subgroup of the additive group of real numbers. Proof: By Lemma 4.1.2, any Archimedean ordered group is Abelian. Let c 6 G + \{1} be fixed. For each g G G + , let Q(g) = {m/n : m,n G Z+ n ^ 0, c m < gn}. Note that 0 G Q(g) for all g G G+, Q{g) f Q+, and r / s < m/n G Q(ff) implies that r/s G Q(#) (since cnT < cms < gns, we have cr < gs by Lemma 2.1.2). Define <j> : G+ —> E by: g t-> sup Q(5). We now show thatis a group homomorphism. Let m / n G Q ( / ) and r / s G Q(g)- Then c m < / " and c r < 5 s We may assume that $ — n. Hence cm+r < fngn = (fg)n Thus Q{fg) 2 QU) + Qis)- Similarly, if m/n > Q{f) and r/n > Q{g), then (m + r)/n > Q(fg). Therefore Q{fg) = Q(f) + Q(g), as desired. Clearly, <j> preserves order. Moreover, it is one-to-one. (If g > 1 then c < gn for some positive integer n; so 1/n G <2()\<2(1).) Consequently, G is order isomorphic (via
Corollary 4.1.3 [Conrad 1959] Every Archimedean right ordered group is isomorphic (as an ordered group) to a subgroup of the additive group of real numbers. Proof: We first show that fgf~l > 1 for all f,g > 1. For if / , g > 1 then (by the Archimedean property) / < gn for some positive integer n. Choose n to be the least such. Now fgf~l < 1 implies that fg
4.1. SIMPLE ABELIAN
LATTICE-ORDERED
GROUPS
57
so / < gn 1 contradicting the minimality of n. Therefore fgf ' > 1 for all/,5>l. If f~lgf < 1 for some f,g> 1, then by the first part of the proof, 9~l = !{f~lgf)~lf~l > 1, which is absurd. Thus G+ o G, and so the right order is two sided. The result now follows from Holder's Theorem. // Corollary 4.1.4 If G is a lattice-ordered group and C(G) = {G, {1}}, then G is order isornorphic to a subgroup of the additive group of real numbers. In contrast, there are non-Archimedean right ordered groups that have no proper convex subgroups. For the group H in Example 1.3.22 is certainly not Archimedean. (If b G Q+ with b > l/£ and B is the upper triangular matrix with l's on the diagonal and b\2 = b then B > D for every diagonal matrix D £ H.) An easy computation shows that H has no convex subgroups: for example, if A € H+ is a non-identity diagonal matrix and c G Q + with (1 - ^ - ) > c£ > 0, then A > C > I where C € H, Cn = 1 and c12 = c. If the positive integer n is chosen so that nc > b, then Cn > B. A somewhat similar result to Corollary 4.1.4 holds even for partially ordered groups whose set of strictly positive elements is semi-isolated; i.e., if gn > 1 for some positive integer n, then g > 1. Corollary 4.1.5 Suppose that G is a partially ordered group whose set of strictly positive elements is semi-isolated. If every convex subgroup of G is trivial, then G is isornorphic (as a right partially ordered group) to either a trivially ordered cyclic group of prime order or a subgroup of the totally ordered additive group of real numbers. Proof: If G has a non-singleton trivially ordered subgroup H, then H = G since H is convex. In this case, G has no non-trivial subgroups and so is cyclic of prime order. If G has no non-singleton trivially ordered subgroups, then every element of G is in G+ U (G+)~1 (otherwise, the non-singleton subgroup generated by an element not belonging to G+(J(G+)~1 would be convex). So G is totally ordered and the result follows. /
58
CHAPTER 4. ABELIAN & NORMAL VALUED
i-GROUPS
Note that the result fails without the extra hypothesis as can be seen by taking the infinite cyclic group generated by g and letting G+ = {" : n > 2}. Then G is a directed Abelian group with only the trivial convex subgroups. But g,g~l $ G+, so G does not satisfy the conclusion of the Corollary. We close this section with an easy but useful result and a consequence. L e m m a 4.1.6 [Hion] Any order-preserving automorphism of a subgroup of B is multiplication by a positive real number. Proof: Let A be a subgroup of K and0. If a4>/c a/c, let m,n G Z+ with a<j)/c<j) > m/n > a/c. Then mc > na, whence m(c ) > n{a<j)). This con tradicts a /ccf> > m/n. Similarly, we get a contradiction if a(j>/c be an order-preserving automorphism of the multiplica tive group of positive rational numbers. Consider the logarithmic map £ : Q+ —> K, an order-preserving isomorphism of Q + onto L — £(Q + ), a subgroup of the additive group of real numbers; so q£ — log 9. Then £~lcj)(. is multiplication by some positive real number r by Hion's Lemma. Hence q ~1 — q1^ for all q € Q + , we deduce that 1/r e Z. Thus r = 1 and so <j> is the identity map. /I
4.2
Normal-valued lattice-ordered groups
Recall from the previous chapter that in any partially ordered group, every value has a unique cover in the set of convex subgroups. We first provide a trivially equivalent condition for a value to be normal in its cover.
4.2. NORMAL-VALUED
LATTICE-ORDERED
GROUPS
59
L e m m a 4.2.1 Let G be a lattice-ordered group and V a value in G. Then V < V* if and only if for all g G V* and f G G+\V, there is a positive integer n such that Vg < V fn Proof: Suppose that V < V* Then as V*/V is isomorphic to a subgroup of M, for all g e V* and / e G+\V, there is a positive integer n such that Vg < Vfn Conversely, assume that V is not normal in V* We first show that f~lVf C V for all / G V* with Vf > V. If not, there are / G V* and g G V'+ such that Vf > V and f~lgf £ V By the condition, there is a positive integer n such that Vf < V(f~1gf)n. Hence V = Vg~n < Vf'1, l a contradiction. We next show that fVf~ C V for all / G V* with Vf > V If not, there are / G V* and g & V+ such that 1// > V and fgf~l 4- V By the condition, V / < Vfgnf~l for some positive integer n. Thus / < hfguf~l for some ft G V. Therefore / < f~lhf g" G V because Z " 1 ^ / C V Consequently, V < Vf < Vf~xhf ■ g11 = V, a contradiction. We have therefore established that f~1Vf = V for all
fe V*- i.e., V
I
If every value is normal in its cover, we call the partially ordered group normal valued. If V is a value in a normal-valued partially ordered group and V* is its cover, then V < V* and V* jV is a partially ordered group whose only convex subgroups are trivial. By Corollary 4.1.5, if the strictly positive elements of V* /V form a semi-isolated set then V*/V is either cyclic of prime order or order isomorphic to a subgroup of E. Since every subgroup of an Abelian group is normal, every Abelian partially ordered group is normal valued. Also, since the set of convex subgroups of an ordered group forms a chain, every ordered group is normal valued. As in Section 3.1, for lattice-ordered groups we need to replace "con vex subgroups" by "convex ^-subgroups" A lattice-ordered group that has every value normal in its cover is said to be normal valued. If V is a value in a normal-valued lattice-ordered group and V* is its cover, then V < V* and V*/V is a lattice-ordered group whose only convex ^-subgroups are trivial. By Corollary 4.1.4, V*/V is order isomorphic to a subgroup of B. As above, every Abelian lattice-ordered group is normal valued and every ordered group is a normal-valued lattice-ordered group.
60
CHAPTER 4. ABELIAN & NORMAL VALUED
£-GROUPS
We now obtain the analogue of Abelian for normal-valued latticeordered groups. Lemma 4.2.2 [Wolfenstein 1968] LetG be a lattice-ordered group. Then G is normal valued if and only if \[f,g}\V(\f\ V \g\). Thus | / | V \g\ # V and there is a value M of | / | V \g\ containing V So M\[f,g]\n > M(\f\ V \g\). Hence f,g E M* and because M*/M is Abelian, it follows that \[f,g}\ G M. Thus M|[/,5]| n < M ( | / | V |5|), a contradiction. Conversely, if V is any value, let g G V and f GV* Then [/, g] G V* If [/>] & ^> t n e n -D> the convex ^-subgroup of G generated by V and [f,g], properly contains V and is properly contained in V* (since it does not contain / by the condition). This contradiction (to V* covers V) establishes that [f,g] G V; so V = V([f,g})-1 = Vg-'f-'gf = Vf~lgf, since g G V Therefore V Vg2f2. Clearly / V g 0 V As in the previous proof, we obtain M D V a value of / V g with Mfg > Mg2f2. But M < M* and M"/M is Abelian; so Mfg — Mgf < Mg2f2, the desired contradiction. Conversely, suppose that V is a value in G and that fg < g2f2 for all f,g G G+ If V were not normal in V*, there would be a G V+ and b G V* with b~lab £ V and b > 1 or b < 1. Then C, the convex ^-subgroup generated by V U Gib^ab) satisfies V C C C V* Hence C — V*. Thus, by the condition and Lemma 2.3.10, every element of (V*)+ can be written in the form xy where i 6 V + and y G G(b~1ab) + . So, if b > 1, then b < x(b~1ab)n+l for some x e V+ and non-negative integer n. Let n be the minimal non-negative integer for which such an
4.2. NORMAL-VALUED
LATTICE-ORDERED
GROUPS
61
x exists. Then Vb < Vx(b-lab)n+1 = V{b~lab)n+l = V(b-lab)n(b^ab) l l 71 and Vb = Va~ b = Vbb~ ab. Hence Vb < V^ab) . Thus there is v E V+ with b < w(6_1a6)"-. This contradicts the minimality of n (since b £■ V, the inequality 1 < b < u(6 _1 a6)° is impossible). Similarly we obtain a contradiction if b < 1. Hence G is normal valued./ Corollary 4.2.4 (i) Any (.-subgroup of a normal-valued lattice-ordered group is normal valued. (ii) Any £-homomorphic image of a normal-valued lattice-ordered group is normal valued. Proof: Clearly the equivalent condition in the lemma holds in any ^-subgroup and any £-homomorphic image of a normal-valued latticeordered group. / Corollary 4.2.5 Every residually ordered lattice-ordered group is nor mal valued. Proof: Since every ordered group is normal valued, the inequality fg < 92f2 holds in every ordered group (for all f,g > 1). Hence it holds in every residually ordered group./ Let A be a root system and A C A. We call A a plenary subset of A if
(i) for all 5 & A and A G A (6 < A => A e A), and (ii) C\{S:SeA} = 0. Now the set of all values in a lattice-ordered group G is a root system (Corollary 3.3.5). If A is a plenary set of values then each g e G\{1} fails to belong to some Vj £ A by (ii). Thus there is a value of g containing Vs. By (i), this value belongs to A; i.e., every g £ G\{1} has a value belonging to A. By the proof of Lemma 3.3.9 we obtain If G is a lattice-ordered group and A is a plenary set of values of G, then / < g if and only if V f < Vg for all values V € A.
62
CHAPTER 4. ABELIAN & NORMAL VALUED (.-GROUPS
If A is a plenary set of values every one of which is normal in its cover, then we will call A a plenary set of normal values. The existence of such a set is enough to guarantee that the lattice-ordered group is normal valued. Lemma 4.2.6 If a lattice-ordered group G has a plenary set of normal values, then G is normal valued. Proof: Let A be a plenary set of normal values of G. If G were not normal valued, then there would be / , g G G+ with fg j£ g2f2 by Lemma 4.2.3. Hence Vfg > Vg2f2 for some V G A. Thus there is a value of / V g containing V, and any such value belongs to A by (i); so we may assume that V itself is such a value. But V*/V is Abelian and f,g G V* Therefore Vfg = Vgf < Vg2f2, the desired contradiction. // Strong hypotheses (a plenary set of normal values) are required in the lemma; as we saw, the lattice-ordered group G in Kenoyer's Example (3.7.1) satisfies Example 4.2.7 [Black Swamp Problem Book No. 4, R. N. Ball] There is a non-normal-valued lattice-ordered group in which every non-identity element has a normal value. We close this section with an important technical result and applica tion. Lemma 4.2.8 Let G be a normal-valued lattice-ordered group with £ideal L. If f,g G G+ with L < Lg -C Lf, then VgVf for some positive integer n since V < V and V*/V is an Archimedean ordered group. But Lgn+l < Lf so Vgn+l < V f < Vgn Hence Vg < V, contradicting that g € G+ // We can now prove the promised easy partial converse to Lemma 3.6.3: Lemma 4.2.9 If G is a normal-valued lattice-ordered group, then every closed value is essential.
4.3. SPECIAL-VALUED
LATTICE-ORDERED
GROUPS
G3
Proof: Let V be a closed value of g G G+ We first note that there is ft G G with ft > 1 such that h < fg for all / G V + (otherwise 9 = g{/\f£v+ /ff)" 1 = V/ e v+ / _ 1 G I 7 ). Thus ft < V+g. We now shew that every value of ft is contained in V If M were a value of ft with M g K, let a G M + \ V By Lemma 4.2.1, there is n G Z+ such that Va" > Vg; so a" A # G V + g. Hence 1 < ft < an A g < a" G M, a contradiction. /
4.3
Special-valued lattice-ordered groups
We now give a deeper application of the results of the previous section. We first need a lemma which uses the fullness of plenary values. L e m m a 4.3.1 Let G be a lattice-ordered group with plenary set of values A. If g £ G+ has only finitely many values belonging to A, then all its values belong to A. Proof: Let Vi,..., Vn be all the values of g that belong to A. If V £ {Vi,...,Vn} were a value of g, let ft G V+\Vi and h, G ^ + \ V (1 < i < n). Let / = Vfci/i and h = AJU ?k. Then / G V\{{J?=lVi) and h G (n?=i VA\V By replacing / by / ( / A h^jmd h by h{f A h)~\ we may assume that / ± h. Also, since g Ah G (fli=i ^ ) \ ^ i w e m a Y further suppose that h < g. Now ft has a value M G A (since A is plenary). So / G M since / ± ft and M is prime. Thus M ^ V, for all i = 1, ...,n. But g > ft so g $ M. Therefore M is contained in some value of g which belongs to A since A is plenary and M G A; i.e., M C Vj for some j G / . This contradiction establishes the lemma. / We call a lattice-ordered group G special valued if every g G G+ can be written as a (possibly infinite) join of orthogonal special elements; these orthogonal special elements are called the special components of g. By Lemma 3.4.3, every finite-valued lattice-ordered group is special-valued. The full Cartesian product n^Li ^ with the cardinal ordering is special (but not finite) valued. Lemma 4.3.2 A lattice-ordered group is special valued if and only if it has a plenary set of special values.
64
CHAPTER 4. ABELIAN & NORMAL
VALUED
i-GROUPS
Proof: If G is special valued, let A be its set of special values. If g € G+, then any value V of a special component of g must contain all the other special components of g. Therefore V is a value of g and so (I A = {1}. Moreover, if M € Y(G) and D V e A, let / € G+ have value M. Since V is special, it is closed (Lemma 3.6.1 (vi)). Hence M is closed (Lemma 3.6.1 (v)). Since the special components of / are pairwise orthogonal and M is closed and prime, it must contain all but one of these special components and be a value for that special element. Thus M € A. Consequently, A is a plenary set. Conversely, suppose that G has a plenary set of special values; so every non-identity element has a value that is special. Let g 6 G+ not be special and V be a special value of g; choose fy special in G with value V Since V < V* by Lemma 3.4.2, it follows that V/V is an Archimedean ordered group. Hence we may assume that V fy > Vg. Let gv — fv A g; so Vgy = Vg and ggvl £ V+ Since gy = fy A g ^ V and V is the unique value of fy, it follows that V is the unique value of gy. Since non-orthogonal special elements have comparable values (under inclusion) by Lemma 3.4.1 (1), {gy : V a special value of g} is a pairwise orthogonal set. Hence gy S W for all distinct special values V and W of g. Therefore 1 < ggvl £ W for all special values W ^ V of g, and so every special value of g other than V is a special value of ggvl Moreover, every value of fvgvx is contained in V, the unique value of fy. Since Vfy > Vg and fvgy1 -L ggy1, every value of ggy1 is incomparable to V by Corollary 3.3.4. Thus the special values of ggy1 are just the special values of g apart from V. If h-v = ggy1 A gy ^ 1 and My is a special value of hy, then g^y 1 £ My. So My C W for some special value W ^ V of g (since the set of special values is plenary). Similarly, My C V since gy $• My. Hence My is contained in two incomparable values of g (since g is not special). We thus get a contradiction to Corollary 3.3.5. Therefore ggvl A gy — I for all special values V of g. Finally we observe that g — \/{gv : V a special value of g}: For if h > gv for all such V and h < g, let M be a special value of g/i - 1 If g € M, then M < Mgh~l = Mh'1 < M, a contradiction. And if 5 £ M, then M C V for some special value V of g. Since ff^y1 A #y = 1 and gv & M, we deduce that ggvl 6 M; so M < Mgh~l = Mgvh~l < M, a contradiction. Hence 5 = V{ffv : V a special value of g}. //
4.4. ARCHIMEDEAN
LATTICE-ORDERED
GROUPS
65
Recall that any special value is normal in its cover (Lemma 3.4.2). Hence, by Corollary 3.6.6, Corollary 4.3.3 Every special-valued lattice-ordered group is normal val ued and completely distributive. Hence so is every finite-valued latticeordered group.
4.4
Archimedean lattice-ordered groups
In this section we first use previous results to deduce two easy lemmata that prove Theorem 4.B Every Archimedean lattice-ordered group is Abelian. Lemma 4.4.1 Every Archimedean lattice-ordered group is residually or dered. Proof: Let G be an Archimedean lattice-ordered group and g, h G G+ with g J_ h. Let / G G+ and c = f~lhf A g. Then 1 < c A fcf~l < g/\h = 1, so c i . fcf-1 Hence cn _l_ fcnf~l. But (cnf~l)+ < cnJcnf~l, so ( c n / _ 1 ) + = 1. Therefore cn < f for all positive integers n. Since G is Archimedean, c = 1. Consequently, gL is normalised by every element of G+ Since (G+) = G, gL
Lemma 4.4.2 Every Archimedean residually ordered lattice-ordered group is Abelian. Proof: This follows at once from Corollary 4.2.5 and Lemma 4.2.2.^ In Lemma 3.6.l(i) we noted that polars are closed subgroups. The converse holds in the Archimedean case. Indeed: Lemma 4.4.3 Let G be a lattice-ordered group. Then G is Archimedean if and only if the only closed convex (.-subgroups of G are polars. To prove this we require two results that are interesting in their own right.
CHAPTER 4. ABELIAN
G6
VALUED
i-GROUPS
L e m m a 4.4.4 Let G be an Archimedean lattice-ordered group, C C G and 1 < d G G 1 1 Then d = Vnez+iCeC{d A |c|"). Proo/; If d A \c\n < g < d for all n G Z + and c G C, then 1 < dg~ < d G C-1"1-; so (ig -1 0 G 1 Therefore there is c0 G C such that / = (dg-^Alcol > 1. N o w d > feC. If/"
Corollary 4.4.5 A lattice-ordered group G is Archimedean if and only if 9= Vn=i{9 A (gnf-1 V 1)") for all f, g G G+ Proof: If / , g G G+ and gu < f for all positive integers n, then 5 ' 7 _ 1 < 1. Hence g = V~=i(s A ( 3 " / - 1 V 1)") implies that 5 = 1; i.e., G is Archimedean. Conversely, suppose that G is Archimedean and that / , g G G + Let C = {(gnrl)+ : n = 1, 2, 3,...}. Then («? A / i ) " / " 1 < hn A ( f f n / ' 1 ) + = 1 for each h G (C-"-)"1" and n = 1,2,3,.... Thus j A /i « / , and so 5 1 /i. Hence 9 G G x x By Lemma 4.4.4 we get 9 = Vn,mez+ A (s 7 '/" 1 V l)m Since ( g " / _ 1 V l ) m < (gkf~l V l)k if k > m,n, we deduce g = V£U(s A (fl'7" 1 V 1)"). // Proo/ of Lemma 4-4-3: Let C be a convex ^-subgroup of an Archimedean £-group G. By Lemma 4.4.4, CL1- = C if C is closed; i.e., all closed con vex ^-subgroups of G are polars. Conversely, suppose that all closed convex ^-subgroups of G are polars and that 1 < / < g in G. Ifft.€ G(/) + then fh G G ( / ) so / / i < g. Let z G (G(/) ± )+ with x < 0. Then z g r 1 / < a; and o ^ " 1 / < / So xg~lf < x A / = 1; i.e., x < / - 1 g . Thus xh = x V h < f~lg < g. Clearly, G = (G(/) ® G(/)- L )- LX Now the closure of any C G C(G) must be a polar by hypothesis and so must equal C11, the smallest polar containing G. Hence g belongs to the closure of G(f) <S> G ( / ) x and so is the supremum of the set of elements xh G G + with x G G ( / ) 1 , h G G(f) and xh < g. Therefore f~lg = 5, and so / = 1. Consequently, G is Archimedean. / Corollary 4.4.6 In any Archimedean lattice-ordered group, the sets of special values and maximal polars are equal. Moreover, if M is a special value, then M is a cardinal summand and G/M is Archimedean.
4.4. ARCHIMEDEAN
LATTICE-ORDERED
GROUPS
C7
Proof: If G is an Archimedean lattice-ordered group, then as special values are closed (Lemma 3.6.1 (vi)), they are polars (by Lemma 4.4.3) and prime. By Lemma 3.6.1(vii), all special values are maximal polars. x 1 Conversely, if G C1- is a maximal polar, then G C L is prime by Lemma 3.6.1(vii). Since every polar is the polar of a convex ^-subgroup, we may 1 1 c cc11 assume that G C e G C(G) C(G). If ccG€ C G with c >> 11, then c g^ cc 1 and C G1 C 1 1 c - by maximality. But if V is any prime subgroup and Hence C = cc 0£ V c: V, then cL1 CCV V. By Lemma 3.6.l(v), V is closed since cl is closed 1 and prime. By Lemma 4.4.3, V is a polar and hence V V == cc 1 by the x maximality of the latter among polars. Consequently, cr is the unique value of c and so is special. If M is special, then G/M is an ordered group since M is prime. M 5 " < Mf M / for all positive integers n, then Moreover, if f,g € G + and M#" n l (9nf~lV eG M and so ffA(( 9 A (5(«/g ":)/n_)+1 ) n ) + £ M. By Corollary 4.4.5, g = Vn 6 z+5 A ( (O5f"//-_11))"n))*+ G M. Thus G / M is Archimedean. Vnez-5 L1 G = = M M® We complete the proof by shewing that G © MM . . Clearly, xx the convex ^-subgroup of G generated by M U MM is £-isomorphic to m Mh for some Mxx, then Md M © ML. If h G G+VM and 1 < < d GM Md m > > M/» positive integer m (since G/M is Archimedean). Therefore (dmf\h)h~l = m l 1 1 1 1 A/i€ M M11; i.e., G = M0M = MM®M d" ^- A 1 G G M. M. So/i/ l GG MeM rt-'Al since ddm/\he e M 1 .. ///
(g r v
A strictly positive element g of a lattice-ordered group G is said to be basic if {/ G G + : / < g} is totally ordered (under the order inherited from G). In any -group, this is equivalent to gL1 being an ordered group: Lemma 4.4.4 gives this immediately in the Archimedean case which is all that we'll need here. In any lattice-ordered group, if g is basic, then g1 is prime: if a, b G + + G ; since g is G with a ± b, then a A g,bb A g < g and a A g ±± b A g; l 1 basic, oa A j = l o r i i A ( / = l ; i.e., o G g or 6 G g By Corollary b A.g Ag 1 3.3.5, 5 is special since any value of g (being prime) must contain g1 So basic elements are a source for special elements. By Corollary 4.4.6, the converse holds in the Archimedean case. Thus Corollary 4.4.7 In any lathee-ordered group, basic elements are special; and in any Archimedean lattice-ordered group, all special elements are basic.
68
CHAPTER 4. ABELIAN & NORMAL VALUED
t-GROUPS
The elements of a maximal pairwise orthogonal set are basic, and when such a set exists we say that the lattice-ordered group has a basis. By the previous corollary and Lemmata 3.6.5 and 3.6.1(vi), we obtain Corollary 4.4.8 An Archimedean lattice-ordered group is completely dis tributive if and only if it has a basis. Relatedly L e m m a 4.4.9 Let G be an Archimedean lattice-ordered group. If every set of pairwise orthogonal elements of G has a supremum in G, then G is strongly projectable. Proof: Since G is Abelian, we revert to additive notation for this proof. We first show that G is projectable. So let f,g£ G+ For each ++ non-negative integer n, let d„ = ((n+2)g-f) ((n + 2)g +— A(f-ng) / ) + A (/ — ng) 2)g. < <{n(n+ +2)g. XJ Note that dn J. d k for any k £ Z Z with > 2, and each d £ k > 2, g^n+ -L dn+k n Hence cc = Mn(2n + 1) l)d^2n and d = \l (2n + 2)d +i exist in G. Then = V„(2n 2)d2n2n+ i 2n n 1 1 // A/\(c + d) £ 9gLL - since c,d £ g-1-1 (polars are closed by Lemma 3.6.1 (i)). (c + x 1 ± We show that // --( /(/A A( c(c + + d)) d ) )£e ^ ,, whence /f £ g £g - ©®g g-1LL-1,, as required. We first establish (by induction on n) that
v„
n
E(r +
I K > [ ( n + l) ((n + 2) 9
f)+] A f > ("9
/)
i
A/.
r-0
The induction step is: ([(n+1) ((n + 2) e - / ) + ] A / ) + (n + 2)dn+l > ([(n + l)((n + 2)g - / ) + ]
([(n+1) ((n +
+ (n + 2)d n+1 ) A /
2)g-f)+} + (n + 2) [((n + 3) 9
+ /)+ A(/-(n+l)5) ])A/
> [(n+2) [(n + 3)ff-/] + ]A[(n+l)[(n + 2 ) 5 - / ] + (n + 2) ( /
(n+1) 9)}+ A /
[(n + 2) ((n + % - / ) + ] A / Thus for each non-negative integer m, 9 A (/
!/ A (c + rf)]) < S A [/
(f + l)d,.)]
(/ A r=0
4.5. HAHN'S
THEOREM
G9
< 9 A [/ - (/ A {mg - /)+)] < g A [/ - (/ A (m - /))] 9 A [2/ - (2/ A mg)] = [{g + 2/) A (g + mg) A 2/] - (2/ A mg) = [2/A(m+l)5]-[2/Am5]. Hence n-l
d))}) < £ n(g ( m + l)g] "(fl A A [/ ([2/ A (m l ) d - [2/ [2fAmg]) A m 5 ]) < 22// ;/' - (/ A (c + d))\) E([2/ m=0 m=0
+ for all n 6e Z 5A + d))] d))} = O 0 by the Archimedean 2 + .. Therefore 0 A [/[ / - ( (/ / AA( c(c + property, as desired. Now let P be any polar of G. Let {{9, £ 7} 1} be a maximal pairwise 3l :: Ji 6 orthogonal subset of P + \ {{I}1 } . Then for any / G G+, f == fi/, ++ fk h% with 1J Si G G g/-. fk G 5^- But {f, : 1 £ 7} is a pairwise orthogonal set U G
We close this section with an unsolved problem: Can every torsion-free Abelian group be ordered in such a way as to make it an Archimedean latticeordered group? Caution: We can certainly embed any torsion-free Abelian group G in a rational vector space V which can be ordered so as to be an {b,:ie: i 6 /1} Archimedean lattice-ordered group. (Let {b; } be a basis of V; V b c define vv = = £E,e/ , € / o0 9* ftb,< >> 0 (g, k C /, 0 t IQ (a €e fQ, /, 7^o0 finite) if and only if <j, Qi > 0 for all i €e IQ.) However, there is no reason why the supremum (in V) of r h two elements of G G may not be a sublattice of G: i.e., G G has to belong to G; V
4.5
H a h n ' s Theorem
Let G be an Abelian ordered group. Then the set of all values of G forms a chain which we denoted by T{G). Each value V in G has a unique cover V and V*/V is an Archimedean ordered group (and so is
70
CHAPTER 4. ABELIAN k NORMAL VALUED
i-GROUPS
order isomorphic to a subgroup of R by Holder's Theorem). We identify each 7 £ T(G) with the pair (Vy, V1) where V7 is the value and V is its cover. Moreover we let R-, be the ordered subgroup of R order-isomorphic to V/V,. We call T(G) the spine of G, {Ry : 7 e T(G)} the H6s of G, and the spine and ribs combined the skeleton of G. In accordance with Example 1.3.16, we let V'(F, Ry) be the set of all functions g from T into R such that supp() = {7 G T : 5(7) ^ 0} is inversely well-ordered (i.e., every non-empty subset of the support of g has a maximal element) and 9(7) G Ry for all 7 £ F. The set V(r, 7?7) is an Abelian ordered group under functional addition ( ( / + g) (7) = f{l) + 5(7) f° r 7 € F) and the order g > 0 if and only if g — 0 or 5(a) > 0 where a is the maximal element of supp(#). T h e o r e m 4.C [Hahn] If G is an Abelian ordered group with spine T(G) and ribs {R~, '■ 7 € T{G)}, then G can be embedded (as an ordered group) in V(V(G), Ry), where R-, is the rational vector subspace of 1 generated
by R, (7 e r(G)). The rest of this section is devoted to proving Hahn's Theorem. Firstly, we recall from Example 1.3.8 that every Abelian ordered group G can be embedded in a divisible Abelian ordered group, G such that T(G) = T(G) and the ribs of G are just the divisible closures of the corresponding ribs of G; i.e., for each 7 G T(G), Ry is the rib of G corresponding to 7. Hence we may assume that G itself is divisible (i.e., a rational vector space). The key lemma in the proof of Hahn's Theorem is: Lemma 4.5.1 [Banaschewski] Let V be a vector space over a field /C, and S(V) be the set of all vector subspaces of V Then there is a function 9 : S(V) -» S(V) such that (I) for each U G S(V), U®U6 = V, and (II) for each UUU2 G S(V), Ux C U2 => Ux6 2 U29. Proof: Let 6 = {(W,6W) ■ W 6 S{V),9W
satisfies
(i) and («) for
S{W)}.
Then ({0},0) G 0 (where 0 denotes the zero map); hence 0 ^ 0 . We define a partial ordering on 0 by: (WU8W]) < (W2,0w2) if and only if Wx C W2 and (Wj n U)9Wx Q (W2 n t / ) ^ for all [/ G ^(V 7 ).
4.5. HAHN'S
71
THEOREM
We now show that 0 satisfies the hypotheses of Zorn's Lemma; specif ically, that the union of any chain in 0 belongs to 0 and is an upper bound for it. Let * be any chain in 0 . Let * , = { [ / : {U,6V) G * } . Then clearly W = \j{U : [ / £ * , } £ S(V). For any T £ S{W), let T9W = \J{{TnU)6u :UeVi}. Since {TC\U)6V G S(U) for all U 6 * x , we have that T9W G <S(VK); hence 6W : S{W) -4 S(W). If T G 5(W) and to G W, then io G £/ for some U G $ i. Thus to = u + y for some u £ T HU and ?y G (TnU)Bn. Hence W = T + r ^ . On the other hand, if v G TnT9w, then TJ G {TnU)9u for some (7 G * i . Since TnU £ S(U) and 6V maps «S(C7) to itself, we have v £ U. Thus v £ (T n U) n (T n [/)0C/ = {0}. Consequently, W = T © T0,y for all T G <S(W), and (i) is satisfied. Now let TUT2 £ S{W) with l i C T2. Then I \ n U C T2 n £/ for all £/ G * i , and so (T2 n f / ) ^ C (T, n [/)0u for all C7 G * i . Hence
{T2nw)6w =\j{{WnT2nU)8u ■. u G *i} = U{(7,2nt/)fly •. (/ e *.} c \J{{T1nu)9u: f/€*i} =u{(w / nr 1 n[/)6i c/ : t/G *,} = ( T , n ^ .
Consequently, (W7,6w) G 1P. A similar computation shows that (W, 0 W ) is an upper bound for ^. By Zorn's Lemma, # has a maximal element, say (W, &w). We now show that W = V, whence the lemma holds with 9 = 0V. If W ^ V, let u0 G V\W and A = W ffi K.v0 £ S{V). Define VR
x
/ t/0w©/Cw o \ {UC\W)9W
ifC/G5(IV) otherwise.
Then 9X : S(X) -4 «S(X) and (W n f/)(9w C (X n l/)0A- for all U £ S(V). We show that (X,9X) £ © contradicting that (W,6W) was maximal in 0 (hence establishing the lemma). If U £ S(W), then U9X + U = U9w + U+ICv0 = X and if?; G U9XC\U, then v = y + kvo for some y £ UQw and k £ K. Since o G U C V7 and 2/ G t/#w C W, it follows that k — 0. Hence y = y G t/ n £/0w = {0}. Therefore X = U® U9XIf C/ G S ( X ) \ S ( W ) , then C/ + U9X contains both W and v0. Hence A = £/ + f/0 x . If v € t/ 0 L/0 X , then v £ {U D W)8W C W/~; so
v £ (U r\W)n{Un
W)9W = {o}.
Hence, in either case, X = U © U9X. So (A, 0 X ) satisfies (i). We leave the similar verification of (ii) to the reader.// We can now prove Hahn's Theorem:
72
CHAPTER 4. ABELIAN k NORMAL VALUED
i-GROUPS
Proof: As already noted, we may assume that G is a divisible Abelian ordered group, and hence a rational vector space. Moreover, every convex subgroup of G is a vector subspace. By Banaschewski's Lemma, there is 6 : C(G) -> C{G) such that, for each 7 G T = T(G), G = V ®j\r>9. Let 7r7 denote the projection of G onto V and p1 : V —» R-, — Ry be the order-preserving homomorphism whose kernel is Vy. For each g G G, define 5(7) = 7r7p7 6 Ry. We first show that g&V = V(T, i ^ ) . Let 5 G G with 5 7^ 0 and let 0 ^ A C supp(). Let L = UlV 7 : 7 G A}. Then L G C(G) so G = L ® LB. Hence 5 = y + z for some y G L and z & L9. If <5 G A, then V*0 2 L0 and g = y6 + zs for some j/4 G V"5 and z^ G V*0. So zs - z £ V69 and hence j/7rA. Since <5 G supp(rj), we have g(5) ^ 0. But Vx C V^, so T/A^ = 0. Hence g((S) = yx^6 = 0. This contradiction establishes that A has greatest element A. Further, let 0 < g G G\ so g G V\Vy for some 7 G I\ Then gp-f > 0 since p 7 preserves order, and 3(7) ^ 0 since g $ V7. The argument just used to show that A was the largest, element of A can be used again to show that 7 is the largest element of supp(). Hence the map g 1—> g is one-to-one and preserves order. It is clearly a group homomorphism thus establishing Hahn's Theorem./ It should be added that the Hahn group so constructed has the same skeleton as the divisible closure of G. To see this, if v G V, let 7 be the maximal element of supp(7;). It is straightforward to prove that H1 — {u G V : u(a) = 0 for all a > 7} is the value of 7; with cover H1 = {u G V : u(a) = 0 for all a > 7}. Moreover, H~>/Hy = R1, so V has the same skeleton as the divisible closure of G. We now show that Hahn's Theorem fails if we replace Ry by just R-,. E x a m p l e 4.5.2 Let {pn : n = 1,2,...} be the sequence of primes in in creasing order and G be the subgroup of Q®Q generated by {(n/pn, l/p„) : n = 1,2,...}. Note that (1,0) = 3(2/3,1/3) - 2(1/2,1/2), so (1,0) G G. Also if (a,0) G G+, then a = Y,{jmj/Pj ■ J G ./} where ^{mj/Pj :j G .7} = 0. Hence pj divides m3 for all j G J. Thus a G Z. Consequently,
4.6. THE CONRAD-HARVEY-HOLLAND
THEOREM
73
C = {(n, 0) : n £ Z} is a convex subgroup of G. If (a, b) € G with 6 ^ 0 , then G((a, 6)) = G since Q is Archimedean. Hence C is the only proper convex subgroup of the ordered group G. Since G/C = Q, the Hahn group with the same skeleton as G is isomorphic (as an ordered group) to H = Z © Q. If there were an embedding > : G -> H, then (1,0)0 = (m, 0) for some m £ Z with m > 0. Let (z n ,g n ) = (n/p„, \/pn)4>. Note that p n + 1 ((n + l)/p„ + i, l/p» + 1 ) - 2(1/2,1/2) = (n,0), so Pn+l(Zn+l,9n+l) - 2(zi,tfi) = ( m n , 0 ) .
Thus z n + i = (mn + 2zi)/p n + 1 . Since n/pn —>• 0 as n —> oo, the integer z n + 1 must eventually be 0. But then 2zx/n = —m £ Z for all sufficiently large n. Therefore Z! = 0, whence m = 0, a contradiction. Hahn's Theorem gives that G can be embedded in Q© Q, which we knew to begin with. For conditions that imply the isomorphism of V(Ti,l) and K(r 2 ,K) (as ordered sets), please see [Kuhlman].
4.6
The Conrad-Harvey-Holland Theorem
We now prove the generalisation of Hahn's Theorem to Abelian latticeordered groups. Recall first from the previous chapter (Corollary 3.3.5) that the set of values of a lattice-ordered group forms a root system. So the resulting spine is potentially more complicated. We saw in Example 1.3.18 that this leads to a more complicated Hahn group. Nonetheless, we can obtain the natural analogue of Hahn's Theorem in this case too. Recall that A a subset of a root system A is plenary if (I) for all 5 e A and A £ A (6 < A => A e A), and (..) |~|{V* : 6 e A} = {0}. Theorem 4.D [Conrad-Harvey-Holland] Let G be an Abelian latticeordered group with spine T(G) and ribs {R-, : 7 £ T(G)}. Suppose that r is a plenary subset of T(G). Then G can be (.-embedded in V(r, Ry), where R-, is the rational vector subspace of 1 generated by Rn (7 £ T).
CHAPTER 4. ABELIAN k NORMAL
7-1
VALUED
t-GROUPS
Proof- Rather than give a modification of the preceding proof, we provide another due to [S. Wolfenstein 1966]. Again, we pass straight to the case that G itself is divisible. For each 7 G T, let V7 be the natural homornorphism of V1 onto VfVT We call a group homornorphism 9 from a subgroup H of G into V = V(F, Ry) a ^-homornorphism if {g6)(j) = gijjy whenever g G HnV1 Let 9 be a ^-homornorphism of a subgroup H of G into V We first shew that for any g e H,we have g G V 7 \K 7 if and only if 7 is a maximal element of supp(#). If g G G\{0}, then g G V'7\V7 for some 7 G I\ Hence {g9){1) = gtp7 / 0 and (g9)(X) = 0 for all A G T with A > 7 (since g G VA)Therefore 7 is a maximal element of supp(##). Conversely, if 7 is a maximal element of supp(<7#), then g £ Vx for all A > 7 and g £ V1. Hence g G V 7 \V 7 . We next prove that any ^/-homornorphism of a subgroup H of G is a lattice homornorphism. If / , y G / / and f -L g. then no maximal element of supp(/#) is com parable to a maximal element of supp(#). Since each ip-y is a lattice homornorphism, we deduce that fO 1 gO. Hence 0 is a lattice homornor phism by Lemma 3.2.2. As in the proof of Hahn's Theorem, we can shew that Zorn's Lemma applies to deduce that there is a subspace W of G which is maximal with respect to the existence of a ^-homomorphism, say 8. To complete the proof of the Conrad-Harvey-Holland Theorem, it is enough to show that W = G. UW ?G, let v0£G+\W and \etX = W®Qv0. If Qv0n(W + V) + {0}, let n G 1 with n > 0 be such that nv0 G (W + V); say, nv0 = w +c with w G W and c G V1. We define v eV by: „
M {1
>
=
I ( M ) ( 7 ) + c^)/n \ 0
if Qu0 n (W + V) jt {0} i{®van(VV + V-<) = {0}.
It is straightforward to check that this is well defined. If Qu0 n {W + V) ± {0} and v{\) # 0 for some A > 7, then nv{\) = {w9){\). Since w9 G V, it has inversely well-ordered support. Hence so does v. That is, v £ V We defineto be the homornorphism of A' into V given by: 4>\W = 9 and v04> = t>.
4.7. ELEMENTARY
THEORY (ABELIAN
l-GROUPS)
75
This is well defined since Wi + miv0 = w2 + m2v0 implies that (m, m2)v0 e W\ so {w2 - wx)6 = (mx - m2)v08 = (mi - m.2)v. Hence W18 + rriiv = w26 + m2v. Finally, we need to show that 0 is a ^-homomorphism. Let g = w + nv0 where w e W and n £ Z and assume that g £ X (IV If Q^o H (W + V7) ^ {0}, then mnvo = wi + c for some positive integer m, 10] £ W and c e V1 Hence mg = mw + Wy + c. Therefore rng - c G W n V 7 . Thus m(g)(j) - ((mw + wi)6)(j) + apy = (mw + wi + c)ip-, — m(gipy). Consequently, (g is a ty-homomorphism. This contradicts the maximality of W; the theorem follows./ For a characterisation of those partially ordered sets T for which V(F,Ry) satisfies the conclusion of Lemma 2.3.10 (where each R-, is a subgroup of K or trivially ordered) and has semi-isolated positive cone (c.f., Corollary 4.1.5), please see [Teller].
4.7
Elementary theory (Abelian ^-groups)
In this section we prove that any two lattice-ordered Abelian groups satisfy the same universal sentences if and only if their maximal pairwise orthogonal subsets have the same size (or both are infinite). Moreover, we show that the universal first order theory of Abelian lattice-ordered groups is decidable but the full first order theory of Abelian latticeordered groups is undecidable. Throughout this section we will assume a certain facility with terms from mathematical logic. We will first assume that the language comes equipped with a con stant 0, a binary relation <, and a ternary relation S(x,y,z) which will be interpreted as x + y — z. T h e o r e m 4.E [Gurevich & Kokorin] Any two non-singleton ordered Abeliar groups satisfy the same universal sentences. Proof: Let G be a non-singleton ordered Abelian group. We show that a universal sentence holds in G if and only if it holds in Z. Since Z can be order embedded in any non-singleton ordered group, every universal sentence that holds in G holds in Z.
76
CHAPTER 4. ABELIAN k NORMAL VALUED
i-GROUPS
We must prove that every universal sentence that holds in Z holds in G. By Theorem l.A, a universal sentence holds in G if and only if it holds in all non-singleton finitely generated subgroups of G and holds in Z if and only if it holds in Q (every finitely generated subgroup of Q is cyclic); so we may assume that G is finitely generated. By Theorem l.A, it is enough to prove the following lemma: Lemma 4.7.1 [Holland, Mekler & Reilly] Any finitely generated ordered Abelian group can be order embedded in an ultrapower of Z. Proof: Let G be a finitely generated ordered Abelian group. Since the divisible closure of G is a finite dimensional rational vector space, G has only a finite number of convex subgroups. By Harm's Theorem, G can be embedded in V(I, R) for some finite totally ordered set / . Since G is finitely generated, so is every subgroup and homomorphic image of G by the Fundamental Theorem of Abelian Groups. Hence for each i € I we can find a finite set of generators {t(i,j) : j 6 J,} C R such that each t(i,j) > 0 and t(i,0) = 1, and G C ({i(i,jj : i 6 I,j € J z }) where Now consider the cardinal product n£Li z - L e t U be any non-trivial ultrafilter on the set of positive integers. For each (i, j) choose a sequence {n(i,j, k) : k = 1,2,...} such that each n(i,j,k) is divisible by each k\, lim n(i, j , k)/n(i, 0, k)
= t(i, j)
k-*oo
and lim n(i,j,k)/n(i',j',k)
= co,
if i' < i.
k—>oo
Then the subgroup of V(I, R) generated by the functions i(i,j) is embed ded in (njJLi Z)/£Y via the map which sends i{i,j) to the ^-equivalence class of the sequence (over k) n(i,j,k) — the specified properties en sure that the map does indeed preserve order. Hence G can be orderembedded in this ultraproduct, too. // For Abelian lattice-ordered groups, we consider the same language but with < replaced by U and L, where U(x,y,z) if x V y = z and L(x,y,z) if x A y = z. In this setting (please see [Chang k Keisler, page 342]), the FefermanVaught Theorem yields:
4.7. ELEMENTARY
THEORY (ABELIAN
t-GROUPS)
77
Given a formula >(xi, ...,x„) of the language, there are formulae IpiiXi, ...,xn),
...,ipm(xu
...,xn)
of the language and a formula cr(y\, ...,ym) of the language of Boolean Algebras satisfying BA \= Vy1...ymVz1...zm[a(y)fc A ^ (y} < z3) ->
nG,M[a]«P(/)N^.,-vi] where 7 ? (/) is the complete Boolean Algebra of subsets of /. The proof actually gives that Vi, ■■■>i>m can all be taken to be universal (existential) if (f> is. Moreover, there is an effective procedure to obtain a. Vi, ■■■,'4>m from 4>. If an Abelian lattice-ordered group G has a set of n pairwise orthog onal non-zero elements but no such set of size n + 1, we will say that G has orthogonal dimension n. If no such finite set exists, we will say that G has infinite orthogonal dimension. T h e o r e m 4.F [Khisamiev] Abelian lattice-ordered groups satisfy the same universal sentences if and only if they have the same orthogonal dimension. Proof: As we saw in the last section of the previous chapter, if G and H have finite orthogonal dimension m, then G and H are subdirect products of m ordered Abelian groups, and contain the cardinal product of m copies of Z. Say, G < T\IZ? d and H < Yl^ZT Hi w i t h e a c h C a n d Ht a non-singleton ordered Abelian group. If <j> is a universal sentence that holds in G then it holds in ITi=i" and so in H by the Feferman-Vaught Theorem. So G and H satisfy the same universal theory. If G and H have infinite orthogonality dimension, then there are cardinals KI,K2 such that
E ^ < G < TiG>
z
78
CHAPTER 4. ABELIAN k NORMAL VALUED
and
^-GROUPS
z z
i
with Gi and H3 non-singleton ordered Abelian groups. But the finitely generated subgroups of £, < K 1 Z and £, < K 2 Z are isomorphic, so they sat isfy the same universal sentences by Theorem l.A. The proof for finite m now demonstrates that G and H satisfy the same universal sentences. For the converse, let
= 0) & ki
m
)^0).
An Abelian lattice-ordered group satisfies
Theorem 4.G A universal sentence holds in all Abelian lattice-ordered groups if and only if it holds in U^o Z. Proof: By Theorem 4.F, a universal sentence holds in all Abelian lattice-ordered groups if and only if it holds in U]=T Z for all positive integres m and in n ^ o 2 - Since each ni=i"Z is ^-isomorphic to an £subgroup of n ^ o Z , the result follows. / If a subset X of the set of all natural numbers can be listed in an algo rithmic manner, we say that X is recursively enumerable. Equivalently, if there is an algorithm to determine if a particular number belongs to the set X. In the event that we also have an algorithm to determine if a number fails to belong to X, then running both algorithms simulta neously on different machines will give us a method to decide whether or not the number belongs to X: If the number belongs to the set, Or lando Gibbons' "Jubilate Deo" is sung; if it doesn't, then Thomas Tallis' "Lamentationes Jerirniah" is sung. If we have both machines available then we say that X is recursive. In contrast, a recursively enumerable set corresponds only to the singing of the ''Jubilate Deo"; if nothing has been sung yet, then we have no idea whether the number will belong to the set or not. And there are recursively enumerable sets that are not recursive [H. Rogers, page 62].
4.7. ELEMENTARY
THEORY (ABELIAN
l-GROUPS)
79
Given any theory, we can ask if there is an algorithm to determine what its theorems are. Usually, we enumerate all statements of the lan guage in an explicit manner and thereby identify the set of statements of the language with the set of all natural numbers. Under this iden tification, we can consider the subset X of those natural numbers that correspond to theorems (i.e., statements which are consequences of the axioms of the theory). If X is recursive then we say that the original problem is decidable; i.e., there is a single algorithm which when pre sented with an arbitrary statement determines whether or not it is a theorem. We can also confine ourselves to sentences of a certain form (e.g., those sentences that have only universal quantifiers (V) present) and ask about the decidability of such sets of theorems. Caution: As we will see in the case of the theory of Abelian latticeordered groups, a nice subset can be decidable even when the full set of theorems is undecidable. T h e o r e m 4.H The universal theory of Abelian lattice-ordered groups is decidable. Proof: Let <j> be a universal sentence. By the Feferman-Vaught The orem we can effectively obtain a,\p\,...,ij)k with ipi,...,tpk universal so that if A3 = {m e Z + : Z \= ipj}, then n , v o z (=if a n c i o n l y li V(Z+) \= o~[Ai,...,Ak]- Each A3 is either Z + or 0, and there is an algorithm to decide which (by a theorem of Pressburger). Also, by [Skolem], "P(Z+) has a decidable theory; so we can determine whether or not V(Z+) f= a[A\,..., Ak]- Therefore we can determine whether or
notlTgoZM-// Corollary 4.7.2 The word problem for Abelian lattice-ordered groups is uniformly soluble; i.e., there is an algorithm which when given an arbitrary finitely presented Abelian lattice-ordered group and a word in its generators, determines whether or not the word is 0 in the given finitely presented Abelian lattice-ordered group. In contrast, we now prove Theorem 4.1 [Y. Gurevich] The theory of Abelian lattice-ordered groups is undecidable.
80
CHAPTER 4. ABELIAN & NORMAL VALUED
£-GROUPS
All proofs of this result arise from interpreting an undecidable theory in the theory of Abelian lattice-ordered groups. Although there are quite short proofs of the theorem (see, e.g., [Burris]), Gurevich's original proof is the most transparent and shows that the V3V theory (in the language of lattices alone) of Abelian lattice-ordered groups is undecidable. The (hereditarily) undecidable theory that we will interpret is the the ory of all symmetric irreflexive relations P on N, the set of non-negative integers. The main steps of the proof are: (1) We define formulae 0(x,y) and P{x,y) such that (1.1) P is symmetric and irreflexive; and (1.2) 6 is an equivalence relation on Gg = {g G G : 0(g,g)} respected byP _ So P induces an irreflexive symmetric relation P on Gg/9. We let G* be the set Gg/Q under this relation P (2) Given M = (N; P), we will build an Abelian lattice-ordered group G{M) such that G{M)* £ M. Then M. = {G(M) : P irreflexive,
symmetric}
will have (hereditarily) undecidable theory. So it remains to establish (1) and (2). We will use the language {+,—,V,A,0} although we could use the restricted language { + , V} and write 0 for the unique z such that z + z = z, -x for the unique element whose sum with x is 0, and x A y for — (-x V — y). We will also write \x\ as an abbreviation for x V —x, as usual. In any lattice-ordered group G, if /, g G G+ then (by Lemma 3.1.5 and our observations in Section 3.5) (/ A )xx = Z 1 1 A gLL and (/ V g)±L — fLL V gLL. Thus the map g i-» gLL is a lattice homomorphism of G+ into V(G). Note that fL± = g11- implies fL = g1 (if h e g^f1, then as / G fLL - g-11, we get / 1 h, a contradiction). For the rest of this section only, let g — {/ G G : fL = gL}, the filet of g. We have just shown that / = g if and only if fx± = g-1-1 With the induced ordering we have (for all / , g > 1), / Ag = f A g and fVg= f + g. Consequently, G = {g ■ g G G} is & distributive lattice and the map g i-> g is a lattice homomorphism. We call G the lattice of filets of G
4.7. ELEMENTARY
THEORY (ABELIAN
(.-GROUPS)
SI
Note that we can write / G g in our language by: (Vx)(|z| A | / | = 1 o
|x|A|| = l).
Hence, with this shorthand, we can express / < g in our language. Construction
of G*
Let L(x) be the formula: x > 0 k (Vy){y < x -4 y < x). Lemma 4.7.3 [ L ( I ) & L ( J / ) ] -> [x < y or y < x or x _L y].
Proof: If x ■£ y k y ■£ i, then 2(x Aj/) = i A i / = i A y < i . Hence 2(x Ay) < x. Similarly 2(x A y) < y, so i A y = 0. / We will write 0(x, y) for x > 0 & y > 0 & (3w)(L(w) k x < w k y < w). Note that if (3x)9(x, x) holds in any Archimedean lattice-ordered group, then that Archimedean lattice-ordered group is ordered. Proof: Since nf = / , hi < h2 can only hold in an Archimedean lattice-ordered group if hi = 0. Thus if L{g) held, then h -£ g for all h 6 G+\{0}; i.e., for all h G G + \{0} either h = g or h _L g. But if h ± g, then h < g + h, a contradiction. Hence g1 — 0 and h — g for all h G G + \ { 0 } . So no two distinct strictly positive elements can be orthogonal. Therefore the Archimedean lattice-ordered group is ordered, and so is isomorphic to a subgroup of R. / L e m m a 4.7.4 \9{x,y) k 9{y,z)} -*■ 9{x,z). Proof: If u, v satisfy the existenial condition in 9(x,y) and 9(y,z) respectively, then u or v satisfies the existential condition for 9(x, z) by Lemma 4.7.3 since y < u,v ( s o u , w £ y, whence u J. v is impossible). / Corollary 4.7.5 9 is an equivalence relation on {x : 9(x,x)}. Let P(x, y) be the formula 9(x,x)
k9{y,y)k^9(x,y)k
(3t > O)(Vu)([(0(u,u) & -.0(u,x) & --fl(u,y)) - » ( * ! tt)] k (Vu)(V«)[(5(i,u) & 0(y,v)) - ► ( f ^ u & t ^ w & t
r«V»)|).
CHAPTER 4. ABELIAN & NORMAL
VALUED
i-GROUPS
Lemma 4.7.6 P is irreflexive and symmetric; moreover, P(x,y) & 6{x,z) -+P(z,y). Proof: Clearly P is irreflexive; it is symmetric since 6 is. The rest of the lemma follows immediately from Lemma 4.7.4: since 6 is symmetric, we have 9(z,x), so 0{z,z) by Lemma 4.7.4. Similarly, ~^6(z, y). Also, by the same lemma, the "t" of P(x,y) satisfies the existential condition in
P{z,y)- II Thus we obtain G* satisfying the required conditions (1.1) and (1.2). Construction
of
G(M).
Let M = (N; P) where P is an irreflexive symmetric relation on the set of non-negative integers N. For each £,i, j € N, let e(£,i,j) be a formal symbol and Ke(f, i,j) be the one-dimensional real vector space with basis e(£,i, j) ordered so that re (£>*>j) > 0 if and only if r > 0 in R. Let 00
E Re(?,i,i)
L(U)
i
o
ordered by £ , r.,e(£,i, j) > 0 if and only if rk > 0 where k = max{j : r3 j= 0}. Let
A(0
oo
n L(U)
i-O
with the cardinal ordering and OO
n May,
A
{=0
i.e., A = FT( n t L(£, *) with each L(£, i) a real vector space ordered so that 0 < e(£, i, 0) < e(£ i, 1) < e(£, i, 2)< .... Observe that L(f, i) n 1,(77, £) = {0} unless £ = r? & i = fc, and so i ( f ) n i f a ) = {0} if f ^ rj. For fc e N let i=k
m,k) £ e(f,i,fc) e A(fl. t=0
4.7. ELEMENTARY
THEORY (ABELIAN
t-GROUPS)
83
Let, B(f,l) = «e((,0J0)+«e(e>l,0)Ci4(0. We define B{£, k) and C(f, k) inductively (on Jfe) by C(f, *) is the Isubgroup of A(§) generated by B(£, k) U R/(£, k), and B(£, k + 1) is the ^-subgroup of A(f) generated by C(£, k) and Re(£, fc + 1 , 0 ) . Note that C(£, fc) = B{Z,kf®Uf(Z,k) and that B(£, fc+1) = C(Ofc)8 Re(f.fc + 1,0). Let oo
B(0={jB(£,k)
and J6N
For £ 7^ 77, let
<7(^) = £e(£,p*+1,0) + ^ e ( , , , p * + 1 , 0 ) d ( ( ) + ^ ) /t6N
where pn denotes the nlh prime. Note that the projection of g(£, //) in L(a,i.) is 0 unless i is a power of p^ or pv and a £ {£,"}• Let G(M) be the ^-subgroup of A generated by B and {^(O 7 ?) : M|=P(0'?)}. In the next four lemmata we will establish that G{M)* = M. L e m m a 4.7.7 Each g G G{M) can be written in the form g = b+
Y;HZ,Tl)9(£,v)
where b e B and (£ < 77 & M (= P{£, n)) for all f, 77 for which A(f, 77) ^ 0. Proof: We need only show that E
A
( ^ ? ) $ ( £ " ) ] + = £ ( A & 1 ) v %(£> »7) K < »?)■
(Since A((,?j) V O / 0 implies that A(C, *?) / 0 whence (f < 7; & M (= P((,,n)), the right-hand side indeed belongs to G(M)). Fix Q , J £ N and project onto L(a, z). If i is not the power of a prime, then the projections of all <J(£, 77) equal 0 and the projections of both sides are equal. Next suppose that 1 — pk0+l for some /?, k. If a = [i then the projection of each #(£, 77) is 0 and the projections of both sides are again equal. Finally assume that a ^ 0. Since the proof is the same in both
84
CHAPTER 4. ABELIAN & NORMAL VALUED
i-GROUPS
cases, we suppose that a < j3. Then the projection of A(£, r])g(£,r]) is 0 unless £ = a and r\ — p. Hence the projection of [ £ A(£, 77)^(0 V)]+ 1S J u s t [\{a,P)e{a,pk0+l,O)] + = [X(a,0) V 0]e(a,p£ +1 ,0). This is the projection of ]C[A(£,?7) V Q]g{£,r)) (£ < 7?) as required. Thus, in all cases, the projections are equal. This establishes the equality. // Lemma 4.7.8 If g G G{M) and G(M) (= L(g), then g G B{Q for some Proof: By the previous lemma we can write g in the form b + £ A(f, J7)s(f,7j) where 6 6 B and (f < 77 & M (= P(0*?)) for all £,r/ for which A(^,T?) ^ 0. We first show that A(£,r/) = 0 for all £,77 6 Z + For, if not, there are a < p with A(a, /?) 7^ 0. Choose /c sufficiently large so that if i = p^ + 1 , then the projection of 6 to L(a,i) is 0. We define g(i) G B to have the same projection in L(a, i) as 5, and to have 0 projection to all other L(£,t). Again, by hypothesis 2g(i) = g(i) < g yet 2g(i) yt g. This contradicts that G(M) \= L{g). Hence g = 6 G B = £ £ L 0 B ( 0 a n d 9 > °- I f Kt),b(v) > 0 with £ < 77, then 26(77) = Kv) < 9 Yet 26(77) ■£ g (a contradiction). Therefore 5 = 6(0 G B ( 0 for some ( H + // Lemma 4.7.9 Lei f,g 6 G(M). TAen G(M) |= 5(/,g) 7/ and on/y 2/ f,g>0 and f. g belong to the same B ( 0 Proo/: By the previous lemma, if G(M) \= 6(f,g), then there is c 6 B ( 0 for some ( £ Z + such that f,g < c. Hence / , 3 G P ( 0 Conversely, let f,g G B ( 0 with / , 0 > 0. Choose t so that / , 3 G B(£,0 and let c = /(£,<)• Clearly G{M) (= L(c) and / , $ < c. // Corollary 4.7.10 U< ez+ (B(O + \{0}) = {9 G G(M) : 9{g,g)}/6. So to show that G(M)' correctly:
= M, it remains to show that P behaves
Lemma 4.7.11 G(M) |= P(e(£,0,0),e(77,0,0)) i/ and on/y if M (= P(^,T?).
4.7. ELEMENTARY
THEORY (ABELIAN
t-GROUPS)
So
Proof: By the definition of P and Lemma 4.7.9 we see that G(M) |= P(e(£, 0,0),e(»7,0,0)) if and only if e # 7 ? & (3* > 0)[t 0 B ( 0 © B ( ) | ) & (VC)[(C * f,i7) -> * ± B(C)]]. Note that any such t must belong to A(£) ® A(rj) since it is orthogonal to fl(C) far all <_#£,»?. If G(M) (= P(e(£, 0,0),e(j7,0,0)) , without loss of generality assume that £ < ??. By Lemma 4.7.7, t = 6 + £ A(a, /J)s(a, /?) where 6 G B and (a <\J5 k M |= P(a,P)) for all a,/? for which A(a,/3) ^ 0; and since 2 G A(?)©A(77), we have that t = 6(f) +b(r}) + \(£,r})g(£,r]). Moreover, •M£>»?) # 0 because t 0 P(f) © B(rj). Hence, by definition, M \= P(f, rj). Conversely, if M (= P{^,rj), then we can take i to be (£, r?) and obtain that G(M) f= P(e(^0,0),e(r?,0,0)). / This completes (2) and hence the proof of the undecidability of the language of lattices for Abelian lattice-ordered groups. We could have replaced B by Q or Z or indeed any Archimedean ordered group. It is possible to show the undecidability of the theory of Archimedean lattice-ordered groups with atomic Boolean Algebras of filets by a more elaborate construction of G(M) and an appropriate Archimedean subgroup, but this requires considerably more ingenuity (see [Gurevich]).
Chapter 5 Archimedean Function Groups In this chapter we use geometric, topological and analytic ideas to better study Archimedean lattice-ordered groups. We show in particular that free Abelian lattice-ordered groups, finitely presented Abelian latticeordered groups, free products of Abelian lattice-ordered groups (in the class of Abelian lattice-ordered groups), and Archimedean lattice-ordered groups, can all be viewed as groups of functions on appropriate spaces.
5.1
Free Abelian lattice-ordered groups
In this section we obtain a concrete realisation for free Abelian latticeordered groups. Let X be a set and A be an Abelian lattice-ordered group. We say that A is the free Abelian lattice-ordered group on X if: (i) there is a one-to-one map 6 : X —► A; (ii) the ^-subgroup of A generated by X9 is A; and (iii) if ip is a map from X into any Abelian lattice-ordered group B, then there is an £-homomorphism x'■A -> B such that 6x — 4>It follows at once from the definition that for any set X there is at most one free Abelian lattice-ordered group on X to within ^-isomorphism. Let n be a positive integer. Consider the projections T\{ : Rn -> E given by: 7Ti(ari,...,xn) = z, (i = l,...,n). Let An be the ^-subgroup of 87
88
CHAPTER 5. ARCHIMEDEAN
FUNCTION
GROUPS
C(R",R) generated by 7ri,...,7r„. T h e o r e m 5.A [K. Baker] An is the free Abelian lattice-ordered group on n generators. Proof: Let w(z) = V, Aj w ; j (z) where z = («i,..., z„) and each w M (z) belongs to the free Abelian group on {zi,...,zn}. If w(tri, ...,7rn) = Q, then for all k = (fci, ...,&„) e Z", we have u/(7Tik, ...,7T„k) = 0. Hence w(ki,...,kn) = 0 for all (ku ...,kn) € Z n By Lemma 4.7.1, VXI...VI„IU(XI,
...,x n ) = 0
holds in every Abelian ^-group. So if B is any Abelian ^-group and ip : [TTX, ...,7rn} —> B is given by 7 ^ = 6j, say (i = 1, ...,n), then V extends to an £-homomorphism of A„ into B in the natural way. Thus An is the free Abelian -group on n generators./ For ease of notation (and because it is all we will need subsequently), we have restricted to the case that n is finite. The same proof shows that if K is any cardinal, then the standard projections from R* to R generate the free Abelian lattice-ordered group on K generators.
5.2
Finitely presented Abelian ^-groups
Let G be a finitely generated Abelian lattice-ordered group with gener ating set (ft, ...,„. Then the map 7r, K-> gi (1 < i < n) extends to a homomorphismof An into G such that An = G. Therefore, there is an ^-ideal L of An such that An/L = G. In the special case that L is finitely generated (as an -ideal), we say that G is finitely presented. Now by Corollary 2.3.9, u>\ = 0 & ... & wm — 0
if and only if
( \f |w,|) = 0.
That is, An/L is a finitely generated lattice-ordered group subject to a single defining relationship of the form W(TTI, ..., irn) = 0. We write An/L as (ni, ...,7rn : w(ir) = 0). If w € An, let Z(w) = {z e R" : w(z) = 0}, the zero set of w, or the variety given by w.
5.2. FINITELY
PRESENTED
ABELIAN
(-GROUPS
89
If w is a group word in iru ..., nn, say k
with each mk e Z, then Z(w) is the hyperplane J^k m,kxk = 0 and Z(wVQ) is just the closed half-space Y.krnk^k < 0. Thus for a general to £ An, say to(z) = Ai Vj Wjj(z) with each w,j(z) a group word, we have Z(K;) is a closed integer polyhedral cone in E"; that is, a finite union of closed convex polyhedral cones in Rn each of which has vertex 0 and is the intersection of a finite number of closed half spaces defined by homogeneous linear forms with integer coefficients. Furthermore, any closed integer polyhedral cone in Rn with vertex 0 is the zero set of some element of An. We will exploit this duality in the remainder of this section and also in the following two sections. Note that each closed convex polyhedral cone is the convex hull of a finite set of points: that is, {A!p (1) -I- ... + A r p (r) : A],...,A r 6 E + } , for some finite set of points { p ' ' ' , . . . , p ' r ' } in E n . Thus Z{w) is just the finite union of the convex hulls of finite sets of points. We now prove: L e m m a 5.2.1 [K. Baker] Let u,v,w G A+. If Z(u) n Z(w) C Z(v) n Z(w), then v\z(w) < muU(™) for some positive integer m. Proof: Clearly it suffices to consider the case that Z(w) is the convex hull of a finite set of points {p (1) , ...,p' r '} in R™. We first consider the special case that both u and v are linear functionals. Let r/> : En —> E2 be defined by ip(p) — (v(p),u(p)). Since u and v are non-negative on Z(w), it follows that ip(Z(w)) is contained in the first quadrant of the plane. Since Z(u) n Z(w) C Z{v) D Z(w), no point of rp{Z(w)) other than (0,0) lies on the n^-axis. Since the linear image tp(Z(w)) of Z(w) is the convex hull in the plane of {^(p(1^), ...,^(p ( r ^)}, there is a positive integer m such that a <mb for all (a, b) £ ip(Z(w)). Hence v\Z(w) < mu\z(wy In the more general case, let f\, ...,/t be the linear functionals which appear in the reduced expressions for u and v. For i,j = l,...,k if i ^ j , let H(i,j) = {p e R" : /,(p) < /j(p)}. Let V be the collection of intersections of any number of sets H{i,j) with Z(w), and £ be the subcollection of V consisting of those closed convex polyhedral cones K\ which are minimal in V with respect to containing some point of Z(w).
90
CHAPTER 5. ARCHIMEDEAN
FUNCTION
GROUPS
Clearly the union of all the K\ G £ is Z(w). On each A^ G £, u and v are linear functional; so by what we have already established, for each K\ G £ there is a positive integer m\ such that V\K1 < m i t t ^ . Since 5 is finite, the maximum of the mi's over all K\ G £ is an appropriate m. // We continue to use our convention: if w € .4 n , then An(w) is the convex ^-subgroup (and so £-ideal) of An generated by w. Corollary 5.2.2 [K. Baker] Let L = An(w) be an (.-ideal of An, and C — C(Z(w),R). Then An/L is i-isomorphic to the i-subgroup of C generated by the restrictions of ni,...,n„ to Z(w). Proof: Clearly, if / - g G L then (/ - g)(z) = 0 for all z G Z(w); i-e., f\z(m) - g\z(w)- Conversely, suppose that h\Z(w) = Q. Then Z{w) C Z(h). By Lemma 5.2.1, h G An{w) = L.// Corollary 5.2.3 [K. Baker] All finitely presented Abehan lattice-ordered groups are Archimedean. The above proofs use the geometry of the situation to obtain results about finitely presented Abelian lattice-ordered groups. This connection was made explicit in [Beynon 1977]. It makes use of the following result [Beynon 1975]: If 0 : Um —> En is a piecewise integral linear homeomorphism from R into K™, then 6 is defined by some ui,...,un G Am; that is, 9(x) = (ui(x), ...,u„(x)) for all x G Rm. m
T h e o r e m 5.B [The Baker-Beynon Duality Theorem]. Abelian latticeordered groups (7T],...,7rn : io(7r) = 0) and (7Ti,..., 7rm : v(ir) = 0) are l-isomorphic if and only if the closed integer polyhedral cones Z(w) and Z(u) are piecewise homomogeneous linear horneornorphic. Proof: Let p be an ^-homomorphism from An into Am. Let 7Ti, ...,7r„ be the natural projections of Rn onto R and tp} = p(nj) G Am (1 < j < n). Define S{p) : Rm -*■ R" b y x ^ (Vn(x),...,Vi n (x)). Note that if S(p) — S(a), then p(ftj) = O(TTJ) for 1 < j < n; so p = o. The map S is therefore one-to-one.
5.3. THE ISOMORPHISM
PROBLEM
91
If C and K are closed integer polyhedral cones in Rm and Kn, re spectively, then let v G Am and w G An be such that Z(u) = C and Z(w) = K. So for all y £ I", y € K if and only if g(y) = 0 for all g G AB(io). Suppose that 0 : Rm -»•ffi"is defined by Ui, ...,u„ e ^4m; that is, 0(x) = (u,(x),...,u n (x)) for all x G Km. Then 6>(C) C if if and only if gd{c) = 0 for all c G C and g G An{w). ^ Let T(0) : An -> A m be given by T(0)(/) = / o f l (/ € >4»). Hence 0(C) C if if and only if for all g G A n (iu), the element T(6)g of >l m , regarded as a function from Rm to 1 is identically 0 on the closed integer polyhedral cone C = Z(v). Bv Corollary 5.2.2, this is equivalent to T(0) G Am(v). Thus if n is the natural map from Am onto Am/Am(v), and v is the natural map from An onto An/An{w), then 0(C) C if implies that An(w) is contained in the kernel of //T(0). Since An is the free Abelian latticeordered group on n generators, there is a unique map T(0) : An/An(w) -» AjjM m (jj) such that /xT(0) = T(6)u. If 0 were another such map and T(0) = fjf), then MT(0) = pT{4>). Hence T(0)(/) - T(0)(/) G Am(v) for all / € 4 . Thus / ( 0 - 0 ) ( c ) = 0 for all c G C and / G 4 n . Therefore (0 —$)(c) = 0 for all c G C, so 0\c = (j>\c- Consequently there is a one-toone correspondence between £-homomorphisms between An/An(w) and i 4 m / i m ( « ) and maps defined by sets of n elements ui,..., un G Am. Since every piecewise homogeneous linear function is definable in this way (as stated above), the theorem follows./ Note that the Galois correspondence just established is a totally ex plicit algorithm to pass between closed integer polyherdal cones with vertex the origin (and such maps) and finitely presented Abelian latticeordered groups (and £-homomorphisrns between them).
5.3
The Isomorphism Problem
In the previous chapter (Corollary 4.7.2) we showed that Every finitely presented Abelian lattice-ordered group has soluble word problem. In contrast, we use the results of the previous section (especially the Baker-Beynon Duality Theorem) to prove
92
CHAPTER 5. ARCHIMEDEAN
FUNCTION
GROUPS
Theorem 5.C [Glass & Madden] The set of ten generator one relator Abelian lattice-ordered groups has insoluble isomorphism problem. Proof: As already noted, the polyhedra we consider in Rn can either be viewed as (topologically bounded) finite unions of finite intersections of half spaces determined by linear forms with integer coefficients, or equivalently as finite unions of convex hulls of finite collections of integer points. Moreover, this equivalence (between presentations) is effective [Weyl]: one can explicitly write down a piecewise linear homeomorphism with all the linear pieces of both the homeomorphism and its inverse having integer coefficients. Any abstract simplicial complex of dimension at most m can be ef fectively realised as a polyhedron in E 2m+1 (see, e.g., [Stillwell, page 22]). Hence, by Weyl's result, any two such realisations are equivalent if and only if the complexes are piecewise linear homeomorphic. Now [Markov] has proved that there is no algorithm to determine whether or not two compact simplicial complexes of dimension 4 are piecewise integral linear homeomorphic. Therefore, there is no algorithm which when given two integral polyhedra in K9 determines whether or not they are piecewise integral linear homeomorphic. So the same con clusion holds for closed convex integer polyhedral cones (with vertex 0) in K10. By the Baker-Beynon Duality Theorem, there is no algorithm to determine whether or not two arbitrary finitely presented Abelian lattice-ordered groups on 10 generators are £-isomorphic.// In contrast, one can certainly determine if a closed polyhedral cone with vertex 0 is piecewise homeomorphic to 0. Since (wi, ...,7rn : w(n) = 0) is £-isomorphic to {0} if and only if Z{w) = {0} if and only if w = 0, by Theorem 4.H we obtain Theorem 5.D There is an algorithm to determine whether or not an arbitrary finitely presented Abelian lattice-ordered group is l-isomorphic to the trivial group {0}.
5.4
Free products of Abelian ^-groups
Let G and H be Abelian lattice-ordered groups. Then L is the Abelian free product of G and H (written L = G *AH) if
5.4. FREE PRODUCTS
OF ABELIAN
i-GROUPS
93
(i) there are ^-homomorphisms a : G —> L and r : H —> L such that the ^-subgroup of L generated by Ga and HT is all of L; and (ii) for every Abelian lattice-ordered group A and ^-homomorphisms 4> : G —> A and ip : H ^ A, there is a unique ^-homomorphism 8 : L —> A such that aO = <j> and T9 = ip. There is a natural way to associate the free product of two finitely pre sented lattice-ordered groups with the external join of the closed integral polyhedral cones associated: Given G = (ici,...,ir„ : w(ir) = 0) and H = (ipi,.-.,ipm : v{i>) = 0), then obviously G*AH = (iru —,nn,tpi, ...,ipm '• w(n) = 0 = v(tp),Trtipj — V'j7ri (1 < i < n, 1 < j < m)). By Corollary 5.2.2, G is ^-isomorphic to the ^-subgroup generated by the restrictions of nu ..., irn to the closed integer polyhedral cone Z(w), and similarly for H. We wish to put the closed polyhedral cones Z(w) and Z(v) together as "freely" as possible. This is the external join Z(w) * Z(v) of Z{w) and Z(v). The purpose of this section is to establish that if we view m, ...,7Tn+m+i as the natural I>rojections on K n + m + 1 , then the restriction of ~K\,..., 7r n+m+i to Z(w) * Z(v) generates an ^-isomorphic copy of the Abelian free product of G and H. Caution: In contrast, the restriction of 7Ti, ...,7r n+m to Z(w) x Z(v) generates an ^-isoinorphic copy of G x H, the direct product of G and H. Let Kj be polyhedra in E n ' for j = 1,2. Then the external join of A'] and K2 is the polyhedron A'|*A'2 = {(Ax,/zyji) £ K"1 xR" 2 xE : x G ^ j
G K2,\,n
> 0,X+JJ, = 1}.
In the special case that K\ is a single point we call K\ * K2 a cone; in the case that K\ consists of exactly two points we call K\ * K2 a suspen sion; in all other cases we say that K\ * K2 is reduced. If a polyhedron P is piecewise linear homeomorphic to Kx * K2 we say that P factors as the external join of ATj and K2. We will use the following lemma: L e m m a 5.4.1 [Morton] Any non-reduced polyhedron factors uniquely as the external join of a ball or a sphere with a reduced polyhedron. More over, the decomposition of any reduced polyhedron as a repeated external join of indecomposable factors is unique up to the order in which they are selected.
94
CHAPTER 5. ARCHIMEDEAN
FUNCTION
GROUPS
Theorem 5.E [Madden] Let G - (iru ...,nn : w{n) - 0) and H = (ipi,...,ipm '■ v{ip) = 0). Then the free product ofG and H is (-isomorphic to the (.-subgroup of C — C(Z(w) * Z(v),R) generated by the restrictions o/7r 1 ,...,7r n+m+1 to Z(w)*Z(V). Proof: For any set A contained in Rp, let A'* = {A(x, 1) : A G R + , x e A } ; so A* is a subset of Rp+1 We can "extend" any function <$>: X —> Y to 0* : A" -> Y* by 0*(A(x, 1)) = A(<£(x), 1). The key point is that if C and K are polyhedra, then (C * K)* is pieccwise homogeneous linear homeomorphic to C'x/C*, the map from R" xK m xExR to Kn x I x K m xK being provided by (x, y, A, fj.) t-> (x, fj, - A, y, A). So the result follows at once by the Baker-Beynon Duality Theorem./
5.5
Kaplansky's Example
The purpose of this short section is to show: Example 5.5.1 There is an Archimedean lattice-ordered group that is not an (.-subgroup of any cardinal product of copies of R. Note that if G is ^-isomorphic to an ^-subgroup of a cardinal product of copies of R and the projection to each coordinate has non-zero kernel, then (since R has no non-trivial f-ideals) the intersection of all maximal ^-ideals of G must be {0}. Let X be an extremally disconnected space and G = D(X) (see Example 1.3.10). Then C(X) is an £-ideal of G. Moreover, if z £ X and Gz = {g e G : g(z) = 0}, then Gz is an £-ideal of G. Note that C(X) + GZ comprises all elements of G that have finite value at z. Let Uz be a meagre open subset of A containing z. Let / be any function from X\UZ into R, and f(x) = oo if x € Uz. Then / e G but / 0 C(X) + G2. So if M is any maximal £-ideal of G, then for each z € X, either there is / z £ M such that fz(z) = oo or M C C{X) + Gz (and so M = C{X) + Gz). Hence, in either case, there are fz G M and an open subset / , of X containing z such that fz(x) > 1 for all x 6 Iz. Since A is compact, there is a finite open subcover {Izl,...,IZn} of A. Let / = fZl V...V/ 2 n . Then / e M and / exceeds the constant function 1. Therefore I belongs to every maximal £-ideal of G. Hence the intersection of all maximal ^-ideals of G is not (0).
5.6. BERNAU'S
THEOREM
95
A similar example is provided by the additive group of all functions on [0,1] of the form g(x) + T,\Zoai/(x ~ h)2 where g is continuous on [0,1]. It is a lattice-ordered group if h > 0 provided that h(x) > 0 for all x £ [0,1] at which h(x) is defined. In place of fz we use l / ( x - z)2 and obtain an interval Iz on which fz{x) > 0. Then l e M for every maximal ^-ideal M.
5.6
Bernau's Theorem
In Example 1.3.10 we mentioned the set D{X) of extended real-valued functions on an extremally diconnected topological space X and noted that under addition and the pointwise ordering D(X) is an Abelian lattice-ordered group. It is clearly Archimedean. We wish to consider this example more closely and show that such lattice-ordered groups are universal for Archimedean lattice-ordered groups. Specifically T h e o r e m 5.F [Bernau] Given an Archimedean lattice-ordered group G there is an (.-embedding i : G —> D(X) the vector lattice of almost finite continuous functions on a Stone space X = S(B) where B is the Boolean algebra of polars in G. Moreover L preserves all joins and meets and LG is a large subgroup of D{X). (The definition of "large" is given later in the section.) The main properties of extremally disconnected spaces that we will need are (a) the closure of every open set is clopen (i.e., closed and open); (b) if A and B are disjoint open subsets of X, then their closures are disjoint; (c) every real-valued function defined and continuous on a dense sub set of A' can be uniquely extended to a continuous extended real-valued function on X that is almost finite; that is, the extension may take the values ±co but is real-valued on a dense subset of X. Note that if / and g are almost finite continuous functions on an extremally disconnected set X, then so is their sum. So the set of all such functions, D(X) is an Abelian group under addition. Define g > f ifg(x) > f(x) f° r all x G A such that f(x) and g(x) are both finite. With respect to this ordering, D(X) is a lattice-ordered group: if g £ D(X)
%
CHAPTER 5. ARCHIMEDEAN
FUNCTION
GROUPS
and 0 is the function 0 : X —> 0, then g V 0 is clearly continuous and almost finite and so belongs to D(X). With respect to this ordering, D(X) is Archimedean. Moreover, / ± g if and only if / and g have disjoint supports. Therefore, f±L — {g G G : supp(g) C supp{f)}. By Theorem 3.B, if G is an arbitrary lattice-ordered group then B, the set of polars of G, forms a complete Boolean Algebra. The set X = S(B) of all ultrafilters on B is the Stone space for B. It is a compact Hausdorff extrcmally disconnected topological space, the basic open sets being the sets S{M) = {U : M eU, U an ultra filter on B}. We will write S(g) for the compact open subset S(g-Ll) of X corresponding to gx±; i.e., the set of all ultrafilters on B containing gLL. If Y C X, we will write cl(Y) for the closure of Y in X; i.e., cl(Y) is the intersection of all closed subsets of X that contain Y Since X is the Stone space for B, we have (I) S(g) — S(\g\) = S(gn) for all non-zero integers n; (II) S(f)nS(g) = S(\f\ A \g\); (III) 5 { | / | M ) = 5 ( | / | V | 5 | ) ; (IV) S(V{\gt\ : i € /}) = d(U{S(|
5.6. BERNAU'S
97
THEOREM
In the proof of Holder's Theorem we let Q(g) = {m/n : m,n G Z + , n ^ 0, cm < gn}. Analagously, for G an Archimedean lattice-ordered group, g e G+ and x e X = S(B) we define : m, n G Z, n > 0, x € d((J{S((mCj - ng) + ) : i £ /})}
g(x) = inf{m/n
where m / ( 0 ) = oo. Note: if g G G+, then for a l i i G / we have S((mci - n^)"1") = 0 if m < 0; hence #(x) > 0 for all x G X. Let Xj be an arbitrary element of S(c;); i.e., x* is an ultrafilter on B containing c/- 1 . Then for all j G /, Cj(xi) = inf{m/n
: m G Z,n G Z+,Xi G c/(jJ{S((mc*-nc.,) + ) : A; G / } ) } .
But \{ k ^ j then ,
v
,,
( 0 [ mck
3J
*
ifm<0 if m > 0
since c^ A Cj = 0. Moreover, S(c,-) n 5(c t ) = 5(0) = 0 if i / j . Hence x t G d((J{S((mc* - nCj)+) : k G /}) whenever m > 0 and i ^ j . Thus if i ^ j then Cj(xi) — inf{m/n
: m,n G Z + } = 0.
If i — j then x, G d(|j{5((mcfc - n Cj ) + ) : k G I}) whenever m > n. Thus Ct(xz) = inf{m/n
: m,n G Z + , m > n} = 1.
Hence c,(x,) = (5«j where Sitj is the Kronecker delta; and d(x) jV
'
=
i 1 if*eS(e,) | 0
otherwise.
If m i / n i > m 2 / n 2 and g e G+, then for each z G / SdmxCt - nxg)+) = S((min 2 Ci - n 1 n 2 ff) + ) 2 5((m 2 n 1 c i - nin2.g) + ) = S ^ m ^ - n 2 5) + ). Thus x G cl{\J{S{{mCi - ng)+) : i G /}) for all m/n > g(x) and x £ c/((j{5'((mci - ng) + ) : i G /}) for all m / n < (x).
98
CHAPTER 5. ARCHIMEDEAN
FUNCTION
Lemma 5.6.2 For f,g £ G+ the functions f,g properties: (i) f + g{x) = f{x) + g{x); (ii) g is continuous on X; (Hi) g(x)
=OJ/I g
GROUPS
satisfy the following
S(g);
(iv) if g = A{9k ■ k £ K}, then g = /\{gk ■ k £ K} in the lattice oS extended real-valued Sunctions on X and g(x) = in}{gk{x) : k € K) for all x in the complement of a set of 1st category. Proof: (ii) Since X is extremally disconnected, cl(\J{S((mcl — ng) + ) : i £ /}) is open for all m,n. Now for any real r, {x £ X : g(x) < r) = \j{cl{[j{S{(mcl - ng)+) : i e I}) : m/n < r}. Hence {x £ X : g(x) < r} is open in X for all r £ K. Similarly {x £ X : g(x) < r} — f]{cl(\J{S((mCi - ng) + ) : i £ /}) : m/n > r} is closed in X. Thus g is continuous for all g € G+ (iii) For any n > 0, S(c;) = S(c{ — ng + ng) C S((ci — ng)+) U S(g). Hence X = 5(g) U U{5((mc ! - np) + ) : i £ I } . So if x $ S{g), then x £ \J{S{(rncl - n§)+) : i € / } for all n > 0. Thus #(x) < 1/n for all n £ Z with n > 0. Consequently #(x) = 0. (i) Suppose that m,\jn\ > f(x) and m2/n2 > g{x). Then x £ d ( U { 5 ( ( m l C i - m / ) + ) : i £ /}) n d(U{5((m 2 c t - n 2 g)+) : i £ /}) = d(U{S((m l C < - m / ) + ) n S((7n2Ci - n2g)+) : i £ /}) = d(U{5((Tnin 2 c, - nin2S)+ A (m2niCj - nin 2 5) + ) : i € I}) since Cj A c3 = 0 if i ^ j . Now for any a, 6 £ G we have (a + 6) + > 2(o A 6)+ = (2a) + A (26)+ > a + A 6+ since (2d)+ =d+\d\>d. Thus c
'(U{^'(( m i n 2 c i
_ n n
\ 2f)+
A (m2niCi - nin2g)+)
: i € J}) C
c
' ( I J { 5 ( ( ( m i n 2 + n i m 2)ci - n i n 2 ( / + #)) + ) : i £ /}). Therefore i £ c/(U{-5(((min2 + nim 2 )cj — n\n2{S + g))+) : i £ / } ) . Hence / 4- (x) < (m-i/ni) + {m2/n2). Consequently, / ( x ) + #(x) >
/Tff(x). This inequality obviously holds if either /(x) = oo or g(x) = oo. Now suppose that mi/ni < /(x) and m2/n2 < g{x). Then / + g(x) > (mi/ni) + (m2/n2) by (IV) and similar computations. Thus / ( x ) + (x) <
/Tg(x).
5.6. BERNAU'S
99
THEOREM
(iv) Since g = f\{gk : k G K}, we have (m,Ci-ng)+ = \/{(rn.cl-ngk) fc G A'} for all m,n. By (II) we have {J{S((mci
- ng)+) : i e 1} = \J{S((ma
+
:
- ngk) + ) : i € / , * € i f } .
If g(i) ^ A{5it(2;) : fc € K}, then there exist positive integers m,n such that #(x) < rn/n < A{<}k(%) '■ k G K}. From the above we see that x e d(\J{cl(\J{S((rnc,
- ngk)+) : k G K}) : i G 7})\
Ui^llmc, - nfl)+) : i 6 /}. This set is nowhere dense and since Q is countable, we deduce that g(x) — /\{gk{x) '■ k G K} for all x in the complement of a set of 1st category. But X is Hausdorff and compact and so is a Baire space. Thus every subset of X that is 1st category has empty interior. Hence if / is continuous on X and f(x) < gk(x) for all x G X, k G K, then f(x) < g(x) for all
x e x. a
Corollary 5.6.3 For all f,g € G + , _____ fVg(x) = max{f(x),g(x)} and f A g(x) =
min{f(x),g{x)}.
Proof: Since (/ - (/ A g))+ A (g - (/ A p))+ = 0, it follows that S((f ~ (/A_ff)) + ) n 5(( ff - (/ A 5 j n = 0. Therefore, for alllx G X, e i t h e r / - ( / A o)(x) = 0 or g-{fAg)(x) = 0. Hence / A g(x) = min{f(x),g(x)}. Since (fVg) - ( / A J ) ( J ; ) + 7~A~g(x) = f~Vg{x) for all x € X, the desired equality holds for / V g(x). // Observe that, by the proof, we have for all x G X, either g+(x) = 0 or g~ (x) = 0. We use this to extend the definition of g from g G G+ to For any g e G, define (x) = 5 + (z) - 5"(^)Corollary 5.6.4 For all g G G the functions g are continuous on X. Moreover, f - g(x) = /(x) - g(x) whenever f(x) - g(x) is defined. If g = \J{gk : k G K} in G, then g = \f{gk '■ k G K) in the lattice of extended real-valued continuous functions on X, and dually for mfima.
100
CHAPTER 5. ARCHIMEDEAN
FUNCTION
GROUPS
Proof: Continuity is obvious. The result about / — g follows from the easily established identity (/ - g) + + f~ + g+ = (f - g)~ + f+ +g~, and that about suprema and infima follows easily from Lemma 5.6.2. / L e m m a 5.6.5 Let G be an Archimedean lattice-ordered group, Then (a) for each g G G, g is almost finite; (b) if g{x) = 0 for all x £ X, then g — 0; (c) supp{g) = S(g) for all g G G; (d) f < g if and only if f < g. Proof: If / < g, then g(x) — f(x) — g - f(x) > 0 by the previous lemma. Also, if x G S{g~) then g{x) < 0; so if g(x) > 0 for all x € S(g~) then S(g-) = 0. Thus g > 0. (a) Let Y = {x € X : \g(x)\ = oo} and V = int{Y). Then Y is closed so V is clopen. Thus there exists M € B such that S(M) C V If M f {0}, let 0 < / € M. Then 0 < / A cx < f for alH_G I and S(f) C K Hence 5 ( / A c,) C V for all i £ I. Now n ( / A c , ) ( i ) < nci(x) = n for all x € S(f A Cj) C 5(Cj). Thus for all x G 5 ( / A C{) C V, (|ff| - ^ ( / A c , ) ) ( i ) = oo. But (|g| - n ( / A c,))- < n(f A c,); so 5((|ff| - n(f A C i ))-) C S{f A c t ). Therefore (| 5 | - n{f A c,))" = 0 and thus n(f A d) < \g\ for alii G / and n G Z + Since G is Archimedean, / A Ci — 0 for all i £ / . This contradicts the maximality of {ct : i G / } . Consequently, M = {0} and V = 0; i.e., 5 is almost finite. (b) If <7(x) = 0 for all x € X, then for all i G / and n G Z + we have (Ci - n{\g\ A Ci))(x) = Ci(x) - (n{\g\ A Cj))(x) = C,(x). But {a-n{\g\ A a))-0 in G suchjthat / A \g\ jt 0 and S{f) C {x G 5( 5 ) : g(x) = 0}. This implies that (/ A |^|) = 0 for all x G X contradicting (b). (d) If/ < g, then ((g^J)-)(x) = ((/^)+)(s) = mai{f^(i),0} = 0 for all 1 G X. Hence (5 - / ) " = 0 by (b), so g - f > 0. // Let H be a subgroup of a lattice-ordered group G. We say that H is large in G if for each g G G+ with 5 / 0, there is /l € if such
5.6. BERNAU'S
THEOREM
101
that 0 < h < ng for some positive integer n; i.e., if 0 < g € G, then HnG(g)?{0}. We will establish two lemmata which immediately imply Bernau's Theorem. Lemma 5.6.6 Let G be an Archimedean lattice-ordered group. There exists a Stone space X and an i-isomorphism L of G onto an i-subgroup LG of D(X) such that (a) i, preserves all suprema and infima that exist in G; (b) L is an i-isomorphism; (c) LG IS a large subgroup of D(X); and (d) if G is divisible, then LG is dense in D(X) as a partially ordered set. Proof: Let A' = S{B) and define t : G -> D{X) by eg = g[x). We have already established (a) & (b); and (d) is an immediate consequence of(c). To prove (c), let 0 < / € D(X). Then there is an open set U and a positive real number e such that f(x) > e for all x £ U. Let n e Z + be such that ne > 1. As before, there are gx e G + and j £ I such that g\ A Cj 7^ 0 and S(g\ A c,) C U. Choose any such j and let g = gy A c,. Now i(g)(x) < L(CJ)(X) < 1 and so 0 < i.g < nf as required. / To complete the proof of the theorem, we need merely prove: Lemma 5.6.7 Let X' be a Stone space and D(X') be the lattice-ordered group of all continuous almost finite functions on X' Let G be an Archimedean lattice-ordered group and g : G —> D(X') satisfy the same properties as L in the previous lemma. Then there is an onto homeomorphism T : X -> X' andO < f e D(X') such that (gg)(Tx) = f(Tx)(ig)(x) for all x € X and g £ G for which the product /(rx)(tg)(x) is defined. Proof: Let M € B and T/>(M) = cl(\J{S(di) : h e M}). Then ip(M) is clopen and so compact. Note that ip{M) = 0 if and only if M = {0}. Clearly, ip(ML) C X\ip{M) where we have used ± to denote the complement in the Boolean Algebra B. If ip(ML) / X\ip(M), then there is a non-empty subset U of X\[ip(M) U rp(ML)}. By the usual argument, there is V C U with V clopen. Let h be the characteristic
102
CHAPTER 5. ARCHIMEDEAN
FUNCTION
GROUPS
function of V. By (c), there is g G G such that 0 < ug < nh for some positive integer n. Hence S(tg) D S(if) = 0 for all / 6 M. Thus g G ML Therefore S(tg) n ^ ( M 1 ) / 0, contradicting that S{tg) C V This shows that tp{ML) = X\ip(M). Now let M, TV G B and 5 G M n N. Then S(iff) C ip(M) n ^(iV), so ?/>(M n / V ) C ^ ( M ) n >(#)• By a similar argument to the one above we obtain that every open subset of tp(M) n ip(N) meets ip(M n N); therefore ip{M n TV) = V(M) n V(-W). Applying these two results to ML and TV1 we obtain that ^ ( M V N) — Tp{M)Uip{N). Hip(M) = ip{N), then ^(AfnJV 1 ) = 0 and MC\NL = {0}. Similarly 1 M n TV = {0} so M = N. Finally, if C is a clopen subset of A', then C* = {g e G : 5(tff) C C} is obviously a polar in G and j//(C*) = C. Thus tp is a Boolean Algebra isomorphism of i? onto the set of all clopen subsets of X. Similarly we can defineX' such that {M) — Ttp(M) for all M € B. The proof of this is obtained by observing that each x € X is the unique limit of the filter base of all clopen subsets of X which contain x. The image of this filter base under <j>il)~x has a unique limit point which we define to be TX. It is easy to prove that r is one-to-one and that both r and T _ 1 are continuous. Now consider the functions QCi (i € I) on X' Define /j on X' by: f M /Uyj
_ f {QC]){y) if V e S^c,) for some j € / \ 0 otherwise.
Then /1 is well-defined on a dense subset of X' (since S(gCi) n S(gCj) = S(gCi A gCj) = S(g(cl A c,)) = 0 if i / j) and is continuous. Since A" is a Stone space, /1 has a unique continuous extension to the whole of X' which we denote by / . Then / 6 D(X') since all pc; are almost finite. Moreover, supp(f) = X' (if not, then by (c) there is g G G such that gg is strictly positive and less than a positive integer multiple of the characteristic function of X'\supp(f); then g A ct = 0 for all i £ I). It remains to show that (gg){Tx)= }{Tx)(ig){x)
(*)
whenever the right hand side of (*) is defined. So let x G X and g e G. Since U{5'(c!);z G / } is dense in X, we may assume that
5.7. THE
SPECTRUM
103
x G ^(cj) for some j G /. It is clearly enough to prove (*) for g > 0. If m/n > {ig)(x), then x G [){S((mCi - ng)+) : i £ I}. Since x £ S(Cj), we have x £ S((mCj - n#) + ). By construction TX £ S(g(mCj - ng) + ) and 0(mcj - ng)(rx) > 0. By the definition of / , (mf - ngg)(rx) > 0, so mf(rx) > n(gg)(rx). Therefore (m/n)f(rx) > {gg){rx); i.e., (tg){x)f(rx) > (gg)(rx). The reverse inequality follows similarly using
r/s < {tg){x). II
5.7
The Spectrum
Let G be an Abelian lattice-ordered group. The spectrum of G, de noted Spec(G) is the set of prime ^-ideals of G other than G. That is, Spec(G) = U{G)\{G}. If A C G, let S(A) = {P 6 Sj?ec(G) : ,4 g P} and ff(A) = 5pec(G)\5(,4) = {P e 5pec(G) : ,4 C P } . We will write 5(y) and H(g) for 5({}) and H({g}) respectively. Note the following obvious facts: (i) 5(0) = 0 and S{G) = Spec{G). (ii) S(f V g)- = S(f) U 5( 5 ) for all / , G G+. (in) 5 ( / A 5 ) = 5 ( / ) n S(g) for all / , S e G + (iv) S(C n £>) = 5(C) n 5(D) for all £-ideals C, P> of G. (v) 5(L) = \J{S(Lt) : i £ 1} where L is the (-ideal generated by the family {L t : i £ / } of ^-ideals of G. By (i), (iv) and (v), Spec(G) becomes a topological space with {S(L) : L £ C(G)} as a basis of open sets. Lemma 5.7.1 Let G be an Abelian lattice-ordered group and L be an (.-ideal ofG. Then L = fl{P G 11(G) : P 3 L } . Proof: If g G G\L, then there is a value V of g containing L. Now V G n(G) since values are prime (Lemma 3.3.1). Hence g $ (~){P G I f ( G ) : P D L } . // Corollary 5.7.2 Let G be an Abelian lattice-ordered group. Then the map L H* S(L) from the set of (-ideals of G into the set of open subsets of Spec(G) is one-to-one.
104
CHAPTER 5. ARCHIMEDEAN
FUNCTION
GROUPS
We close this short section with the following observation: Lemma 5.7.3 Let G be an Abelian lattice-ordered group and C\ and C2 be prime (.-ideals of G. If C\ and C2 are incomparable (with respect to inclusion) then there are disjoint open sets Ui and U2 with Cj € Uj for J' = l,2. Proof: Let c € C 1 + \C 2 and d £ C£\CV Let / = d - (c A d) and g = c— (cAd). Then f & C\ and g & Ci because c A d € C\ n C%. Since f -L g and C\ € S(f), G2 S S(g), the required disjoint open sets are furnished by S{f) and S{g). //
5.8
Hyperarchimedean ^-groups
A lattice-ordered group is said to be hyperarchimedean if all of its £homomorphic images are Archimedean. Example 5.8.1 An Archimedean lattice-ordered group that is not hyper archimedean. Let G = rinez+ K a n d L be the £-ideal of all eventually constant sequences in G; so G/L is an ^-homomorphic image of G. Let f,g £ G be defined by f(n) = 1 and g(n) = n for all n 6 Z+ Then / + L
5.8. HYPERARCHIMEDEAN
^-GROUPS
105
Proof: (i) => (ii) Let P / G be a prime £-idcal of G. Then G / P is an ordered Archimedean group and hence has no non-trivial convex ^-subgroups. Thus P is maximal. (ii) =>• (iii) For each g G G, if g ^ 0, let P 9 be a value of g. By assumption, P 9 is maximal, so by Corollary 4.1.4, G/Pg is £-isomorphic to a subgroup of R. Hence G is ^-isomorphic to an f-subgroup C of all real-valued functions on X = G\{0}. If 0 < f,g G C and for each integer n there is xn G X such that 5(x„) > nf(xn) > 0, then g does not belong to the prime ^-ideal of C generated by any cover of any value of / . Thus the values of / are not maximal, a contradiction. (iii)=^ (i) It is enough to show that C is hyperarchimedean. Let 4> be an ^-homomorphism from C onto L and 0 < f(j> -C gcj> with 0 < / , g G C. By hypothesis, for some positive integer n we have g(x) < nf(x) for all x G supp(f). Hence nfAg = (n+l)fAg. So n(f(j>)Ag= (n+l)(fcf))Ag<j>. Thus / ^ = 0. Therefore L is Archimedean. / By using property (iii) we immediately obtain: Corollary 5.8.3 Every £-subgroup of a hyperarchimedean lattice-ordered group is hyperarchimedean. Corollary 5.8.4 // G is a lattice-ordered group of real valued functions each with finite range, then G is hyperarchimedean. \ o t e that (ii) can be rephrased as: Spec(G) comprises the set of maximal ^-ideals of G. By Lemma 5.7.3, we deduce Corollary 5.8.5 If G is a hyperarchimedean lattice-ordered group, and C i , C 2 G Spec(G) are distinct, then there are disjoint open sets Ui,U2 C Spec(G) with C3 G U3 (j = 1,2).
Chapter 6 Soluble Right Partially Ordered Groups & Generalisations In this chapter we study partially ordered groups that are nilpotent and soluble (as groups). This is distinct from imposing the properties in the category of partially ordered groups since we are not requiring that the subgroups that comprise the lower central or derived series be convex. If G is a nilpotent lattice-ordered group with each term in the lower central series a convex £-subgrDup, we will say that G is l-nilpotent. Similarly we can define ^-soluble.
6.1
Nilpotent lattice-ordered groups
We will use the term "locally nilpotent" in the group-theoretic sense; that is, every finitely generated subgroup is nilpotent. As we saw in Example 1.3.23, every nilpotent torsion-free group can be made into an ordered group. The purpose of this section is to prove that for any lattice ordering, such groups are already residually ordered. Indeed: T h e o r e m 6.A [Kopytov 1975] Locally nilpotent lattice-ordered groups are residually ordered. To prove the theorem, we require a lemma. An element g £ G + \{1} is called a weak order unit if g is not orthogonal to any / £ C\{1}; i.e., 5 X = {1}107
108
CHAPTER 6. SOLUBLE RIGHT P.O. GROUPS
L e m m a 6.1.1 Let G be a lattice-ordered group with (-monolith TV ^ {1}. Then (i) every non-identity element of d(G) + is a weak order unit in G, (ii) Ci(G) is an Abelian ordered group, and (iii)ifNnh(G) ± {1}, then TV = {g G G : (3c G Nn&(G))(\g\ < c)} and TV fl Ci(G) is Archimedean. Proof: (i) Let C = (i(G) and suppose that g G C+, f € G with / , g / 1 and / l j . Since j 6 C, we have / G gL < G and g G c ^ 1 < G. Since g1 D gL1 — {1}, this contradicts that TV is non-trivial. (ii) Clearly, C is an ^-subgroup of G. Hence C is an Abelian ^-group in which every strictly positive element is a weak order unit. Thus C is totally ordered. (iii) Let L = {g G G : (3c G NnC)(\g\ < c)} C AT. Since TV n C < G, it follows that L is an £-ideal of G. By the minimality of TV, L = TV (since L D TV n C ^ {1}). If -<4 were a non-trivial proper convex subgroup of TV n G, then L = {5 G G : (3a G A)(\g\ < \a\)} D A would be a nontrivial £-ideal of G contained in TV. This is a contradiction. By Lemma 4.1.1, TV n G is Archimedean./ We are now ready to prove the theorem. Proof: Since every locally nilpotent lattice-ordered group is a subdirect product of subdirectly irreducible locally nilpotent lattice-ordered groups, it is enough to prove the theorem for an arbitrary subdirectly irreducible locally nilpotent lattice-ordered group G. Since a latticeordered group is residually ordered if and only if all its finitely generated ^-subgroups are (by Lorenzen's Theorem), we may also assume that G is finitely generated (as a lattice-ordered group). Let gl} ...,gn be genera tors for G. Since G is a distributive lattice (Lemma 2.3.5), each element of G can be written in the form Vt Aj fij where each f\j is in the sub group A generated by {gi,..., gn}. Now A is a finitely generated subgroup of G and hence is nilpotent; so A has non-trivial centre. But any ele ment of (\(A) clearly commutes with any element of the form just given. Therefore C = Cx(G) D Ci(^) # W Since G is subdirectly irreducible, it contains a unique minimal nontrivial £-ideal, the ^-monolith, TV. Let b G TV with 6 / 1 . Let H = {A, b), the subgroup of G generated by A and 6. Then H is nilpotent and b G Nr\H
6.2. THE ENGEL
109
CONDITION
for some h G H (and [b,h] e NnQiH) as NnH1 (since c" 1 = g_1[6,#-1]). Now 5 _1 b# = 6c so g~nbg" = 6cn for all integers n. Since b G AT, by the lemma there is 2 e j V f l C such that z _ 1 < b < z. Since TV n C is an Archimedean ordered group, z < c m for some positive integer m. Then 6cr" > 1; so = bcjn >Y T h e refore b > 1. But c = [6,5] 6 C, so [ft"1, ff"1] = g-mbgm 65[6,9] (65) _1 = c. Thus 1 ^ [fe-1,ff_1] G TV n C. By the same argument, 6"1 > 1, a contradiction. Hence N DC — N. If G were not residually ordered, then, by Corollary 3.8.2, there are f,g G G with g J_ f~lgf. Since AT = TV n C is a totally ordered Abelian group and 2V < G, G(p) n J V = {1}. Thus G TV1, so JV 1 is non-trivial; it is an £-ideal since TV is an ^-ideal. Since N is the ^-monolith, it follows that NL D N. This is the desired contradiction./
6.2
The Engel Condition
We continue to write [f,g,g] for [[f,g],g], [f,g,g,g] for [[f,g,g],g], etc. A group G is said to be c-Engel if [/, g, ■ ■ ■, g] = 1 for all / , g G G. Clearly c
every nilpotent class c group is c-Engel. It is unknown if the converse is true for torsion-free groups, or even if it is true for right ordered or right orderable groups (when c > 4). However: T h e o r e m 6.B [Medvedev 1988] Locally c-Engel lattice-ordered groups are residually ordered. Corollary 6.2.1 [Kim & Rhemtulla] Locally c-Engel lattice-ordered groups are nilpotent. In order to prove the theorem, we will make use of the following easy lemmata:
110
CHAPTER 6. SOLUBLE RIGHT P.O. GROUPS
Lemma 6.2.2 Let G be a group and f,g € G. If {f,g 'fg, ...,g cfgc} freely generate a free Abelian subgroup of rank c+1, then G is not c-Engel. Proof of Lemma: Since [/, g, ■ ■ ■, g] can be written as a product of the c
elements f±l,{g~1fg)±\..-,(g~cfgc)±1 and f±l occurs only once in the _1 product (/ occurs if c is even, / if c is odd), we get [/, g, ■ ■ ■, g] / 1. / c
Recall that C(g), the centraliser of g, is the subgroup of elements that commute with g. Also recall that a subgroup H of a group G is isolated if (/" 6 H for some g € G and integer n ^ 0 implies that g € H. The following lemma is reminiscent of Corollary 2.1.5. Lemma 6.2.3 If G is a torsion-free c-Engel group and f € G, then C(f) is isolated. Proof: Suppose that [/, gn] = 1 for some integer n / 0; without loss of generality, n > 0. We must prove that [f,g] = 1. Now [[/, ],<7n] = [ / " V / 3 , . 9 n ] = [3-1/3>ffn] = 9~l[f,9n]9 = !• Continuing we obtain that [[/,g,---,gj,ff n ] = 1. Since G is c-Engel we have [/,g, ■ ■ ■,g] = 1; c
c-l
so we may assume that [/, g,g) = 1 without loss of generality. Now i = [/,"] = [f,9}{g-V,9}g)-(g-{n-l)[f,9}yn-i)
= \f,g]...\f,g]
= [/,.g]n
Therefore, [/, g] = 1 since G is torsion-free. / Proo/ of Theorem: Let G be a locally c-Engel lattice-ordered group. If G is not residually ordered, then c > 1 since Abelian groups are residually ordered. We will use orthogonality to obtain a contradiction. Since G is not residually ordered, by Corollary 3.8.2 there are f,geG such that g~*fg A / = 1 with / > 1. Since (/, g) is c-Engel, by Lemma 6.2.2 there is a positive integer m = m(f,g) with 1 < m < c — 1 such that {f,g~lfg,..,g~rnfgrn} is a maximal pairwise orthogonal set (and so generates a free Abelian group of rank m + 1). Choose / , g £ G so that m(f,g) is maximal and let d = m + 1 > 2. Let w = f A g~dfgd Then 10 ^ 1 since f,g~lfg, ■■■,g~dfgd cannot be pairwise orthogonal (by the maximality of m). Since (w,gd) is c-Engel, there is c0 € Z with
6.2. THE EN GEL
111
CONDITION
0 < c0 < c - 1 such that z = [w,gd,--^gd]
/ 1 but [z,gd] = 1. By
CO
Lemma 6.2.3, [z, g] = 1 Now z can be written as a non-trivial product of elements u;*1 conjugated by gd,g2d,...,gc°d, and these are all orthogonal to w±l conjugated by gd+\g2d+1, ■■.,gcod+1. Hence z 1 g~lzg with z ^ 1. This contradicts [z,] = 1. Thus G is residually ordered. / In trying to extend the above proof to right ordered groups, the stum bling block is to obtain an appropriate element z (in the absence of or thogonality). As is customary, we write G' for 71(G), the subgroup generated by all commutators of elements of G. The derived series for G is defined inductively as the series G\ G",... where Gin+1) = [G (n) , G (n) ] is the com mutator subgroup of G generated by all [/, g] with / , g e G*n) A group G is soluble class m if G' m ' = {1}. In order to prove the Corollary we will need two further facts: L e m m a 6.2.4 If G is a finitely generated c-Engel group, H is a normal subgroup of G and G/H is cyclic, then H is finitely generated. Hence all the terms of the derived series of G are finitely generated. Proof: Let g € G be such that Hg is a generator of G/H. Since G is finitely generated, there are hi, ...,/ifc S H such that G = (g,hi,..., hk) and H is the normal subgroup of G generated by all conjugates of h\,..., hk- For each x,y £ G, an easy induction on r establishes that the subgroup generated by x, [x, y), [x, y, y], ...[x, y, ■ ■ ■ ,yj is precisely the subr
group generated by x,y~lxy, y~2xy2,...,y~rxyT Since G is c-Engel, the subgroup generated by {y~nxyn : n € z} is precisely the subgroup gener ated by x,y~lxy,y~2xy2, ...,y~f-e~x^xyc~1. Hence for each i — 1,..., k, the subgroup H% of G generated by {g'nhlgn : n € Z} is finitely generated. Let L be the finitely generated subgroup generated by this finite union (i — 1,..., k) of elements of H. Then g~lLg — L and since hi,..., hk € L, it follows that NG{L), the normaliser of L in G, is all of G. Hence L = H so H is finitely generated. Since G is finitely generated, G/G' is a finitely generated Abelian group. By the Fundamental Theorem of Abelian Groups, there is a finite sequence of normal subgroups LQ D Li D ... D La such that L0 — G,
112
CHAPTER 6. SOLUBLE RIGHT P.O. GROUPS
Li = G' and Li/Li+i is cyclic for i = 0,..., d— 1. Each Li is c-Engel so by induction using the first part of the lemma, each Lt is finitely generated. Therefore G' is finitely generated. Since G" is c-Engel, G" is finitely generated by what we have just shown. Continuing in this way we get that all the terms of the derived series are finitely generated. // The second fact is far deeper and beyond the scope of this book: Lemma 6.2.5 [Zelmanov] If G is a torsion-free c-Engel locally nilpotent group, then G is nilpotent of class ip{c) for some function ip. Since soluble c-Engel groups are locally nilpotent [Robinson, Part II page 64], Zelmanov's result implies that torsion-free soluble c-Engel groups are also nilpotent of class ip(c). Proof of Corollary: By Medvedev's Theorem, we need only prove: Lemma 6.2.6 Every finitely generated ordered c-Engel group is nilpo tent class ip{c). Proof: Let G be a finitely generated ordered c-Engel group, L0 ~ G and L\ be the union of all proper convex subgroups of G. By Lemma 3.1.2, the convex subgroups of G form a chain under inclusion; hence L\ is a convex subgroup of G. Moreover, L\ ^ G since G is finitely generated. Now L\ <\ G and L0/Li has no non-trivial proper convex subgroups. By Lemmata 4.1.1 and 4.1.2, L 0 /Li is Abelian. Therefore L\ 3 G', so L\ is finitely generated (see the proof of Lemma 6.2.4). Since it is convex, it is clearly isolated. Repeating the process we obtain a sequence {L, : i £ N} of finitely generated isolated normal subgroups of G such that l o 3 t , 3 L2 3 t 3 3 ... with each LJLi+i orderisomorphic to a subgroup of K. Moreover, by Hion's Lemma (4.1.6), conjugation by elements of G are equivalent to multiplication by real numbers; so G" acts trivially on each Lt/Ll+1 by conjugation. Thus G' C C(Li/Li+i). Consequently, each G'/{G' 0 Li) is a torsion-free cEngel locally nilpotent group. By Zelmanov's result each G'/(G' n Lx) is nilpotent of class ip(c). Let L = (~}{LX : i £ N}. Then G'/{G' n L) is also nilpotent of class ip(c) being a subdirect product of G'/(G' fl L,) (i £ N). Moreover G'/(G' n L) is finitely generated since G' is (Lemma 6.2.4). Since finitely generated nilpotent groups satisfy the descending chain condition on isolated normal subgroups [P. Hall], there is an integer
6.3. THE WORD
PROBLEM
113
(isuch that G'C\L = G'nLd. But G/(G'nL) = G/{G'C\Ld) is soluble and c-Engel. Hence (as noted above) it is nilpotent class tp(c). Thus G/L is also nilpotent class tp(c) and finitely generated. By the descending chain condition again, L = L^ for some positive integer d', and G/Ld> is nilpotent. Since Ld' is finitely generated (by Lemma 6.2.4) and hd> = Lji+i, it follows from the definition that Ld> = {1}. Consequently, G is nilpotent class tp(c). // If a locally c-Engel group G is right orderable (as opposed to lat tice orderable), then it is not known if G is nilpotent. However, locally 4-Engel right orderable groups are nilpotent (please see Section 6.8); the statement in [Kopytov & Medvedev, page 57] for all locally Engel right orderable groups is as yet unproved and arose from a simple mis understanding in a conversation (in English) between those authors and Rhemtulla.
6.3
The Word Problem
The purpose of this section is to use Theorem 6.A and the solubility of the word problem for nilpotent groups to prove T h e o r e m 6.C [Kopytov 1982] Every lattice-ordered group that is finitely presented as a nilpotent class n lattice-ordered group has soluble word problem. Proof: Let G be a lattice-ordered group that is finitely presented as a nilpotent class n lattice-ordered group. So G is the quotient of the free nilpotent class n lattice-ordered group on xi,...,xm by the £-ideal generated by to(x), say (see Corollary 2.3.9). Let u(x) be an element of the free lattice-ordered group on xi, ...,x m . Then u(x) = 1 in G if and only if Vx(w(x) = 1 -> u(x) = 1) is a theorem of nilpotent class n lattice-ordered groups. Since the class of nilpotent class n lattice-ordered groups has a finite set of axioms, its set of theorems is recursively enumerable. Hence this yields an algorithm to determine if u(x) = 1 holds in G. To complete the proof of the theorem, we need an algorithm to de termine if u(x) 7^ 1 in G. Since any nilpotent class n lattice-ordered
114
CHAPTER 6. SOLUBLE RIGHT RO.
GROUPS
group is residually ordered (Theorem 6.A), if u(x) ^ 1 in G, then there is an ordered nilpotent class n £-homomorphic image B of G in which u(x) ^ 1. Let bi,...,bm be the images of x\,...,xm\ so &i, ...,6 m generate B and tu(b) = 1 j^ u(b). Without loss of generality, w(b) > 1. Since B is ordered, every ele ment of B can be expressed in terms of bu ...,bm using only the group operations. Since B is finitely generated (as a group) and nilpotent class n, it has finite Hirsch length, and so only finitely many convex subgroups. We therefore obtain {1} = Be <...< B{ <\B0 = B where each Bt/Bl+i is a finitely generated free Abelian (Archimedean) ordered group and hence can be order-embedded in E (by Holder's The orem). Let Ci = Bi/Bl+i and {c%0 : j £ Jt} be a basis for d with ctJ < c t J + 1 . Choose bij G Bi\Bt+i so that Bl+ibliJ = chj (0 < i < I — 1 : j G Jx). Note that each element of B can be written uniquely in the form aa...a^\ where
ai = b'SA...b?j?i ( 0 < t < / - l ) . Next observe that if ui(x) = Vp A, wPtQ(x) where each u;Pi9(x) is a group word, then w(b) — 1 in B is equivalent to: for each p, there is q such thai wP,q(b) < 1 and for some p0, wP0:q(b) > 1 for all q. Similarly, u(b) > 1 is equivalent to: for some p0, upo>q(b) > 1 for all q. Now let ipi be an embedding of C, into E. Since 9, defined by ^i(Qj) = ipi(cij)/N is also an embedding if TV is an arbitrary positive real number. we may assume that ip^aj) < 1/2 for all i,j, 0 < i < I — 1, j G J, and that {^(Cjj) : 0 < i < £ - 1, j £ Jt} is a set of algebraic numbers linearly independent over Q. Let kij be 1 plus the maximum of the absolute values of the expo nents of btJ in the normal form for the various group words wpJh) and u P) ,(b) occurring in w(b) and u(b), respectively. Let M be a positive integer much bigger than YliZo £j€./, Kj- The map 4>% : Ci -* E given by (j>i(cij) = M'ipi(cij) is an order embedding of Ci into E. Define >: B —> E by e-i e-i 4>(a0...ae-i) mij4i{cij) M'mijViCcjj)
EE
1
where a, = bil' ...b^!
11
EE
t=0 j € J,
6.3. THE WORD
115
PROBLEM
Observe that (f>(w(b)) = 0 < »(u(b)) and that {4>(bid) : 0 < t < £ - 1, j € J,} is a rationally independent set of algebraic numbers. By the above, this is just a finite system of inequalities and strict inequalities each of the form
£
kij(bij)
< 0 > 0 > 0.
Conversely, let (*) be this system with each >(&;,_,) replaced by yitJ. Any solution {yitj : 0 < i < £ — 1, j € Jt} of (*) in K that satisfies no non-trivial linear equation with integer coefficients in absolute values less than twice the sum of the absolute values of all the ktys appearing in (*) will define a map from C, into K. Hence it induces a possibly new order on each Cl and thus on B. Let B' be this new ordered group (isomorphic to B as a group). Then w(b*) = 1 < w(b') in B*. Since B* satisfies tu(b*) = 1, it is an ^-homomorphic image of G\ and in Bm, u(b') ^ 1. But the solution set in R of any finite set of inequalities and strict inequalities is a polyhedron in Rk (c.f., Section 5.2) and so contains points {yhJ : 0 < i < £ - 1, j e J , } in the set of algebraic real numbers satisfying no non-trivial linear equation with integer coefficients in absolute values less than twice the sum of the absolute values of all the kij's appearing in (*) (since {>(&t,j) : 0 < i < £ — 1, j € Jt} is a solution). Hence if u(x) / 1 in G, there is an m generator nilpotent class n group B" with a finite system of quotients each of which is free Abelian of finite rank (at most m) and isomorphic to a subgroup of the algebraic real numbers such that iu(b*) = 1 / u(b*) in B" Since the set of m-generator nilpotent class n groups with such quotients is recursively enumerable and the set of inequalities and strict inequalities in the algebraic real numbers is effective, we obtain an algorithm that will show if u(x) -^ 1 in G; viz: enumerate all such group words and attempt to solve the corresponding system of inequalities and strict inequalities in each. If there is a solution in any of them, then u(x) ^ 1 in G. Since both the set of words equal to 1 in G and those not equal to 1 in G are recursively enumerable, G has soluble word problem./ Caution: A lattice-ordered group may well be finitely presented as a nilpotent class n lattice-ordered group but not as an arbitrary latticeordered group.
116
CHAPTER 6. SOLUBLE RIGHT P.O. GROUPS
By Theorem 5.C, the isomorphism problem for finitely presented nilpotent lattice-ordered groups is insoluble. However, the conjugacy problem for this class remains open. As the following example shows, however, it is possible for two non-conjugate elements of a finitely pre sented nilpotent class 2 lattice-ordered group to be conjugate in every ordered ^-homorphic image. Example 6.3.1 Let TV be the free 2 generator nilpotent class 2 group; say N = (a,b,c : [a,b] = c, [a,c] — 1 = [b,c]). Order N so that K c « ( ) « o . Note that, in N, anbm = cmnbman\ so every element of TV can be written uniquely in the form c£bman Let G — {(clbjak,cmbnak) : i,j, k, m, n € z} C N x N. Then G is a lattice-ordered group under the order inherited from the cardinal product, and f,g 6 G where / = (b,b) and g — (b,c~lb). An easy computation shows that no element of the form (clb>ak,cmbnak) conjugates / to g; so they are not conjugate in G. However, if: G —> H with H an ordered group, then the kernel of is a prime £-ideal of G. Since (b, 1) A (1,6) = (1,1) in G, we must have (6,1)^ = 1 or (1,6)0 = 1. In the former case, {a,a)<j> conjugates f<j> to g4>, and in the latter case, (1,1)0 does.
6.4
Weakly Abelian ^-groups
A lattice-ordered group G is said to be weakly Abelian if / - 1 | ; ? | / < |g| 2 for all f,g G G. Clearly, polars in weakly Abelian lattice-ordered groups are normal. So, by Lorenzen's Theorem, every weakly Abelian lattice-ordered group is residually ordered (and hence normal-valued by Corollary 4.2.5). Even more obviously, any Abelian lattice-ordered group is weakly Abelian. We now show T h e o r e m 6.D [Reilly 1983] Any locally nilpotent lattice-ordered group is weakly Abelian. Proof: By Theorem 6.A, it is enough to show that every ordered locally nilpotent group is weakly Abelian. Clearly, an ordered group is weakly Abelian if every finitely generated subgroup is weakly Abelian. Hence it is enough to prove the theorem for an arbitrary ordered nilpotent group G of nilpotency class n, say. We argue by reductio ad absurdum.
6.4. WEAKLY
ABELIAN
117
(-GROUPS
Let f,g e G with g > 1 and f~lgf > g2. Then f-mgfm positive integers m. Using this, it is easy to see that f-myjnym
^-^-m^m/m
=
> #2'" for all
> f-mg-lg2-fm
>
/-v m "'r = (rmgrr" > [g,n2""1 > i. Suppose that
r m [g,/ m ,--;,/ m j/ m > [g.fv^r] 2 " 1 "' > i. k
k
We prove that
> [g,/ r o ,---,/ m j 2 "" t "' > i
rm[9,r,•■-,/"']/" k+i
if k
+■
fc+i
1 < m. Let h
\9,
/m---,n k
Since
/-
m
/i/
m
> h2m~K
/-m[/i,/m]
> 1, we have >, /"] > 1; so
r> m
/
m
(r hf )
cm
1 (/T l h ^)
-m
2m-k-l
m
> [hj ]
>
rm h
2m-Jt-l
2m
-*-'
r
> 1,
as desired. Putting m — n + I and k = n, we deduce that
Is,
r
+
i )■ * * " 1
\
fn+x J
} + 1.
n
This contradicts that G is nilpotent class n. Thus G is weakly Abelian./ We now give a useful equivalent to weakly Abelian. L e m m a 6.4.1 A lattice-ordered group is weakly Abelian if and only if |[/,s]| <. f for all f & G+ andgeG. Proof: Clearly this condition implies that G is weakly Abelian. Conversely, suppose that G is weakly Abelian and / £ G+ and g EG. Since G is residually ordered, it is enough to prove that the condition holds if G is ordered. Let V be a value of / and V* be the cover of
118
CHAPTER 6. SOLUBLE RIGHT P.O. GROUPS
V. Since G is weakly Abelian and / e V*, we have that g~l fg S V*. But G is ordered and so normal valued (Corollary 4.2.5). Hence V < V* and V'/V is order isomorphic to a subgroup of E. So conjugation by g induces an order-preserving automorphism of a subgorup of E. By Hion's Lemma (4.1.6), the induced automorphism is multiplication by some positive real number r Since G is weakly Abelian it follows that g~nfgn < f2 f° r a ^ integers n. Hence r = 1. Thus Vg~lfg = Vf; so [f,g]eV. Therefore |[f,g]\ « f. // In Example 1.3.24, we ordered the free group on more than one gen erator via the lower central series so that this above equivalent condition held. Hence Example 1.3.24 provides an example of a weakly Abelian ordered group that is insoluble. Since W = Z Wr Z is metabelian but not orderable (Lemma 2.4.6), W is not residually orderable (the direct product of orderable groups is orderable — see Example 1.3.14). So we have an example of a soluble lattice-ordered group that is not weakly Abelian. In contrast, the inter section of the terms in the lower central series of Z wr Z is {1}; hence Z wr Z can be ordered so as to be a weakly Abelian o-group. For the same reason, so can the free metabelian group on two generators.
6.5
Divisibility
In Chapter 1, we mentioned that one of the main outstanding questions in the subject is whether every ordered group can be embedded in a divisible ordered group. The technique of using free products with an amalgamated subgroup does not work since the amalgamation property fails (Theorem 2.E). Indeed, it is still unknown whether every soluble ordered group can be embedded in a divisible ordered group (soluble or otherwise). However T h e o r e m 6.E [Bludov & Medvedev] Let G be an ordered group with convex normal Abelian subgroup N. If G/N is also Abelian, then G can be order embedded in a divisible metabelian ordered group. Proof: Let G be a metabelian ordered group with convex Abelian normal subgroup N such that G/N is Abelian. Let W0 = N Wr G/N, and 0 : G -» W0 be the map g »-> (g, Ng) where g is defined as follows:
6.5.
119
DIVISIBILITY
Let T be a transversal for N in G; i.e., 1 £ T and for each / € G, there is a unique tf £ T such that Nf = Ntf. Let (g){Nf) = tfgt]*. If g £ G+\N, then clearly g> 1. Ug £ AT+, then Nfg = Ngf = Nf since G/N is Abelian; so tfg = tf for all f £ G. Thus (|)(iV/) = t , . ^ 1 > 1 since g > 1. Hence 0 is an order embedding of G into W0. Now iV can be embedded in a rational vector space V which can be ordered so as to extend the order on N; similarly, G/N can be order embedded in an ordered rational vector space U. So W0 can be order em bedded in W — V Wr U which is a divisible metabelian group. However, W and W0 are not orderable (Lemma 2.4.6). Let I = {w £ W : a product of conjugates of w is the identity}, the strong isolator of the identity. Clearly, I is a normal subgroup of W and W/l is divisible. Let ip be the induced embedding of G(j> in W and 9 = <j>ip.
If g £ G\N then clearly g £ I. If g £ N with g > 1, then as we noted already (g)(Nf) > Nf for all f £ G. Hence, {gO){u) > 1 for all u £ U and (g6)(u) > 1 if u € G/N. Consequently, g £ I. Finally, for any subset X of W/I, let N(X) be the normal subsemigroup of W/I generated by X. If xu ...,xn £ W/I, then by the definition of / we can find ei, ...,e n G {+1, - 1 } such that
N(x\\...,x^)n(i(Ge)+//)
= 0.
By Lemma 2.2.1 W/I is an ordered group and 9 an order embedding. /
Corollary 6.5.1 Every metabelian orderable group can be embedded in a divisible metabelian orderable group. Proof: Let G be a metabelian orderable group with derived subgroup [G, G\. Then [G, G] is a normal Abelian subgroup of G and G/[G, G] is Abelian. Let N be the isolator of [G,G\; that is, N = {g £ G : gn £ [G, G] for some positive integer n). Then clearly N < G, and G/N is a torsion-free Abelian group. Moreover, since G is an orderable group, N is Abelian by Corollary 2.1.5. So the hypotheses of the theorem apply./
120
6.6
CHAPTER 6. SOLUBLE RIGHT P.O. GROUPS
Conrad right orders
A right ordering on a group G is said to be a Conrad right ordering if and only if for every g G G"\{1}, the value V of g is normal in its cover V" and V'/V is an Archimedean ordered group. This is the obvious analogue of a lattice-ordered group being normal-valued. In analogy with normal-valued lattice-ordered groups (Lemma 4.2.2): In any Conrad right ordered group G we have |[/, g]\Vf > Vfg~l since V*/V is an Abelian ordered group. Hence g2f2 > fg. // We postpone the proof of the converse direction of Theorem 6.F to Section 8.1. We also note, in passing, a useful technical tool.
6.6. CONRAD RIGHT
ORDERS
121
Lemma 6.6.2 Let G be a right ordered group. Then the following are equivalent: (1) G is Conrad right ordered; (2) V/, g£G,lg for some positive integer n; (3) V/, g G G+, gnf > g for some positive integer n if g > 1; (4) for every g G G+, {/ G G : \f\ gk for all integers k. Therefore gn > gn+my~1 for all non-negative integers m. But y < gn so ygm < gn+m. Hence ygmy~i < gu for all positive integers m. This contradicts (2). // n
We can now prove Lemma 6.6.2. Proof: (2) => (1). Let V be a value and f,g € {V*)+\V We first note that gk > f for some positive integer k. (Otherwise C — {c G G : (3n)(|c| < gn)} is a convex subgroup of G by (2) and the lemma just proved. But / £ C, so C C V\ yet g G C\V, a contradiction.) If h G V, then V is a value of fh~x. Hence {fh~x)n > f for some positive integer n and we choose n least such. Thus f(fh~l)f~l > 1. (If f(fh~l)f~l < 1 1 1 1, then fifh' ) < f < {fh' )"; so / < (//i" )"- 1 contradicting the minimality of n.) Therefore f(hf~l)f~l < 1; i.e., / / i / " 1 < / . So, m m l applying this to /i we get that fh f~ < f for every integer m. If fhf~l $ V and fhf~l > 1, then fhmf~l > / for some positive integer m by what we first shewed, contradicting the previous sentence; and if fhf~l < 1 then we get the same contradiction from —m. Consequently fV.f~l Q ^'■ N ° w 9k > I f° r some positive integer k. By (2) there is a positive integer r such that fgkrf~l > gk. Hence fgf~l G V*+\V; i.e., / ( V ^ V ) / - 1 C V*\V. Thus f~lVf C V Therefore V < V This establishes (1).
122
CHAPTER 6. SOLUBLE RIGHT P.O. GROUPS
(1) => (3). If f,g G G+ and gnf < g for all n G Z + , then £ 2 G V, the value of gf~l Since V < V*, we have {gf~l)~lg2{gf~1) G V Hence 1 2 l l l fg~ g gf~ < gf~ ] i.e. / < g~ < 1, the desired contradiction. (3) => (2). First observe that a, b > 1 implies that {ab)n > ba for some n G Z + (otherwise (a6) m < 6a for all m G Z + ; so (6a)m6 < a{ba)mb = (ab)m+1 < ba, contradicting (3)). Hence if 1 < / < g, then fgnf~l = (/ • gr1)" > {gf"l)f = 9 for some n G Z + (2) =» (4). Suppose that g G G+ and | / | , \h\C , so we may assume that / $ (h). If fh > 1, then clearly |//i| 1, then hfnh~x > f for some positive integer n by (2). Hence / / i < hfn < fn+1 <§: g. If ft < 1, then / i ' 7 / i > fh > 1, and so / " = (//i)(/i _ 1 fh)n(fh)~l > h~lfh > fh for some positive integer n by (2). Thus, in either case, fh<^gby the previous lemma. If / / i < 1, then by applying the argument of the previous paragraph to h~x and / " ' (in place of / and h) we obtain \fh\ = h~lf~l (2). Let 1 < / < g and suppose that fguf~l < g for all integers n. Then f j& g. (Otherwise g < fg = fgj~x ■ f -C g by (3).) So fm > g for some positive integer m. Thus / < 5 g for all f,g G G with Kf f in condition (2) of Lemma 6.6.2 then we get an ordered group: If G is a right ordered group and for all f,geG+ with g > 1 there is a positive integer n such that fgnf~l > g, then G is an ordered group.
6.7. LOCAL
NILPOTENCY
123
Proof: We must shew that G+ is normal in G. Let g > 1 and / £ G Then {fgf~l)n — fgnf~1 > g > 1 for some positive integer n; so _1 /s/ > 1. If / - ' 5 / < 1, then / " V 7 > 1; so / ( / " W r / " " 1 > l l f g f > 1 for some positive integer m. That is, "m > 1. Thus <7_I > 1, a contradictio i. Hence G+ O G as required. / +
We conclude this section with an important application: Theorem 6.G Every Conrad right ordered c-Engel group is mlpotent class yj(c). Proof: The proof of Lemma 6.2.6 applies verbatim with Conrad right ordered in place of ordered. /
6.7
Local nilpotency
We saw earlier (Theorem 6. A) that every locally nilpotent lattice-ordered group is residually ordered and hence normal-valued (Corollary 4.2.5). In light of Theorem 6.F W3 would expect: Theorem 6.H [Ault] and [Rhemtulla] Every right order on a locally mlpotent group is a Conrad right order. We need a lemma. Although a stronger version (due to Mal'cev) will be needed in the next section, the following weak form (due to B. H. Neumann) suffices here and is easily established. L e m m a 6.7.1 Let G be a nilpotent group and g.h £ G. Then there are a,b € S, the subsemigroup of G generated by g and h, such that ah — bg. Proof: The proof is by induction on the nilpotency class. If the class is 1, the group is Abelian and we can take a = g and b = h. Assume that the lemma is true for groups of nilpotency class n and let G have nilpotency class n + 1. It is enough to establish the equality for the subgroup of G generated by g and h; so we will assume that G = (g, h). Let H = {gh, hg). Now [g, h] € Q(G) n H and H/{C,n{G) n H) is cyclic since gh = hg[g, h]. Hence the nilpotency class of H is at most n. Thus
124
CHAPTER 6. SOLUBLE RIGHT P.O. GROUPS
there are c,d £T, the subsemigroup of H generated by gh and hg, such that cgh = dhg. Since T C 5, we get the desired result by letting a — eg £ 5 and b — dh £ S. // We can now prove the theorem. Proof: We use Lemma 6.6.2. To obtain a contradiction, assume that f,g £ G + but g " / < g for all positive integers n. Since it suffices to shew that a contradiction arises in the nilpotent subgroup generated by / and g, we may assume that G is generated by / and g. By the previous lemma, there are a, b £ S C G + , the subsemigroup of G generated by gf and g such that a y / = bg. But (/)m < g for all positive integers m (by induction). Hence g < bg — agf < g, the desired contradiction. /
6.8
4-Engel right ordered groups
As promised in Section 6.2, we now prove: Theorem 6.1 [Longobardi & Maj] Every 4-Engel right ordered group is nilpotent. The proof requires two key lemmata. Lemma 6.8.1 [Longobardi, Maj, & Rhemtulla] If G is a right ordered group and G+ contains no non-Abelian free subsemigroup, then the right order on G is Conradian. Proof: Let f,g £ G+ with g > 1. If / > g, then g2f > f > g. If / < g, then by hypothesis, there are positive integers p, g and Ti, Si, rrij, n3 € Z + (1 < i < p; 1 <j0 such that
9r7Sl-3r'fp = /""f.../'""j'" If g > gnf for all n £ Z + , then (if si > 0)gr,fs' = (grif)fSl~l < gf6'-1 < ■■■ < df < 9- Continuing in this way, we obtain that grifS)...gr''fSp < g (since sp > 0). Since fmigni.../m« > 1, it follows that fmign> ...fm"gn" > n n g i Therefore g > g « > g, a contradiction. By Lemma 6.6.2, the right order on G is Conradian. //
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125
If N is a nilpotent group and u0, v0 € N, write ul+l = utv, and vi+\ = ViUi for all i e Z ^ j O } . If N has nilpotency class c, then u c + ] = u c+ i [Mal'cev, 1958]. Our second lemma is purely group-theoretic and is a partial converse. Lemma 6.8.2 If G is a torsion-free 4-Engel group and f,g € G, then u s(/> g) = Vs(f,g). Hence any free subsemigroup of a torsion-free 4-Engel group is Abelian. From it we can immediately deduce the theorem: Proof of Theorem 6.1: By Lemmata 6.8.2 and 6.8.1, any 4-Engel right ordered group is Conrad right ordered. Since we may clearly confine our attention to finitely generated 4-Engel Conrad right ordered groups, the theorem follows immediately from Theorem 6.G. / We will deduce Lemma 6.8.2 from three easy results. Lemma 6.8.3 Let G be torsion-free c-Engel group and / , g S G. If (fm,g) is nilpotent class n for some positive integers m,n, then (f,g) is nilpotent class n. Proof: Let C = Ci((/, )) and C{m) = Ci((/ m ,» By Lemma 6.2.3, it is easy to see that C(m) = C D (fm,g)- Since {f,g)/C is torsion-free and ({fm,g)C)/C = {fm,g)/C{m) is nilpotent class n- 1, we obtain (by induction on n) that (f,g)/C is also nilpotent class n — 1. Hence (/, g) is nilpotent class n. // Lemma 6.8.4 Let G be a 4-Engel group and f,g € G. If h — then {h.g~lhg) is nilpotent class 2.
f~lgf,
we obtain Proof: Since \f~\g] = f\g,f\f~\ [/Iff. /I/" 1 .*?, 3, ] = Thus [f' ,g,g,g,9] = 1> conjugating by / gives [g,f,h,h,h] = 1. Hence 1 = [g~lf~lgf, h, h, h] = h'1^1, h, h, h]h; so [g~\ h, h, h] = 1. Therefore [g[h,g]g~l, h, h] = 1, and so 1 = \h,g,g~lhg,g~lhg] = \h~lg~ihg,g~lhg,g-lhg). Hence [h~l,g~lhg,g~ihg] = 1. Consequently, [h-\g-lhg]€C{g-1hg). The same argument with g~] and h ' in place of g and h, respectively, gives [h,ghrlg-l\ G Cighg1'). Thus [h-\g-lhg]-1 = [g-'hgjr'} e l C{h). Hence {h,g~ hg] = [g^hg.h,-1] and so {h,g-'hg} £ C{(h,g-lhg)). Consequently (h,g~1hg) is nilpotent class 2. / l
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Corollary 6.8.5 If g and h are conjugate elements of a torsion-free 4Engel group G, then the subgroups (h, g~mhgm) are nilpotent class 2 for all positive integers m. Proof: By Lemma 6.8.4, (gm, h-mgmhm) is nilpotent class 2 for all positive integers m. By Lemma 6.8.3, so is (g,h~mghJn) (m G Z + ). // Proof of Lemma 6.8.2: Let G be a 4-Engel torsion-free group and f,g eG. Since ui(f,g) = u0(fg,gf) and vi(/,#) = v0(fg,gf), we must show that u4{fg,gf) — v^(fg,gf). Now fg and gf are conjugate since fg = g~1(gf)g. So by Mal'cev's result it is enough to shew that if g and h are conjugate elements of G, then (g, h) is nilpotent class 4. We therefore assume that g and h are conjugate in G. The proof that (g, h) is nilpotent class 4 proceeds by the Commutator Calculus. Specifically, we will repeatedly use (without mentioning) the following identities: \xy,z]=y
l
[x,z]y[y,z]
and
[x,yz] = [x,z]z
l
[x,y\z.
By Corollary 6.8.5, the subgroup ([h3,g],g) is nilpotent class 2. Hence
lh3,g} = h-l\h2,g}h[h,g} = h-2lh,g}h2 hr^gjh
[h,g} =
h-2[Kg]h2[h,g][h,g,h)[h,g} = [h,g}[h,g,h2}[h,g][h,g,h][hig] = [h,g][h,g,h]2[h,g}[h,g>h}[h,g] = [h, gf[h,g, h}3 since ([h,g],h) is nilpotent class 2 by Corollary 6.8.5 (so [h,g,h] is cen tral). Since [h,g,h] is central, \h,g,h]r = [[/i, ]r,/i] by an easy induc tion on r. Hence [h3,g] = [h,g]*[h,g]-i(hr1[h,g]h)' = (h-^h, g}h)3 -1 3 Consequently, ((/i [/i, g]h) ,g) is nilpotent class 2. By Lemma 6.8.3. so is (h~i[h,g]h,g); thus ([h,g],hgh-1) is nilpotent class 2. Therefore [h,g, hgh~l] commutes with hgh~x. Since [h, g, hgh~x\ — [[h, g],g[g,h~r]] [[h, g], [g, ^ _1 ]] • [g, ^ _ 1 ] _ 1 [^, g, g][g, h~l], the latter commutes with hgh~l But [[h, g], [g, h-1}} = [[h, g], h[h,g]h-1] = [[h, g], [h, g][h, g, h~1}} = 1 since {[h,g],h) is nilpotent class 2. Therefore [g,h~i}~1[h,g,g}[g,h~l] com mutes with hgh~l\ i.e., the conjugate of [h,g,g] by g~1hgh~l commutes with hgh~x Conjugating both by hg~l gives that h~1g[h,g,g]g~1h com mutes with g. By Corollary 6.8.5, [h, g,g, g~1} = 1 since (g,h~1gh) is nilpotent class 2. Thus h~l[h,g,g)h (= h~1g[h,g,g]g~lh) commutes with g, as does
6.9. LOCAL
INDICABILITY
127
[h,g,g]- So \h,g,g,h] = [h,g,g\~lh~l[h,g,g)h commutes with g and h~l[h,g,g]h commutes with h~lgh (since [h,g,g] commutes with g). Also [h, g, g] commutes with h~lgh since {g, h~lgh) is nilpotent class 2. There fore [ft, g,g, ft] commutes with g and h~lgh, and so with g and [h,y\. By Corollary 6.8.5, {[h,g2],h) is nilpotent class 2; so [h,g2,h] £ Ci«[ft,(?2],ft)). But [ft,52] = [h,g]g-l[h,g]g=[h,g]2[h,g,g\ = [h,g,g][h,g]2 because {g,h~lgh) is nilpotent class 2. Hence {Vlig,g]\h,g}2,h\ commutes with ft. Thus [ft,g,g, ft] [[ft, g)2, ft] = \h,g]-2{h,g,g,h][h,g}2[[h,g}2,h] = [[h,g,g][h,gf,h] commutes with ft. Therefore [h,g,g, ft] commutes with ft since [[ft,]2,ft] = [h,g,h]2 does (([h,g],h) is nilpotent class 2). Consequently, [h,g,g,h] £ £i{{h,g}). Since [h,g,g,g] = 1 we obtain [h,g,g] 6 ^{(h,g)). We can get [g,h,h] G CaCC*,5)) by a similar argument. Hence [, /i] € Q{(h,g)). Therefore (.9, ft) is nilpotent class 4, as desired. /
6.9
Local indicability
We explore two group-theoretically interesting consequences in the last two sections of this chapter. Although both are of independent interest (and the results in this section are deep), they will not be needed in the rest of the book. A group G is said to be locally indicable if for every non-trivial finitely generated subgroup H of G, there is a homomorphism of H onto the infinite cyclic group. We first note L e m m a 6.9.1 [Burns & Hale] Every locally indicable group is right orderable. Proof: Let G be locally indicable. By Lemma 2.2.2, if G were not right orderable, then there would be #1, ...,gn £ G such that 1 € S(g\', ■••,<#') for all choices of ± 1 for elt ...,e„. Among all such choices in G, choose one with n least. Let H be the subgroup of G generated by c/i, . , „. Since G is locally indicable, there is a normal subgroup N of H such that II/N is infinite cyclic. By relabelling the subscripts, we may assume that gi,...,gm e N and gm+i,...,gn & N, and that rn < n. By the minimality of n, for some assignment of ±1 to r]i,...,r]m we have 1 £
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S{g1l,...,g^). Also, H/N is orderable so by Lemma 2.2.2 again, there is an assignment of ± 1 to rjm+u ..., T]n so that N n S(gim+1, ■■■, g%) - %■ Hence 1 0 S{g[\ ...,g^n), the desired contradiction. / The converse is false as was shown independently by Bergman and Tararin. E x a m p l e 6.9.2 [Tararin] Consider the right ordered group A(Q) from Example 1.3.20. Let L comprise all those g € A(Q) such that the re striction of g to any bounded interval is piecewise linear. Then L is a subgroup of A(Q). Let C = {g G L : (Va)((a + l)g = ag + 1)} which is a right ordered group under the inherited ordering. It can be shewn that there are finitely generated subgroups of C for which Z is not a homomorphic image; i.e., C is right orderable but not locally indicable. Another example of this phenomenon is the quotient of the free group on x, y, z by the normal subgroup generated by the elements x~1yz, x2y~3 and z7y~3. For details, please see [Bergman]. So we can ask "Which right ordered groups are locally indicable?" Obviously, we can either restrict the class of right orderings or the class of groups. We will prove two fundamental results; we begin with the easier. Theorem 6. J A group is locally indicable if and only if it can be Conrad right ordered. Proof. Let G be a Conrad right ordered group and g\,...,gn 6 G. Let Vj be the value of g^ in this right ordering (j = l,...,n). Since the values of a right ordered group are totally ordered under inclusion (Lemma 3.1.2), we may assume that Vi C ... C Vn. Then Vn< V* and Vn* contains H = (g-i,...,gn). Now V*/Vn is Abelian, so H/(H D Vn) = (Hf\ V*)/(HP\Vn) is a finitely generated Abelian group. By the Fundamental Theorem of Finitely Generated Abelian Groups, Z is a homomorphic image of H/(H n Vn) and hence of H. Thus G is locally indicable. Conversely, by Lemma 6.9.1 every locally indicable group is right or derable. Hence by Corollary 6.6.4 and Theorem l.A, it is enough to prove that every finitely generated (locally) indicable group can be Con rad right ordered. If there were a counterexample, there would be one
6.9. LOCAL
INDICABILITY
129
with a minimum number of generators, say n. Let G = (<7i, ...,„) D e such a counterexample and TV <\ G be such that G/N = Z. As before, we may assume that gu ...,gm € TV and gm+i, •••i5n £ TV. Clearly, m < n so by the minimality of n, there is a Conrad right order on TV. Define g € G to belong to P if and only if either g e TV+ or g g" TV and Ng is a positive integer (under the isomorphism between G/N and Z). Then P is a subsemigroup, P n P _ 1 = {1} and G = P U P " 1 ; i.e., G is right ordered with respect to P Moreover, TV is convex in this right ordering. Let f,g eG with 1 < / < g (in the right ordering given by P). If g G TV, then by Corollary 6.6.4 fg2f~l > g since TV is Conrad right ordered; and if g £ TV, then Nfg2f~1 = TVg2 > TVp since TV < G and G/N is Abelian. Hence, in either case, fg2f~l > g. By Corollary 6.6.4, G was not a counterexample. / By the theorem and Tararin's example preceding it, we obtain Corollary 6.9.3 There is no Conrad right order on A(Q) or A(S). We need some definitions to state the other major result. A group G is said to be a radical group if there is an ordinal K and a family {Ga : a < K} of normal subgroups of G with Go = {1} and G = \Ja
Theorem 6.K [Longobardi, Maj & Rhemtulla] Every NFS-radical right orderedable group is locally indicable. Corollary 6.9.4 [Tararin], [Chiswell and Kropholler] Let G be a radicalby-finite group. Then G is right orderable if and only if it is locally indicable.
130
CHAPTER 6. SOLUBLE RIGHT P.O. GROUPS We will require six easy lemmata to prove Theorem 6.K.
For ease of exposition, we will assume that all free subsemigroups are on sets with more than 1 free generator (and so are non-Abelian). Our first result is an immediate corollary of Theorem 6.J and Lemma 6.8.1 Lemma 6.9.5 [Longobardi, Maj, & Rhemtulla, 1995] If G is a right ordered group and G+ contains no free sub semigroup, then the order on G is Conradian and G is locally indicable. As we saw in Example 1.3.20, if Q is any totally ordered set, then we can right order A(Q) as follows: Fix a G fi and let ~< be any well-ordering of Q with least element a. Declare / > 1 if / ? / > 0 where /? is the least element (under -<) of {6 G Q : 5f ^ 5}. Hence any subgroup G of A(il) can be right ordered. Letting Ga = {g G G : ag = a}, we have: Lemma 6.9.6 If G is a subgroup of A(Q,) and a £ fi, then there is a right order on G in which GQ is convex. If G is a right ordered group (with positive cone P) and A C G, we will denoce by conp(A) the convex subgroup of G generated by A. Lemma 6.9.7 Let G be a right ordered group and A be a subgroup of G. If A is bounded above in G, then there is a right ordering P on G in which conp(A) ^ G. Proof: If the right ordering on G were Conradian, this would follow immediately from Lemma 6.6.2 (with the same right ordering). So the onus is to prove the result when the original right ordering is not Conra dian. Let G+ refer to the given right ordering. Let A = {g G G : (3a G A){g < a)} and A = {A/ : / € G}. Then A C G (since A is bounded above in G) and each A/ is a convex subset of G and A/ C Xh if h & A/. We can therefore totally order A by A/ < Xh if Xf C Xh. Now G acts on A via right multiplication and preserves the ordering on A. Further, A C Gx since Xa = A for all a G A (A is a subgroup). Hence L C G\ where L is the kernel of this action (the lazy subgroup). By Lemma 6.9.6, there is a right order on G/L in which G\jL is convex. We extend this order to G via: put g 6 P if either g G LC\ G+ or Lg > L in this right order on G/L. Then P is the set of positive elements of a right order on
6.9. LOCAL
INDICABILITY
131
G in which G\ C G is convex. Since conp{A) C G\, the lemma follows. // We extend this result slightly: Lemma 6.9.8 Let G be a right ordered group and A be a normal sub group ofG. If A is bounded above, then there is a right order P onG and a proper normal subgroup L of G which is convex under P and contains A. Proof: Note that L in the above proof has the desired properties since A/a = A ( / a / " 1 ) / = Xf for all / € G and a € A. // Lemma 6.9.9 Let G be a right ordered group and f € G with / > 1. // (/) is neither bounded above nor below, then g~lfg > 1 for all g 6 G. Proof: Let g 6 G. If g < 1, then g~l < fn for some n e Z + ; so (g~ fg)n = g~lfng > fng > 1. If g > 1, then / " " < g" 1 for some n € Z + ; so s" 1 /"*, > f~nfng = g > 1. // l
Finally we need an application of Theorem 2.2.2: Lemma 6.9.10 [Tararin] Let {G : i £ 1} be a chain of subgroups of a group G. Assume that there is a right order Pt on G in which Ci is convex (i € I). Then there are right orderings P and Q of G in which Uig/C) and Hie/ Ci are, respectively, convex. Proof: Let C = \JieICz. By Lemmata 2.2.2 and 2.2.3, it is enough to show that for all gi,-,gn S G\C there are ei,...,e n € {±1} such that C n S(g\\ ...,<&n) = 0- But this clearly follows since C = \Jiel C, and for a l i i e / and g\,-..,gn € G\Ci there are ei,...,e„ € {±1} such that
c,nw,...,«£») = 0.
The proof for flie/ C» is similar. / We can now prove Theorem 6.K.
Proof: Let G be a non-trivial NFS-radical group. We must show that Z is a homomorphic image of every non-trivial finitely generated subgroup of G. Since any subgroup of an NFS-radical group is also an NFS-radical
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group, we may assume that G itself is finitely generated and just prove that Z is a homomorphic image of G. By Lemma 6.9.10, the finitely generated group G has a proper normal subgroup that is convex in some right ordering (and maximal with this property). Since the quotient of an NFS-radical group is also an NFS-radical group, we may further assume that there is no non-trivial normal subgroup of G that is convex in any right ordering. Since G is an NFS-radical group, there is a largest normal subgroup A of G that contains no free subsemigroup. For this proof only, we will use C(G) to denote the set of relatively convex subgroups of G other than G. Let a = {A n K : K £ C(G)}. If C G a and {Kt : i G / } is a chain of proper relatively convex subgroups of G with C = A n Ki G a for alii 6 I, then C = A n f]iel Kx. Since n i£ /AT I is also a relatively convex subgroup of G (by Lemma 6.9.10), there is a mimimal K' G C(G) such that A D K* = C. For each C G a, let <SC = {K' € C(G) : An K* — C with K* minimal such}; and let 93 = U{93c : C G a } . By Lemma 6.9.10, if {Lt : i £ 1} is a chain in 93, then L = Uie/ -^i is a relatively convex subgroup of G that is proper (since G is finitely generated). Moreover, if T is relatively convex and A n T = A n L, then by the minimality of L, we get that T C\ L{ — L, for all i £ I; i.e., T D L. Thus L G 03. Therefore <» contains an element Go that is maximal in <8. Let < be a right order on G with respect to which Go is convex and let
6.10. TWO SIDED RIGHT
ORDERS
133
unbounded in G for every right order on G by Lemma 6.9.8, the same is true of (a) for all a G A\A0. We use this to show that A0 < G. Let / G ^4+\>lo and g £ G. Then (/) is neither bounded above nor below in A, so the same is also true in G. Hence h~1fh > 1 for all ft € G by Lemma 6.9.9. If g~lfg G AQ, then / > g~lJng for all n G Z + Thus c„ = }g~lf~ng > 1 for all n G Z + Let m € Z + be such that m f > cugfg~^ Since c m G A\Ao, the subgroup (cm) is unbounded; so gcmg~l > 1 by Lemma 6.9.9. Therefore ff/ff_1/"mfl5_1 > 1; i.e., 9f9~l > fm > gfg~l, a contradiction. Consequently, A\A0 is invariant under conjugation by elements of G, and so A0 <3 G. By Lemma 6.9.8, A0 — {1} since G contains no non-trivial normal subgroups (which are convex in any right ordering). Therefore the right order < ^ is Archimedean and A is isomorphic to a subgroup of JR. Since A is unbounded in G, the conjugating action of G on A is order-preserving. Let L be the kernel of this action. By Hion's Lemma (4.1.6), G/L is a torsion-free Abelian group. Since G is finitely generated, we deduce that Z is a homomorphic image of G (as desired) unless L = G. Hence we may assume that A C Ci(G). Since G/A is an NFS-radical group, it has a normal subgroup NjA that is an NFS-group. So for any x, y G N, there are semigroup words u(x, y),v(x, y) such that u(x, y) — v(x, y)a for some a G A. Thus uv — vu since A C (i(G). So TV is an NFS-group containing A. Hence TV = A by the maximality of A. Consequently, G = A and every right order on G is Conradian. Therefore G is locally indicable by Theorem 6.J. /
6.10
Two sided right orders
In this section we establish Theorem 6.L [Darnel, Glass & Rhemtulla] If G is a right orderable group and every right ordering on G is two sided, then G is Abelian. We will deduce the theorem from four lemmata. To avoid symbols for the various "<" that we consider, we will identify the right order with its set of strictly positive elements, and use P in place of G+ to avoid ambiguity.
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Lemma 6.10.1 If G is a right orderable group and every right order ing on G is two sided, then the convex subgroups of G under any right ordering are all normal in G. Proof: Let P be a right order on G and V be a value in G (under P). Then P < G by hypothesis. For any g G G, g~lVg is also a value and V Q g~lVg or V 3 9~lVg by Lemma 3.1.2; without loss of generality, assume the former and the containment is strict. So V* C g~^Vg. Define Q = (Png-'Vg) U ( P n ( G ^ V * g ) ) U (P~l D ( t f - ' V ^ V r 1 ^ ) ) . Then a routine verification establishes that Q is a right order on G. It is not two sided since g~lcg G <5_1 for any c G Q fl (V*\V). // Recall that if H is a subgroup of a group G, then the isolator sub group of H in G is the subgroup of G generated by {g G G : 5" G if / o r some n G Z, n > 0} Since every Abelian group is embeddable in a vector space over the field of rational numbers, the minimum of the dimensions of these vector spaces is known as the rank of the Abelian group. For example, Z has rank 1, Z © Z has rank 2 and E has rank 2N°. Lemma 6.10.2 Let G be a right orderable group and D be a torsion-free Abelian normal subgroup of G. If g~iPg = P for every g G G and right order P of D, then every isolated subgroup of D in G is normal in G. Proof: Let a G D, a / 1. Denote the isolator of (a) in D by 1(a). If 1(a) is not normal in G, then b = g~lag $ 1(a) for some g G G. Let I(a, b) be the isolator of (a, b) in D. Then I(a, b) has rank 2. Thus there exists an order P on D such that a, 6 _1 G P Since 6 G g~lPg, this order fails the hypothesis of the lemma. / Lemma 6.10.3 Let A be a torsion-free Abelian group of rank 1 and P be a total order on A. If A is the group of all order-preserving automor phisms of A, then A is a free Abelian group. Proof: Since A has rank 1, it follows that P is an Archimedean order and we may assume that A is order-isomorphie to a subgroup of Q. By Hion's Lemma (4.1.6), A is isomorphic to a subgroup of the multiplicative group of positive real numbers: to each A G A there corresponds an
6.10. TWO SIDED RIGHT
ORDERS
135
?"A G K+ such that X(a) - rxa for all a G A. Then rA £ Q since a G A C Q. Thus A is isomorphic to a subgroup of the multiplicative group of positive rationals, and so is a free Abelian group. // Recall that a subset X of a group G is G-invariant if g~lxg G X for all x e X, g eG. We will call a pair (V,V) a jump if V is a value (under the right order) and V" is its cover. Lemma 6.10.4 Suppose that P is a two-sided right order on G and rank(V'/V) = 1. // [G,V*] C V for all values V, then G is Abelian or there is a non-two-sided right ordering on G. Proof: Suppose the lemma were false. Let a,b e G with x = [a, b] / 1. We may clearly assume that a,b G P and that a < b (interchange a and b if necessary). Let V be the value of x under P; so [a, b] $. V and P = PV/V is a two-sided order on G/V with all convex jumps under P central with factors of rank 1 (V < G since [G, V] C [G, V'} C V). It is enough to show that there is a right order Q o n G = G/V which is not two sided, for then Q = {g G G : g G P n {geG:gePnVor(gtVkVgeQ)} V or (g # V & Vg G Q)} is a right order on G in which V is convex. So if Q were G-invariant, then Q would be G-invariant and hence two sided. We may thus replace G by G if necessary and assume that {1} is the value of x under P Let B — conp(b) be the convex subgroup of G (under P) generated by b; i.e., B is the intersection of all convex subgroups of G (under P) that contain b. Then a G B since 1 < a < b. Now any right order on B can be extended to a right order on G since G/B is orderable. It therefore suffices to produce a right order on B that is not two sided. So we may assume that B = G without loss of generality. Let C be the value of b (under P). Then C* = G and rank(G/C)=l by hypothesis. Thus every element of G/C(b) has finite order. Hence for every element g G G there is a least positive integer n such that gn = cbr for some c G G and r G Z. Let Pi = {<7 G G : c G P or (c = 1 & r > 0)}. Observe that if TO > 1 and gm = Cibs, then m = dn for some positive integer d and grn = {cbT)d = c\bTd where C] is a product of conjugates of
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c. Since P is two sided and c £ P, we have C\ G P and Ci = 1 if and only if c — 1. So <; £ P] if and only if 1. It is easy to verify that Pi n Pf 1 = {1} and that P U P{1 = G. We next shew that Pi is a right order on G. Let gug2 G P . Then #"J = c,-6r-> with nhr2 G Z, c,- G P n C and nj > 0, with r, > 0 if Cj — 1 (j = 1,2). If C\ = c% = 1, then flii32 G /(&)• Since G is orderable, 1(b) is a torsion-free Abelian group. Hence (ch^)" 1 " 2 = & ri " 2+r2ni and #i#2 G Pi- If c,c2 / 1, say d < c2, let V be the value of c2 under P By hypothesis, [G, V*] C K so that [<7i",22] £ ""• Since F is convex and normal in G, [31,32] G V Thus Vr(<7lfl2)nin2 = Vg?in2g2in2 Now K^" 1 2 = V(ClbT')n2 = Vc3br'n> where C3 = 1 if C\ — 1, C3 G P if c\ G P and cj and c:i have the same value. Similarly, Vg""17 = l/c467'2"1 where c4 and c2 are likewise related Therefore V(3i3 2 ) ni " 2 = Vc3c4br>n2+r'n> It follows that gxg2 G p . If C\ > c2 we similarly deduce that gig2 G P\. Consequently, P\ is a right order on G. However, {1} is the value of b under Pj with {1}* = 1(b), and 1(b) is not normal in G since [a, b] = x # 1(b). By Lemma 6.10.1, there is a right order on G that is not two sided. / We are now ready to prove the theorem. Proof: Let G be a right orderable group in which every right order is two-sided. By Lemma 6.10.1, every convex subgroup of G under any right ordering P is normal in G. If V is any value with respect to P then every right order on G/V is two sided; so every isolated subgroup / of G with V C / C V is normal in G by Lemma 6.10.2. So by refining the right order P on G if necessary, we may assume that V'/V has rank 1 for every value V for P We next show that [G, V*] C V by reductio ad absurdum. Let C be the centraliser of V'/V in G, a free Abelian group by Lemma 6.10.3. Therefore there is a subgroup H D C such that G/H is infinite cyclic (C / G by our assumption). Thus G - H(f) for some / € G. Let P n H be the G-invariant order on H obtained by restriction. Define Q = {hfn : l ^ / i G P n H or ( / t = l & n G Z + )}. Clearly Q n Q~l = {1} and Q U Q" 1 = G. Moreover QQ = Q since Q H H — P D H is G-invariant. Hence Q is a right order on G. It is two sided by hypothesis, {1} is the value of / under Q and (/) — {1}* By Lemma 6.10.1, [H, )] C H D {/) = {1}. But / G \C so that
6.10. TWO SIDED RIGHT
ORDERS
137
[V, (/)] ± {1}. This contradiction establishes that [G, V] C V; that is, every "jump" under P is central. Lemma 6.10.4 now applies and implies that G is Abelian. /
Chapter 7 Permutations Cayley's Theorem is of fundamental importance in group theory. It allows one to study groups by studying groups of permutations. In this chapter, we derive a similar theorem and study such groups; we will cover some of the consequences in the next chapter. For a more extensive account, please see [Glass 1981].
7.1
The Cayley-Holland Theorem
In Chapter 1, we introduced a very important class of lattice-ordered groups (Example 1.3.19). We now show that this is a universal class; i.e., every lattice-ordered group can be embedded in a member of the class. T h e o r e m 7.A [Holland 1963] Every lattice-ordered group G is l-isomorphic to an ^-subgroup of A(fl) for some totally ordered set f2 (of cardinality at most that of G). Proof: If G — {1}, the result is trivial; so assume that G ^ {1}. For each g £ G\{1}, let Cg be a value of g. By Corollary 3.3.7, H{Cg), the set of right cosets of Cg, is a totally ordered set (where Cgf < Cgh if and only if / < ch for some c G Cg). Totally order G\{1} arbitrarily; denote this total ordering by -<. Let Q = \J{Tl(Cg) : g G G\{1}}, totally ordered by: Cgf < C^k if and only if g -< h or (g = h and f < k). Note that Q, has cardinality at most that of G. Consider the map cf>: G -»• A(Q) given by (Cgf)(h) = Cg(fh). 139
140
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For each h € G, h£ A(fl). Moreover, since C/, is a value of h if h ^ 1, it follows that Chh ^ C/, if h ^ 1. Thus <> / is one-to-one. Trivially,
1; so / is not orthogonal to h. Therefore 0 is an £-homomorphism of G into A(Q).// If G is an ^-subgroup of A(i1) and a G ft, let GQ = {g £ G : ag — a}, the G-stabiliser of a. Clearly GQ is a prime (convex £-) subgroup of G. If H is a subgroup of ^4(ft), we say that H is a transitive permutation group (on ft) if for all a,/? G ft there is h € / / such that a/i = /?. That is, aH, the or6zi of a under H, is all of ft; aliter, ft is an //-orbit. A lattice-ordered group that is £-isomorphic to an ^-subgroup of <4(ft) for some linearly ordered set ft will be said to have a representation or, more accurately, a faithful representation on ft. A lattice-ordered group that is ^-isomorphic to a transitive ^-subgroup of ,4 (ft) for some linearly ordered set ft will be said to have a transitive representation or, more accurately, a faithful transitive representation on ft. Corollary 7.1.1 A lattice-ordered group G has a faithful transitive rep resentation if and only if it contains a prime subgroup C such that core(C)
= {!}• Proof: If G has a faithful transitive representation, then let a € ft. Then Ga is the desired prime subgroup. Since g~[Gag = Gp where 0 = ag, we get f}{Gp : 0 € ft} = {1}; so Ga is such a C. Conversely, given such a C, consider the map cj> given by: (Cg)(f>) = Ggf for / , g £ G Then <j> is an £-group homomorphism with kernel C\{g~lCg : g £ G} = {1}. So G<j> is transitive on ft = 71(C) and has trivial kernel. //
Corollary 7.1.2 Any lattice-ordered group whose only (.-ideals are the trivial ones has a faithful transitive representation. Corollary 7.1.3 Any subdirectly irreducible lattice-ordered group has a faithful transitive representation.
7.1. THE CAYLEY-HOLLAND
THEOREM
141
Proof: Let G be subdirectly irreducible with ^-monolith L\ let g € L with g > 1. If C is a value of 5, then the intersection of all conjugates of C is an £-ideal not containing g. By the definition of the /?-monolith, V\{f-lCf : f € G} = {1}. // Corollary 7.1.4 /In Abelian lattice-ordered group has a faithful transi tive representation if and only if it is an ordered group. Corollary 7.1.5 Any ordered group has a faithful transitive representa tion. Of course, this is just the right regular representation g 1-4 g where fg = fg for all f,g€G. Instead of totally ordering G\{1} in the proof of the Cayley-Holland Theorem, we could instead have taken the cardinal ordering on r W \ { i ) A(K(C9)) or even Ylg€G\{i}(G/Lg,ll(Cg)), where Lg = f]{f^lCgf : / G G}. Thus we obtain: Corollary 7.1.6 Every lattice-ordered group G is a subdirect product of transitive lattice-ordered groups. In Example 1.3.20, we saw how to right order the group A(Q)\ in deed, by taking all well-orderings of Q, we clearly obtain that the latticeordered group A(fl) can be order embedded in the cardinal product of right ordered groups. Since the cardinal product of right ordered groups can be order embedded in a right ordered group (Example 1.3.14) and any subgroup of a right ordered group is right ordered (Example 1.3.5), we obtain: Corollary 7.1.7 Every lattice-ordered group can be order embedded in a right ordered group; indeed, as a lattice-ordered group, it is a subdirect product of right ordered groups. We next give an application of the Cayley-Holland Theorem: T h e o r e m 7.B [Weinberg 1967] Every lattice-ordered group is t-isornorphic to an t-subgroup of A(F) for some totally ordered field F
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In order to prove the theorem, recall the definition of a group ring: Let G be a group and R be a ring. We write R[G] for the set of formal sums Y,geGrgx9> where rg € R and rg = 0 for all but a finite number of elements g £ G. It is an Abelian group under addition and a ring under polynomial multiplication: {Y.g&Grgxg){T,g^G sg%3) = E 9 eG tgx9 where tg = Y.f£G(rfsf-,g)If G is an ordered group and R is an ordered ring (e.g., a subring of R), then R[G] is an ordered ring when J2geGrgx9 > 0 if rf > 0 where / = max{g 6 G : rg ^ 0} (i.e., / is the maximal element of the support of £ 9 6G r 9 x 9 )Proof of Theorem l.B: By the Cayley-Holland Theorem, it suffices to prove that for any chain ft, there is an ordered field F such that A(Q) can be ^-embedded in A(F). Given ft, let G be the free Abelian group on ft: so G comprises all formal sums £ Q £ n naa where na £ Z and all but a finite number of na are 0. We extend the order on ft to G via: YLatn n a a > 0 if np > 0 where 0 = max{a e fl : n a ^ 0}; so G is an ordered Abelian group. We identify A € ft with its natural image in l^A e G; this identification preserves order. We form the group ring Q[G], an integral domain. Let F be its field of quotients ordered by: ab~l > 0 if and only if ab > 0 in Q[G]. We identify h E G with its image l/ l i ft in Q[G] and a € Q[G] with its image a l _ 1 € F These identifications preserve order. Define: A(Q) -> /1(G) by: ( £ n a a ) (/<£) ( / <=£$) =»$ »/ / ) a£fi
aefi
for / € i4(fi). Clearly, /<# G A(G) and /> extends / . An easy verification shows that-> A(Q[G}) by:
(£g E^ 9 ( / * ) sx»)(/^) =g(zG »6G »6G
g(zG
for / e ,4 (ft). Again an easy verification establishes that ip is an £embedding of A(tt)
7.1. THE CAYLEY-HOLLA
ND THEOREM
143
L e m m a 7.1.8 Let A be a chain and Q a subchain of A. If every element of A(Ci) can be extended to an element of A(A), then A(Q) can be lembedded in A(Q).
Proof: Let £(Q) be the set of all left segments in Q; i.e., A € £(fl) if A = {6 € 0. : 6 < a] for some a G Q. Let C(A) be the cut in A determined by A; i.e., C(A) = { A G A : A < A < Q\A}. For a fixed A0 G A, let A 0 = {a G 0 : a < A0}; so A0 G C(A 0 ). If A! G £(Q) with A 0 ^ Ai, then A 0 D Aj or A 0 C Aj. In the first case there is T G A 0 \ A i ; hence C(A\) < r < C(A 0 ). In the second case we similarly obtain that C(A 0 ) < C(Ai). Thus {C(A) : A G £(0)} inherits a total ordering in the natural way. We define an equivalence relation £ on £(Q) by: A\£ A 2 if C(Ai) and C(A 2 ) are isomorphic as ordered sets. Fix a member of each £ class; if A were chosen and A l 7 A 2 also belong to this class, let 7r AAl be an isomorphism between the ordered sets C(A) and C(Ai) with 7rA]A the identity, and 7rAliA2 = 7rAJAl7rAiA2; so nAuA = irA]Ar Clearly, 7rA,,A3 = "'Al,A37rA2,A3 f° r a 'l Aj, A 2 , A3 belonging to the same £ class. Let A G £(Q) and /* G .4(A) be an extension of / G A(Q). If A G C(A), then ax < A < a2 for all aY G A and Q 2 G fi\A. Thus c ^ / < AT < a 2 / f o r all such Qi,a 2 . Hence A/* G C ( A / ) , so C(A)/* C C ( A / ) . Similarly C ( A / ) ( / * ) - ' C C(A), so C(A)f = C(Af). Consequently, A / £ A for all A G g(Q). We can now define 8 : A(Q) -> .4(A) as follows: for A G A and r; G -4(Q), let A(##) = A7rA|Ay where A is the unique element of £(n) such that A G C(A). Now for all A G A and f,g G A(Q), if A G C(A), then A(/6>) G A / ; so \(f6)(g$) = XnAAfTiAfAfg = A?rA,A/9 = A(/#)0. Hence # is a group homomorphism. H f < g, then a 0 / < ^ofl for some Q0 G Q. Since A / C A3 for all A G /(fl) we obtain A(/<9) < A(p0). If A 0 = {a G 0 : a < a 0 } , then A 0 / C A0g; so C{A0f) < C{A0g) and hence ao(fO) < a0(g9). Thus 9 preserves order. Finally, an easy computation shows that A ( / V 1) = A / U A for all A G £(ll). Therefore TTA,A(/VI) = TA,A/ V 1. Consequently, (/ V 1)0 = fO V 1. Thus 8 is an ^-embedding. //
144
7.2
CHAPTER
7.
PERMUTATIONS
Amalgamation
As promised in Chapter 2, we now use the Cayley-Holland Theorem to prove that the class of lattice-ordered groups fails the amalgamation property. (For the definition of the amalgamation property, please see Section 2.4.) Theorem 7.C [K. R. Pierce] The class of lattice-ordered groups fails the amalgamation property. Proof: Let Hi = ( 2 ® Z ) V z where (TO,n,0) conjugated by (0,0,1) is (n,TO, 0). Then Hi is a (metabelian) lattice-ordered group. Let a — (1,0,0), b= (0,0,1), c = (0,0,1) and G = ((a) 0 (b))^ (c 2 ), an Abelian ^-subgroup of Hi. So A = (a) and B = (b) are prime subgroups of G. Thus TlG(A) and TlG{B) are totally ordered sets (Corollary 3.3.7) and so is fl — 7lc{A) U TZQ(B) if all the cosets of A are less than all the cosets of B; that is, f2 = ( Z ® Z ) i U (Z ^ ~ z ) 2 with the first copy of ( Z ^ Z ) less than the second. Since A n B = {1}, (G92,Q) is an £permutation group where $2 ■ G —> A(U) is given by: {Xf){g82) = Xfg (X e {A,B};f,g € G). Let h G A(Q) be given by: (vl/)ft = Af and ( 5 / ) / i = Bfc2 (f e G). Let H2 be the ^-subgroup of A{9.) generated by G02 and h, and let #i : G —> Hx be the inclusion map. We now prove by reductio ad absurdum that there does not exist a lattice-ordered group L with £-embeddings T/>, : Hi -> L (i = 1,2) such that (7#ii/>i = <7#2T/>2 for all 5 G G. So suppose that such L,ipi,ip2 did exist. Let C be the convex tsubgroup of H2 generated by b92 and d — h(c292)~l Since B(b92) = B and Bd = B, it follows that C C // 2 n A(Q) fl . Now 5(a0 2 ) = Ba ^ B, so a#2 0 C- Hence there is a value P of a92tp2(= aBiipi) con taining Ct/>2- Since c~lbc = a and b9{ipi = b92ip2 e Cip2, it follows that crpi $ P Since P is a value, it is prime; hence (by Corollary 3.3.7), A = TIL{P) is a totally ordered set. Let x '■ L —> A(A) be de fined by: (Px)(yx) = Pxy (x,y G L). Then x is an f-homomorphism and (Lx, A) is a transitive ^-permutation group. Since G is Abelian, Bfb = Bbf = Bf for all f e G. Thus supp(b2) C 71{A). Hence 602 A /i = 1 (in A(Q)). Therefore b92ip2X A hip2x = 1 (in A(A)). Now (Pah)(b0iTpiX) = P{cb)i>x = P(ac)^i = P(o*ifc)(cfc) 9^ Pqfe, as a#iV>i = a02i/)2 0 P Hence (Pcipi)(hip2x) = Pcipx. Consequently, PaPi = P(cA)(hihx) > P(hip2x) = P{dc292)ip2X = P{al>i)'\ as ah Z
7.3. CONVEX BLOCKS AND
P and dt(>2 G Ctp2 Q P Thus Pc^ contradiction since cipi > 1. /
7.3
145
CONGRUENCES > P(apx)2
This is the desired
Convex blocks and congruences
Let ft be a totally ordered set and G be a subgroup of A(Q). We say that an equivalence relation C on ft is a convex congruence of (G, ft) if all the equivalence classes of C are convex and for all g G G, agCflg whenever Note that there are two trivial convex congruences; they are the equiv alence relations aUP for all a, /? G ft and aS0 if and only if a = j3. If C is an equivalence relation on ft, we write aC for {/? G ft : aC/3}, the equivalence class of C that contains a, and ft/C for {QC : a G ft}, the set of all C-classes. If C is a convex congruence of (G, ft), then ft/C inherits a total order in a natural way: aC < j3C if a < r for all a € o.C and r G /?C. There is a natural action of G on ft/C given by: aCg — agC (g G G,a G ft), and the natural map ft —> ft/C is an order (^-)homomorphism. We denote the lazy subgroup of this action by L(C); that is, L(C) = {g G G : aC = aC / o r a^/ a G ft}. Let ft be a totally ordered set and G be a subgroup of ,4(ft). We say that A is a convex block of (G, ft) if A is a convex subset of ft and for all g G G, either A# = A or Ag D A = 0. The following lemma takes full advantage of the order present. L e m m a 7.3.1 Let (G, ft) be a transitive group of order-preserving per mutations of a totally ordered set ft. If A} are non-disjoint convex blocks o/(G,ft) (j = 1,2,), then A] C A 2 or A 2 C Aj. Proof: If the conclusion fails, then without loss of generality there are exj G Aj (j = 1, 2) such that a! < Aj n A 2 < a 2 . Let 6 G A, n A 2 . Let j e G map 6 to a i ; then Ai = A^ Hence A 2 j n A 2 ■£ 0, so A 2 # = A 2 . This contradicts that 0^ G A 2 #\A 2 . / Note that if (G, ft) is transitive, then {Ag : g G G} forms a set of equivalence classes of a convex congruence C(A) of (G, ft), and any
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convex congruence of (G, Q.) that has A as a convex block must equal C(A). Thus, if (G, Q) is transitive, there is a one-to-one map of the set of convex blocks containing a fixed point (say S) into the set of convex congruences of (G, fl). Conversely, if C is a convex congruence of (G, Q), then the class 5C is a convex block of (G, fi). Hence: L e m m a 7.3.2 If G is a subgroup of A(Q) for some totally ordered set Q, 5 e f i and (G,Q) is transitive, then there is a one-to-one correspondence between the set of convex blocks of (G, fi) containing 6 and the set of convex congruences of (G, fi). Hence the set of convex congruences of (G, Q) forms a totally ordered set under inclusion. If (G, Q) is not transitive, the lemma can fail: If A is a convex block, let C\ be the convex congruence whose classes are either Ag (g e G) or singletons. If there is a non-singleton convex subset A of Q that is disjoint from U{A<7 : g € G}, let C2 be the convex congruence whose classes are either A3 or Ag (g € G) or singletons. Then C\ ^ C2Given a subset A of Q and a subgroup G of A(Q), let G& be the (pointwise) stabiliser of A. That is, G& = {g £ G : b~g = 6 for all 5 £ A}. Clearly GA is a convex subgroup of G that is an ^-subgroup if G is an ^-group. If A = {<5}, a singleton, then the definition of GA agrees with the one we gave in the previous section for Gj. In distinction, we let G(A) be the setwise stabiliser of A. That is, G(A) = { € G : Ag = A}, the set of elements of G that map A onto itself. Clearly G(A) is also a subgroup of G that is an ^-subgroup of G if G is an £-group; moreover, GA Q G(A>- Observe that if C is a convex congruence of (G, fi), then
L(C) = n{G(ac) : a e n>.
Let A be a convex subset of fi and suppose that G is transitive on A; i.e., for all <$i, <52 6 A, there is g 6 Gsuch that 5xg — <52. Then ( G ( A ) | A , A) is transitive. If C is a convex congruence of (G, Q) and A is a union of C-classes (so L(C) is contained in G(A>), then let G(A) = G ( A ) / L ( C ) . Note that G(A) is naturally isomorphic to G ( A ) | A .
7.4
Primitive permutation groups
Let (G, Q) be a transitive group of order-preserving permutations of a totally ordered set Q. If the only convex congruences are the trivial ones
7.4. PRIMITIVE
PERMUTATION
GROUPS
147
then we say that (G, ft) is primitive. In this section we obtain a complete classification of primitive ^-permutation groups. To avoid trivial exceptions, we will assume that ft always denotes an infinite totally ordered set. A transitive group G of order-preserving permutations of a totally ordered set ft is said to be doubly transitive if for any a j , a 2 , f t , ft G ft with Q] < a 2 and ft < ft, there is g G G such that a3g = 0j for j = 1,2. More generally, a permutation group (G,ft) is said to be n-transitive if for all ct\ < ... < an and ft < ... < 0n in ft, there is g G G such that Q3y = ft for all j G {l,...,n}. For example, the permutation group (A(H),R) is doubly transitive. Indeed, since R is afield, there is a straight line joining ( a i , f t ) to (a/ 2 ,ft). L e m m a 7.4.1 If (G, ft) is a doubly transitive I'-permutation group, then it is n-transitive for all positive integers n. Proof: By induction on n. If a t < ... < an+i and 0\ < ... < 0n+\ in Q. let g € G map a,- to 03 (j — 1,..., n). If 0n+i < an+\9, let f eG be such that axf = /?„ and a n + i / = ft+iThen Oj{f A g) = m'mlajf^jg} = 0: for j = l , . . . , n + 1. If an+ig < 0n+\, let h € G be such that ctn/z = ft and a n + 1 /i = 0n+iThen a}(g V h) = max{acjg,QJ/I} = ft for j = 1,..., n + 1. Hence (G, Q) is (n + l)-transitive./ If the "£"-hypothesis is dropped, the conclusion is invalid. For each positive integer n, there are groups (G, ft) of order-preserving permuta tions that are n-transitive but not (n + l)-transitive. However, doubly transitive permutation groups ("£" or otherwise) are always primitive. L e m m a 7.4.2 If (G, ft) is a doubly transitive group of order-preserving permutations of a totally ordered set ft, then (G, ft) is primitive. Proof: If C were a non-trivial convex congruence, let a, ft S £ ft be distinct with 0 € aC and 5 # aC. Without loss of generality, a < ft If a < S, there is g G G such that ag = a and 0g — 6. By the former equality, we have aCg = aC (since C is a convex congruence of (G,ft)); so 6 = 0g G aC, a contradiction. If S < a, then there is g G G such that
148
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ag = 6 and fig = a. By the latter equality, aCg = aC and a contradiction again ensues. // Let Q be a totally ordered set. As in the construction of the real line from the rationals by Dedekind cuts, there is a natural way to complete fl: A non-empty proper subset A of Q is called a left cut of fi if (i) a G Q, S G A and a < 5 imply that Q G A, and (ii) A has no greatest element or, for some /? G ft, A = (1^ = {a e
n-.a
of GJ We now prove the theorem:
7.4. PRIMITIVE
PERMUTATION
GROUPS
149
Proof: Suppose that (G,fi) is primitive. We first show that G„ is a maximal convex subgroup of G for all a G Q. So assume that 5 G il. If C is a convex subgroup of G and G 5 C C C G, then aC is a convex block of (G, Q). For if ac^ < P < ac2 for some ci, c2 6 C, let # 6 G be such that ag = P (since (G, Q) is transitive on £2). Then /? = 5[(ci V j ) A c2j and cj! A c2 < (ci V g) A c2 < c2. Since C is a convex ^-subgroup of G (by the previous lemma), (cj V 5) A c2 € C Thus aC is a convex set. If A 6 aCf D aG, then A = a c i / = ac 2 for some Ci,C2 G C. Hence c c \f 2' ^ GQ C G, so / G C. Therefore aCf = aC and aC is a convex block of (G, f2). By primitivity, aC = Q. Now if ft e G, then ac — ah for some c 6 G. Hence ftc-1 6 G„ C G, so ft € C. Consequently, C = G as desired. More generally, let a £ Q. Let G be a convex subgroup of G with GQ C G C G. If G C \ G Q , then a
CHAPTER
150
7.
PERMUTATIONS
ft = ft = Z or ft is dense (if (G, ft) is primitive). Hence, in the latter case, 57G is dense in ft if 57 g ft. If 5 € ft\ft, define C by: aC0 if there is no g G G such that ag lies between a and 0. Then C is a convex congruence. It has more than one class since no point of ft below 57 can belong to the C-class of a point of ft above 57. Therefore C is the singleton congruence; i.e., aG is dense in ft. Conversely, if ft = ft = Z, then (G, ft) is primitive by the previous corollary. Suppose that C / U is a convex congruence of (G, ft). So some C-class has a supremum in ft, say 57. If fii,{fa were distinct points in this C-class, then there is no g G G such that ag lies between 0\ and 02. Thus C = S as 57G is dense in ft. / We now have two kinds of primitive ^-permutation groups: doubly transitive ones, and Archimedean ordered groups (under the right regular representation). We seek others. Example 7.4.6 Let ft = R and G = {g € A(R) : [a + l)g = ag + 1 for all a G R}. Clearly (G, ft) is a transitive ^-permutation group that is neither doubly transitive nor ordered. However, it is primitive: If a < 0, there is g G G such that a < ag < 0 < 0g so (a, 0) is never a block of (G, R). We generalise this example. A subset A of ft is said to be coterminal in ft if for every a G ft there are Si, 62 G A such that <5j < a < 62We let CA,QAG)
denote the centraliser of G in A(Q.).
Let (G, ft) be a transitive ^-permutation group. Suppose that there exists 1 < z G A (ft) such that CA,QJG) = (z) and {azn : n G Z} is coterminal in ft (for any a G ft). Then (G,ft) is said to be periodic with period z. Our goal in the rest of this section is to show that every primitive ^-permutation group is either doubly transitive, the right regular repre sentation of an Archimedean ordered group, or periodic. Theorem 7.E [McCleary 1972/3] The Trichotomy Theorem for Primitive ^-permutation Groups / / (G, ft) is a primitive £-permutation group, then either
7.4. PRIMITIVE
PERMUTATION
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(1) (G, ft) is doubly transitive; (2) G is an Archimedean ordered group and (G, ft) is the right regular representation; or (3) (G, ft) is periodic; if z is the period and a G ft then the restriction of Ga to ft n (azu,azn[~1) is doubly transitive. We achieve the proof with the aid of three lemmata. Let G be a subgroup of A(Q) and a G ft. We let Fix(Ga) - {a G ft : Ga C Ga}- So, if (G, ft) is a primitive ^-permutation group, then F~ix{Ga) = {67 e ft : Ga = Ga} by Theorem 7.D. L e m m a 7.4.7 Let (G, ft) be a primitive l-permutation group and Z = CA(nAG). Let a G ft, z G Z and a — az. Then a € Fix(GQ). Con versely, i / a G Fix(G0) and z is the unique member of A(Q) such that (ag)z = ag (g G G), then z £ Z. Proof: Let g G Ga,z G Z and 5 = az. Then TTg = azg = agz = QZ = a, so 5 G Ga- Hence a G Fix(Ga)Conversely, since 5 £ Fix(Ga), we have that GQ = GQ. Hence the map z0 : ft —> ft given by (a<7)zo — &9 {g & G) is well-defined. Let /? G ft and / G G. Let 5 G G be such that a# = ft Then /?/z 0 = (agf)z0 = agf = 0zof. Thus ZQ commutes with every element of G. If ft < ft in ft, let g G G + map ft to ft. Let /1 G G with a / ! = ft. Then a fig = ft > ft = a / i , so /iff/f 1 0 GQ = G^. Hence 5 / , .9 / a / i . Now ftz0 = 5/i < a/1.9 = ft-zo> s o -zo is one-to-one and preserves order. Therefore it extends uniquely to an element z of A (ft) which must still commute with G; i.e., z G Z.fj
Corollary 7.4.8 Let (G, ft) be a primitive l-permutation group, a G ft and Z = CA,QJG). Then the map z ^t az is an ordermorphism of the set Z onto Fix(Ga). Thus Z is an ordered group. Note that if G is an Archimedean ordered group, then (G, G) is prim itive and Z = Z if G is cyclic, and Z = R otherwise. If (G, ft) is doubly transitive, then Z - {1}. (For if 1 ^ g € .A(ft), let a G ft be such that a ^ a; let / G GQ\Gag- Then gf ^ fg so g ^ Z.) In contrast, periodic permutation groups are non-Abelian and have non-trivial centralisers.
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Suppose that A is an orbit of Ga; that is, A = {Sg : g G Ga} for any S G A. We have already seen that A is a convex subset of Q. We can thus totally order the set of Ga orbits in the natural way: Aj < A 2 if and only if 5\ < S2 for some (all) 53 e A , (j = l,2). The paired orbit, A' of A is defined by A' = {ag : ag'1 G A}. Lemma 7.4.9 If (G, D.) is a transitive (-permutation group, a £ (] and A is a Ga-orbit, then so is A', and the map A H-> A' is an antiorderrnorphism of the set of Ga orbits onto itself. Proof: Let 0 G A'. Then 0 - af for some / G G with a / " 1 G A. If ag G A' with ag-1 G A, there is h G Ga such that ag~l = af~lh. Hence f~xhg G Ga and 0f~lhg — ag. Moreover, if 8 = 0k for some k G Ga, then 6 = afk and a(fk)~l = af~l G A. Thus A' is a Ga orbit. If A i , A 2 are distinct Ga orbits, let a^" 1 G A)\A 2 . Then ag G A i \ A 2 (for if a s = a / G A 2 with a / " 1 G A 2 , then ag'1 = ( a / " 1 ) / ^ 1 G A 2 since fg~l G Ga). Hence A[ / A 2 . Since they are distinct. A', and A 2 are disjoint. If A] < A 2 , let ag~l G A ; (j = 1, 2). Then g^lg2 V 1 G Gn so msLx{ag],ag2} = ag^l V g f 1 ^ ) e A't. But ar/2 G A 2 so ag 2 < a g t . Thus A 2 < A',. Since A" = A, the proof is complete. // If A is a Ga orbit and a < A then we call A a positive orbit of GQ. In Example 7.4.6, Ga has a first positive orbit, namely (a, a+ 1); its paired orbit is (a - I,a). Both are non-singletons and the only point of Q between them is a. We now show that this phenomenon is quite general. Lemma 7.4.10 If (G,Q) is a primitive (.-permutation group that is nei ther doubly transitive nor regular, then for each a G f2 Ga has a first positive non-singleton orbit Aj and the only point of Q between L\\ and Aj is a. Proof: Since (G, ft) is not regular, Gn ^ {1}. non-singleton orbit A. If A' were a singleton, say Go C G j C G since {0} is a GQ-orbit. Let 6i,S2 G / G GQ with 8if = 62. If 6vg - a, then ag~lfg since ag~l,ag~lf G A, it follows that ag,af~lg G 1 l l 0g~ f~ g = 0, so g~ fg G Gp\Ga. This contradicts
Hence Ga has a {0}, then clearly A be distinct and = 62g ^ a. But A' = {/?}. Hence the maximality of
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Gn (Theorem 7.D). Therefore A' is not a singleton, and we may fur ther assume that A is a negative orbit. Consequently, supA G ft and we denote it by 5. Since A is a Ga orbit, it follows that Ga C G^; hence GQ = Gs- Choose gQ G G so that ag^ G A and let A! be the convex subset of ft generated by ay0Ga. We now show that Ax is the first positive Ga orbit. Since ft is not ordermorphic to Z and ag^1 < <^> the interval (a,ag0) is non-empty. Let 0 G ft D (a, ago). By Corollary 7.4.5, aG is dense in ft; so we may pick h 6 G+ such that a < ah < ft Then ah'1 < a so there is / 6 G+ = G± such that ah~l < ag^1 f'. Thus go~lfh A l e Ga. Now (ag0)(go~lfh A 1) < afh = ah < 0 < ag0, so 0 6 A]. Therefore (a,ago) C A]. Since a g" Ai, we have that a is the infimum of Ai as required. / We now complete the proof of McCleary's Trichotomy Theorem for primitive ^-permutation groups. Proof: Assume that (G, ft) is a primitive ^-permutation group that does not fit conclusion (1) or (2). By the previous lemma, for each a e ft Ga has a first positive orbit A! and it is not a single point. Since (G,ft) is not doubly transitive, there is 0 £ ft such that A] < 0. Let a = sup(Ai) G Fix(Ga). By Lemma 7.4.7 and Corollary 7.4.8, we obtain z G CA,QJG) with z > 1 such that az = a. For each n G Z, let A n = ft n (azn-\azn). Note that Aj = (a,az) and A n = A ^ " " 1 Since (z) C CA,m(G), each A n is a Ga orbit. Moreover, A n is paired with A_ n + 1 . Since U{A n : n G Z} is a convex block of (G,ft), it must be all of ft by primitivity. Hence {azn : n G z} is coterminal in ft for some (and hence all) a G ft. Consequently Fix(GQ) = {azn : n G z } , so (z) = CA,ftJG) by Corollary 7.4.8. Therefore (G, ft) is periodic. Finally we prove that (G Q |A, A) is doubly transitive, where A = 0Ga. Firstly, by periodicity, GQ is faithful on A. To prove double transitivity, it is enough to show that if /5i < 02 < 03 in A, then there is g G G Q nG/3, such that 02g = 0z- There is h G G+ such that 02h = 03. There is m 6 Z such that A = ( a 2 r a , a ^ m + 1 ) . Now a*™-1 < /? 3 z -1 < a z m < ft < ft/i < 02h = 03. Thus there is / G G^3 such that ft/ = ft/i. Let 5 = hf~l VI G G^,. Then 02g = ft and Q.g = m a x { a / i / " ' , a } = m a x { a / " ' , a } = a. Consequently, (G Q |A, A) is doubly transitive. / It is worth noting that (G Q |A, A) contains an element of bounded support. For if we choose g\ G G such that azmgx < azm and ft
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/?!, then gi V 1 G G„ and fixes all points in the non-empty set A D {azm,azmgi l). Therefore g\\& ^ 1 and is bounded below in A. But g+IAz - 1 is bounded above in A z _ 1 and so, by periodicity,
ag = 0. Corollary 7.4.11 [Holland 1965] If fl is a totally ordered set and ,4(0) is transitive, then A(Q.) is either the right regular representation of a subgroup o/R or is n-transitive for every positive integer n. McCleary has developed a similar beautiful theory for intransitive ^-permutation groups that relies on certain "natural" blocks and congru ences; it leads to a very similar result in the intransitive primitive case. The interested reader should consult [Glass 1981, Chapters 3 & 4] or [McCleary 1976].
7.5
Primitive components
Let (G,Q) be a transitive group of order-preserving permutations of the totally ordered set Q, with convex congruences C C AC. Let A be a K. class. Then (G( A ),A) is a transitive action in the natural way and C — C|A x A is a convex congruence of (G( A ),A). Thus (G( A ). A / C ) is a transitive action. If L denotes its lazy subgroup (i.e., L = {g G G(A) : Afg = A / for all f e G}), then G ( A ) = G ( A ) / L Si G ( A ) |A.' So (G(A), A / C ) is a faithful transitive action called the component of (G, ft) relative to (C,fC) at A. We first show that the component is independent of the particular AC class chosen. Lemma 7.5.1 Let (G,fi) be a transitive group of order-preserving per mutations of the totally ordered set Q. with convex congruences C C /C. Let Ai, A2 be K. classes. Then the components of (G, D.) relative to (C, AC) at A] and A2 are order-isomorphic; indeed, £-isomorphic if (G, fi) is an ^-permutation group.
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Proof: There exists g € G such that A^g = A 2 . Clearly, g~ 'G(A,) = = G(A 2 )- Let C[ = C|A; x A, (i = 1,2), and L2 = .(/ ~xL\g where L\ is the lazy subgroup of (G(A,), Ai/C'J. Then L2 is the lazy subgroup of (G ( A 2 ),A 2 /C 2 ). For {, e A h let (^CJ)^ = (M2 and {LJ)ip = ^ " V j ( / 6 G ( A l ) )- Then V : G ( A l ) = G ( A 2 ) and 0 : A./Cj 2 A 2 /C 2 . Now toClM^/W = <5 l5 C 2 L 25 -'/5 = *i/flCj = [ ( ^ C l ) ^ ! / ) ] ^ Hence (G ( A l ) , Ai/Cl) = (G(A 2 ), A 2 /C 2 ) as desired. // C(A 1 9 )
From now on we will therefore drop all reference to A in the compo nent of a transitive action relative to a pair of congruences. We will also write A/C for A/(C n (A x A)) if A is a union of C classes. If C C K. are convex congruences of a transitive group of orderpreserving permutations (G, Q), then we say that K. covers C if there is no convex congruence V of (G,f2) such that C C V C /C. L e m m a 7.5.2 Let (G, Q) be a transitive group of order-preserving per mutations with convex congruences C, K, such that C C fC. Then K, covers C if and only if the component of (G, fi) relative to (C, K.) is primitive. Proof: Let A b e a C class. Any convex block of (G(A), A/C) has the form A/C where A C A is a block of (G (A ), A). If g € G and AgflA / 0, then Ag n A jt 0 since A C A. Because A is a convex block, Ag = A. Hence g € G(A)! SO Ay = A. Thus A is a block of (G, 0) and the lemma follows from Lemma 7.3.2. / We close this section with an application of the Trichotomy Theorem. Let (G, Q) be an ^-permutation group. We say that suprema are pointwise in G if g — V,e/ ft in G implies ay = sup{agl : 1 £ 1} for all a e f2 (the supremum being in Q). Corollary 7.5.3 >i(f2) is completely distributive for all totally ordered sets fi. Indeed, suprema in A(Q) are pointwise. In particular, A(il)a is closed for all a € £1. Proof: We confine our proof to the transitive case (for which we have the necessary theory); McCleary's results are needed for the intransitive case. It clearly suffices to prove: g — V,e/ft in A(Q) implies that ag = sup{agi : i € / } for all a e fi. For reductio ad absurdum, assume that
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ag > P > cagi for some a € Q and alii G I. If / € A(Q)+ and supp(f) C (P,ag), then g > gf~l and 5gf~l > 5g{ for all t G i and 8 £ fi. Hence / = 1 and a
7.6
The Wreath product
In Chapter 1 we introduced the notion of the Wreath product G Wr H when H was an ordered group (see Example 1.3.27). We wish to extend this definition to include the possibility that H is a lattice-ordered group. We will prove some easy consequences of the definition. By the Cayley-Holland Theorem, a lattice-ordered group H is lisomorphic to an ^-subgroup of A(Q) for some linearly ordered set Q. We will identify H with its ^-isomorphic image in A(Q). Let B = Ylaen G(a), where each G(a) is an ^-isomorphic copy of G. We define the Wreath product of G and H to be the group W — B » H with the following partial ordering: (b, h) € W+ if and only if h e H+ and b(a) > 1 for all a G fl for which ah = a. With respect to this partial ordering, it is easily checked that W is a lattice-ordered group. In the case that H is an ordered group and H acts on Q — H by the right regular representation, the lattice-ordered group just constructed is the Wreath product in Example 1.3.27. Caution: The actual Wreath product used depends on the represen tation of H, unlike the group case (where only the right representation is considered). The difficulty for lattice-ordered groups is that except when H is an ordered group, there is no obvious "unique" representation to choose; the right regular representation does not yield a lattice-ordered group. The first theorem we wish to establish is that, as in groups, the
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PRODUCT
157
Wreath product is universal for splitting extensions of lattice-ordered groups under special circumstances. Let G be a lattice-ordered group and L a prime ^-ideal of G. We say that G is a lex extension of L if I < c for all ( € L and c € G+\L. If G is a lex extension of L, then L is closed: if {; : z € / } C L + and c = Viergi & L, then L ^ {1} and c > c^ _1 > , for all i £ I and g £ L + ; this contradicts that c = Vie/9'Theorem 7.F Let G be a lattice-ordered group with prime (.-ideal L. If G is a lex extension of L, then G can be (-embedded in L Wr G/L. Proof: Let T be a right transversal for L in G; that is, 1 € T and for each g £ G, there is a unique tg e G such that Lg = Ltg. Let B = \\{L : t € T } . We map G into W = L Wr G/L = B x G/L by: 50 = (#, Lg) where («) = tyf,"1 G L. Then {g<£){h<j))^ = (b,Lgh) where 6(t) = g(t)h(ttg). Hence 6(f) = tgt^gttght~gh = {gk)(t) for all < G T. Therefore, 0 is a group homomorphism. If / < g in G, then either Lf < Lg (in which case f< gcf>) or Lf — Lg. In the latter case, Ltf = Ltg for all t € T since L < G; i.e., £(/ = i (9 for all t e T. Thus /(*) = £"/ = t/^" 1 < tgt^gl = (£) for all t G T. Consequently, / 0 < gcj> and 0 is an order-preserving embedding. Finally, if / i . g in G, then since G is a lex extension of L, we have / , g 6 L. Since L is normal in G, it follows that Ltf — Lt — Ltg for all t. e T. Therefore f(t) = tft'1 L tgt~l = g(t) for all t e T; i.e., /> 1 g(p. Consequently, 4> is an ^-embedding. / For the more general case, we need to pass to the permutation group setting. Theorem 7.G Let C be a congruence of a transitive (-permutation group (G, Q). If A is a C-class, then (G, fi) can be (-embedded in {W, A) = (G ( A ) ,A) Wr(G/L(C),fi/C). Proof: For each C-class A, choose cA e G such that AcA = A and c A = 1- Let 4> : Q -> A = A £~ (12/C) be given by a = (ac~l,aC). Clearly, is onto. It is also order-preserving: If a < f) either aC < 0C (in which case acf> < p<j) trivially), or aC = [3C. In this latter case, act) = {acZc,aC) = (ac$,0C) < (/?c^,/3C) = 0
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Let g i-> g be the natural map of G onto G/L(C), and for each C-class A, let „ = c ^ c ^ l A G G"(A). Let i/> : g i-4 ( { ^ } , g ) . We now show that i/> is an ^embedding. _If / < g, then a / C < agC for all a e O. Therefore / < 9- I f Aj = Ag, then afA = acAfc~A-; = acAfc~A\ < agA for all a G ft; so /?/> < gil>. Further, if /?/ < fig, then either /?C/ < fiCg, or /j/jc < fig0c. Hence ftp < gip. The same argument shows that (/ V l)ip — fip V 1. If ft € G, then {fh)A = c^/Zic"^ = c^c^c^nc^ = fAhAj. Thus {fh)ip = (fip)(hip), so ip is an ^-embedding. Now {ag)