The Project Gutenberg EBook of Groups of Order p^m Which Contain Cyclic Subgroups of Order p^(m-3), by Lewis Irving Neikirk Copyright laws are changing all over the world. Be sure to check the copyright laws for your country before downloading or redistributing this or any other Project Gutenberg eBook. This header should be the first thing seen when viewing this Project Gutenberg file. Please do not remove it. Do not change or edit the header without written permission. Please read the "legal small print," and other information about the eBook and Project Gutenberg at the bottom of this file. Included is important information about your specific rights and restrictions in how the file may be used. You can also find out about how to make a donation to Project Gutenberg, and how to get involved.
**Welcome To The World of Free Plain Vanilla Electronic Texts** **eBooks Readable By Both Humans and By Computers, Since 1971** *****These eBooks Were Prepared By Thousands of Volunteers!*****
Title: Groups of Order p^m Which Contain Cyclic Subgroups of Order p^(m-3) Author: Lewis Irving Neikirk Release Date: February, 2006 [EBook #9930] [Yes, we are more than one year ahead of schedule] [This file was first posted on November 1, 2003] Edition: 10 Language: English Character set encoding: TeX *** START OF THE PROJECT GUTENBERG EBOOK GROUPS OF ORDER P^M ***
Produced by Cornell University, Joshua Hutchinson, Lee Chew-Hung, John Hagerson, and the Online Distributed Proofreading Team.
GROUPS OF ORDER pm WHICH CONTAIN CYCLIC SUBGROUPS OF ORDER pm−3 by
LEWIS IRVING NEIKIRK sometime harrison research fellow in mathematics
1905
2
INTRODUCTORY NOTE. This monograph was begun in 1902-3. Class I, Class II, Part I, and the selfconjugate groups of Class III, which contain all the groups with independent generators, formed the thesis which I presented to the Faculty of Philosophy of the University of Pennsylvania in June, 1903, in partial fulfillment of the requirements for the degree of Doctor of Philosophy. The entire paper was rewritten and the other groups added while the author was Research Fellow in Mathematics at the University. I wish to express here my appreciation of the opportunity for scientific research afforded by the Fellowships on the George Leib Harrison Foundation at the University of Pennsylvania. I also wish to express my gratitude to Professor George H. Hallett for his kind assistance and advice in the preparation of this paper, and especially to express my indebtedness to Professor Edwin S. Crawley for his support and encouragement, without which this paper would have been impossible. Lewis I. Neikirk. University Of Pennsylvania, May, 1905.
3
GROUPS OF ORDER pm , WHICH CONTAIN CYCLIC SUBGROUPS OF ORDER p(m−3)1 by lewis irving neikirk Introduction. The groups of order pm , which contain self-conjugate cyclic subgroups of orders pm−1 , and pm−2 respectively, have been determined by Burnside,2 and the number of groups of order pm , which contain cyclic non-self-conjugate subgroups of order pm−2 has been given by Miller.3 Although in the present state of the theory, the actual tabulation of all groups of order pm is impracticable, it is of importance to carry the tabulation as far as may be possible. In this paper all groups of order pm (p being an odd prime) which contain cyclic subgroups of order pm−3 and none of higher order are determined. The method of treatment used is entirely abstract in character and, in virtue of its nature, it is possible in each case to give explicitly the generational equations of these groups. They are divided into three classes, and it will be shown that these classes correspond to the three partitions: (m−3, 3), (m − 3, 2, 1) and (m − 3, 1, 1, 1), of m. We denote by G an abstract group G of order pm containing operators of order pm−3 and no operator of order greater than pm−3 . Let P denote one of these operators of G of order pm−3 . The p3 power of every operator in G is contained in the cyclic subgroup {P }, otherwise G would be of order greater than pm . The complete division into classes is effected by the following assumptions: 2
I. There is in G at least one operator Q1 , such that Qp1 is not contained in {P }. II. The p2 power of every operator in G is contained in {P }, and there is at least one operator Q1 , such that Qp1 is not contained in {P }. III. The pth power of every operator in G is contained in {P }.
1 Presented
to the American Mathematical Society April 25, 1903. of Groups of a Finite Order, pp. 75-81. 3 Transactions, vol. 2 (1901), p. 259, and vol. 3 (1902), p. 383. 2 Theory
4
The number of groups for Class I, Class II, and Class III, together with the total number, are given in the table below:
p>3 m>8 p>3 m=8 p>3 m=7 p=3 m>8 p=3 m=8 p=3 m=7
I
II1
II2
II3
II
III
Total
9
20 + p
6 + 2p
6 + 2p
32 + 5p
23
64 + 5p
8
20 + p
6 + 2p
6 + 2p
32 + 5p
23
63 + 5p
6
20 + p
6 + 2p
6 + 2p
32 + 5p
23
61 + 5p
9
23
12
12
47
16
72
8
23
12
12
47
16
71
6
23
12
12
47
16
69
Class I. 1. General notations and relations.—The group G is generated by the two operators P and Q1 . For brevity we set4 Qa1 P b Qc1 P d · · · = [a, b, c, d, · · · ]. Then the operators of G are given each uniquely in the form ! y = 0, 1, 2, · · · , p3 − 1 . [y, x] x = 0, 1, 2, · · · , pm−3 − 1 We have the relation 3
3
Qp1 = P hp .
(1)
There is in G, a subgroup H1 of order pm−2 , which contains {P } self-conjugately.5 The subgroup H1 is generated by P and some operator Qy1 P x of G; it then 2 contains Qy1 and is therefore generated by P and Qp1 ; it is also self-conjugate in H2 = {Qp1 , P } of order pm−1 , and H2 is self-conjugate in G. From these considerations we have the equations6 (2)
2
2
Q1−p P Qp1 = P 1+kp
m−4
,
2
(3)
βp p P α1 , Q−p 1 P Q1 = Q1
(4)
bp a1 Q−1 1 P Q1 = Q1 P .
4 With J. W. Young, On a certain group of isomorphisms, American Journal of Mathematics, vol. 25 (1903), p. 206. 5 Burnside: Theory of Groups, Art. 54, p. 64. 6 Ibid., Art. 56, p. 66.
5
2. Determination of H1 . Derivation of a formula for [yp2 , x]s .—From (2), by repeated multiplication we obtain [−p2 , x, p2 ] = [0, x(1 + kpm−4 )]; and by a continued use of this equation we have [−yp2 , x, yp2 ] = [0, x(1 + kpm−4 )y ] = [0, x(1 + kypm−4 )]
(m > 4)
and from this last equation, [yp2 , x]s = syp2 , x{s + k 2s ypm−4 } .
(5)
3. Determination of H2 . Derivation of a formula for [yp, x]s .—It follows from (3) and (5) that p βk α1p − 1 m−4 α −1 2 p p , α1 1 + p (m > 4). [−p2 , 1, p2 ] = β 1 α1 − 1 2 α1 − 1 Hence, by (2), α1p − 1 2 p ≡ 0 (mod p3 ), α1 − 1 αp − 1 2 βk αp1 − 1 m−4 p p +β 1 hp ≡ 1 + kpm−4 α1 1 + 2 α1 − 1 α1 − 1 β
(mod pm−3 ).
From these congruences, we have for m > 6 αp1 ≡ 1
(mod p3 ),
α1 ≡ 1
(mod p2 ),
and obtain, by setting α1 = 1 + α2 p2 , the congruence (1 + α2 p2 )p − 1 (α2 + hβ)p3 ≡ kpm−4 α2 p3
(mod pm−3 );
and so (α2 + hβ)p3 ≡ 0
(mod pm−4 ),
since (1 + α2 p2 )p − 1 ≡1 α2 p3 6
(mod p2 ).
From the last congruences (α2 + hβ)p3 ≡ kpm−4
(6)
(mod pm−3 ).
Equation (3) is now replaced by 2
2
βp −p 1+α2 p . Q−p 1 P Q1 = Q1 P
(7) From (7), (5), and (6)
[−yp, x, yp] = βxyp2 , x{1 + α2 yp2 } + βk x2 ypm−4 . A continued use of this equation gives [yp, x]s = [syp + β
(8)
s 2
xyp2 ,
xs +
s 2
{α2 xyp2 + βk
x 2
ypm−4 } + βk
s 3
x2 ypm−4 ].
4. Determination of G.—From (4) and (8), [−p, 1, p] = [N p, ap1 + M p2 ]. From the above equation and (7), ap1 ≡ 1
(mod p2 ),
a1 ≡ 1
(mod p).
Set a1 = 1 + a2 p and equation (4) becomes bp 1+a2 p Q−1 . 1 P Q1 = Q1 P
(9) From (9), (8) and (6)
# 2 (1 + a2 p)p − 1 p2 bp, (1 + a2 p) , [−p , 1, p ] = a2 p "
2
2
and from (1) and (2) 2
(1 + a2 p)p − 1 bp ≡ 0 a2 p
(mod p3 ),
2
2
(1 + a2 p)p + bh
(1 + a2 p)p − 1 p ≡ 1 + kpm−4 a2 p
(mod pm−3 ).
By a reduction similar to that used before, (10)
(a2 + bh)p3 ≡ kpm−4
(mod pm−3 ).
The groups in this class are completely defined by (9), (1) and (10). 7
These defining relations may be presented in simpler form by a suitable choice of the second generator Q1 . From (9), (6), (8) and (10) 3
[1, x]p = [p3 , xp3 ] = [0, (x + h)p3 ]
(m > 6),
and, if x be so chosen that x+h≡0
(mod pm−6 ),
Q1 P x is an operator of order p3 whose p2 power is not contained in {P }. Let Q1 P x = Q. The group G is generated by Q and P , where 3
Qp = 1,
m−3
Pp
= 1.
Placing h = 0 in (6) and (10) we find α2 p3 ≡ a2 p3 ≡ kpm−4
(mod pm−3 ).
Let α2 = αpm−7 , and a2 = apm−7 . Equations (7) and (9) are now replaced by 2
m−5
Q−p P Qp = Qβp P 1+αp
(11)
m−6
Q−1 P Q = Qbp P 1+ap
,
.
As a direct result of the foregoing relations, the groups in this class correspond to the partition (m − 3, 3). From (11) we find7 [−y, 1, y] = [byp, 1 + aypm−6 ]
(m > 8).
It is important to notice that by placing y = p and p2 in the preceding equation we find that8 b≡β
a≡α≡k
(mod p),
(mod p3 )
(m > 7).
A combination of the last equation with (8) yields9 (12)
x 2 m−6
[−y, x, y] = [bxyp + b2 x(1 + ayp
yp2 ,
) + ab
x 2
ypm−5 + ab2
x 3
ypm−4 ]
(m > 8).
m = 8 it is necessary to add a2 y2 p4 to the exponent of P and for m = 7 the terms a(a + abp ) y2 p2 + a3 y3 p3 to the exponent of P , and the term ab y2 p2 to the exponent of Q. 2 The extra term 27ab2 k y3 is to be added to the exponent of P for m = 7 and p = 3. 7 For
2 3
m = 7, ap2 − a 2p ≡ ap2 (mod p4 ), ap3 ≡ kp3 (mod p4 ). For m = 7 and p = 3 the first of the above congruences has the extra terms 27(a3 + abβk) on the left side. 9 For m = 8 it is necessary to add the term a y xp4 to the exponent of P , and for m = 7 2 the terms x{a(a + abp ) y2 p2 + a3 y3 p3 } to the exponent of P , with the extra term 27ab2 k y3 x 2 for p = 3, and the term ab y2 xp2 to the exponent of Q. 8 For
8
From (12) we get10 [y, x]s = ys + by (x + b x2 p) 2s + x 3s p p, xs + ay (x + b x2 p + b2 x3 p2 ) 2s + (bx2 p + 2b2 x x2 p2 ) 3s + bx2 4s p2 pm−6
(13)
(m > 8).
5. Transformation of the Groups.—The general group G of Class I is specified, in accordance with the relations (2) (11) by two integers a, b which (see (11)) are to be taken mod p3 , mod p2 , respectively. Accordingly setting a = a1 pλ ,
b = b1 p µ ,
where dv[a1 , p] = 1,
dv[b1 , p] = 1
(λ = 0, 1, 2, 3; µ = 0, 1, 2),
we have for the group G = G(a, b) = G(a, b)(P, Q) the generational determination: ( G(a, b) :
µ+1
Q−1 P Q = Qb1 p 3
m+λ−6
P 1+a1 p
m−3
Qp = 1,
Pp
= 1.
Not all of these groups however are distinct. Suppose that G(a, b)(P, Q) ∼ G(a0 , b0 )(P 0 , Q0 ), by the correspondence C=
Q, Q01 ,
P P10
,
where 0
0 m−6
Q01 = Q0y P 0x p
,
and
10 For
m = 8 it is necessary to add the term P , and for m = 7 the terms x
na 2
a+
P10 = Q0y P 0x ,
1 axy 2s [ 31 y(2s 2
− 1) − 1]p4 to the exponent of
ab 2s − 1 a3 s 2 p y − 1 2s yp2 + y − (2s − 1)y + 2 yp3 2 3 3! 2 a2 bxy 2 s 3s − 1 3 a2 b s(s − 1)2 (s − 4) + p + y− 3 2 2 2 4!
s 3 yp 3
o
with the extra terms n bk o s 2 27abxy y − (2s − 1)y + 2 3s + x(b2 k + a2 )(2y 2 + 1) 3s , 2 3! for p = 3, to the exponent of P , and the terms ab 2s − 13 y − 1 2s xyp2 to the exponent of Q. 2
9
with y 0 and x prime to p. Since m−6
Q−1 P Q = Qbp P 1+ap
,
then −1
1+apm−6
bp
Q0 1 P10 Q01 = Q0 1 P 0 1
,
or in terms of Q0 , and P 0 y + b0 xy 0 p + b02 x2 y 0 p2 , x(1 + a0 y 0 pm−6 ) + a0 b0 x2 y 0 pm−5 + a0 b02 x3 y 0 pm−4 = [y + by 0 p, x + (ax + bx0 p)pm−6 ]
(m > 8)
and (14) (15)
by 0 ≡ b0 xy 0 + b02 x2 y 0 p (mod p2 ), ax + bx0 p ≡ a0 y 0 x + a0 b0 x2 y 0 p + a0 b02 x3 y 0 p2 (mod p3 ).
The necessary and sufficient condition for the simple isomorphism of these two groups G(a, b) and G(a0 , b0 ) is, that the above congruences shall be consistent and admit of solution for x, y, x0 and y 0 . The congruences may be written 0 0 2 b1 pµ ≡ b01 xpµ + b0 1 x2 p2µ +1 (mod p2 ), a1 xpλ + b1 x0 pµ+1 ≡ 0
y 0 {a01 xpλ + a01 b01
x 2
0
0
2 x λ0 +2µ0 +2 } 3 p
pλ +µ +1 + a01 b0 1
(mod p3 ).
Since dv[x, p] = 1 the first congruence gives µ = µ0 and x may always be so chosen that b1 = 1. We may choose y 0 in the second congruence so that λ = λ0 and a1 = 1 except for the cases λ0 ≥ µ + 1 = µ0 + 1 when we will so choose x0 that λ = 3. The type groups of Class I for m > 811 are then given by 1+µ
(I) G(pλ , pµ ) : Q−1 P Q = Qp
m−6+λ
P 1+p
3
m−3
, Qp = 1, P p
=1
! µ = 0, 1, 2; λ = 0, 1, 2; λ ≥ µ; . µ = 0, 1, 2; λ = 3 Of the above groups G(pλ , pµ ) the groups for µ = 2 have the cyclic subgroup {P } self-conjugate, while the group G(p3 , p2 ) is the abelian group of type (m − 3, 3).
11 For m = 8 the additional term ayp appears on the left side of the congruence (14) and G(1, p2 ) and G(1, p) become simply isomorphic. The extra terms appearing in congruence (15) do not effect the result. For m = 7 the additional term ay appears on the left side of (14) and G(1, 1), G(1, p), and G(l, p2 ) become simply isomorphic, also G(p, p) and G(p, p2 ).
10
Class II. 1. General relations. 2 There is in G an operator Q1 such that Qp1 is contained in {P } while Qp1 is not. 2
2
Qp1 = P hp .
(1)
The operators Q1 and P either generate a subgroup H2 of order pm−1 , or the entire group G.
Section 1. 2. Groups with independent generators. Consider the first possibility in the above paragraph. There is in H2 , a subgroup H1 of order pm−2 , which contains {P } self-conjugately.12 H1 is generated by Qp1 and P . H2 contains H1 self-conjugately and is itself self-conjugate in G. From these considerations13 p 1+kp Q−p 1 P Q1 = P
(2)
Q−1 1
(3)
PQ=
m−4
,
α1 Qβp . 1 P
3. Determination of H1 and H2 . From (2) we obtain (4) [yp, x]s = syp, x s + k 2s ypm−4 (m > 4), and from (3) and (4) βk αp1 − 1 m−4 αp1 − 1 p βp, α1 1 + p . [−p, 1, p] = α1 − 1 2 α1 − 1
A comparison of the above equation with (2) shows that αp1 − 1 βp ≡ 0 (mod p2 ), α1 − 1 βk αp1 − 1 m−4 αp − 1 p α1 1 + p + 1 βhp ≡ 1 + kpm−4 2 a1 − 1 α1 − 1
(mod pm−3 ),
and in turn αp1 ≡ 1
(mod p2 ),
α1 ≡ 1
(mod p)
(m > 5).
Placing α1 = 1 + α2 p in the second congruence, we obtain as in Class I (α2 + βh)p2 ≡ kpm−4
(5)
(mod pm−3 )
12 Burnside, 13 Ibid.,
Theory of Groups, Art. 54, p. 64. Art. 56, p. 66.
11
(m > 5).
Equation (3) now becomes β 1+α2 p Q−1 . 1 P Q1 = Q P
(6)
The generational equations of H2 will be simplified by using an operator of order p2 in place of Q1 . From (5), (6) and (4) [y, x]s = [sy + Us p, sx + Ws p] in which Us = β
s 2
xy, n Ws = α2 2s xy + βk 2s x2 +
s 3
x2 y
1 1 + αk s(s − 1)(2s − 1)y 2 − 2 3!
s 2
o m−5 y x p .
Placing s = p2 and y = 1 in the above 2
2
2
2
[Q1 P x ]p = Qp1 P xp = P (x+h)p . If x be so chosen that (mod pm−5 )
(x + h) ≡ 0
(m > 5)
Q1 P x will be the required Q of order p2 . Placing h = 0 in congruence (5) we find α2 p2 ≡ kpm−4
(mod pm−3 ).
Let α2 = αpm−6 . H2 is then generated by 2
Qp = 1,
m−3
Pp
= 1. m−5
Q−1 P Q = Qβp P 1+αp
(7)
.
Two of the preceding formulæ now become (8) [−y, x, y] = βxyp, x(1 + αypm−5 ) + βk x2 ypm−4 , [y, x]s = [sy + Us p, xs + Ws pm−5 ],
(9) where
s 2
Us = β
xy
14
and
Ws = α 14 For
s 2
xy + βk
s x 2
2
m = 6 it is necessary to add the terms
12
+
ak 2
n
s 3
x2 yp
(m > 6).
s(s−1)(2s−1) 2 y 3!
−
s y 2
o
p to Ws .
4. Determination of G. Let R1 be an operation of G not in H2 . Rp1 is in H2 . Let Rp1 = Qλp P µp .
(10)
Denoting Ra1 Qb P c Rd1 Qe P f · · · by the symbol [a, b, c, d, e, f, · · · ], all the operations of G are contained in the set [z, y, x]; z = 0, 1, 2, · · · , p − 1; y = 0, 1, 2, · · · , p2 − 1; x = 0, 1, 2, · · · , pm−3 − 1. The subgroup H2 is self-conjugate in G. From this15 b1 a1 R−1 1 P R1 = Q P ,
(11)
m−5
d1 c1 p R−1 1 Q R1 = Q P
(12)
.
In order to ascertain the forms of the constants in (11) and (12) we obtain from (12), (11), and (9) [−p, 1, 0, p] = [0, dp1 + M p, N pm−5 ]. By (10) and (8) Rp1 Q Rp1 = P −µp Q P µp = Q P −aµp
m−4
.
From these equations we obtain dp1 ≡ 1
(mod p)
d1 ≡ 1
and
(mod p).
Let d1 = 1 + dp. Equation (12) is replaced by m−5
1+dp e1 p R−1 P 1 Q R1 = Q
(13)
.
From (11), (13) and (9) [−p, 0, 1, p] =
ap1 − 1 b1 + Kp, ap1 + b1 Lpm−5 a1 − 1
in which K = a1 b1 β
p−1 X
ay 1 2
.
1
By (10) and (8) p −λp R−p P Qλp = P 1+aλp 1 P R1 = Q 15 Burnside,
Theory of Groups, Art. 24, p. 27.
13
m−4
,
and from the last two equations ap1 ≡ 1
(mod pm−5 )
and (mod pm−6 )
(m > 6);
a1 ≡ 1
Placing a1 = 1 + a2 pm−6
(m > 6);
a1 = 1 + a2 p
a1 ≡ 1
K≡0
(mod p)
(m = 6).
(m = 6).
(mod p),
and16 ap1 − 1 b 1 ≡ b1 p ≡ 0 a1 − 1
(mod p2 ),
b1 ≡ 0
(mod p).
Let b1 = bp and we find ap1 ≡ 1
(mod pm−4 ),
(mod pm−5 ).
a1 ≡ 1
Let a1 = 1 + a3 pm−5 and equation (11) is replaced by (14)
m−5
bp 1+a3 p R−1 1 P R1 = Q P
.
The preceding relations will be simplified by taking for R1 an operator of order p. This will be effected by two transformations. From (14), (9) and (13)17 h c1 y m−4 i −c1 y m−4 i h p = 0, (λ + y)p, µp − p , [1, y]p = p, yp, 2 2 and if y be so chosen that λ+y ≡0
(mod p),
R2 = R1 Qy is an operator such that Rp2 is in {P }. Let Rp2 = P lp . Using R2 in the place of R1 , from (15), (9) and (14) h ax m−4 i ax m−4 i h p = 0, 0, (x + l)p + p , [1, 0, x]p = p, 0, xp + 2 2 16 K has an extra term for m = 6 and p = 3, which reduces to 3b c . This does not affect 1 1 the reasoning except for c1 = 2. In this case change P 2 to P and c1 becomes 1. 17 The extra terms appearing in the exponent of P for m = 6 do not alter the result.
14
and if x be so chosen that ax m−5 p ≡0 2
x+l+
(mod pm−4 ),
then R = R2 P x is the required operator of order p. Rp = 1 is permutable with both Q and P . Preceding equations now assume the final forms m−5
Q−1 P Q = Qβp P 1+ap
(15) (16)
R
−1
bp
PR=Q P
1+apm−4 m−4
R−1 Q R = Q1+dp P cp
(17) 2
,
, ,
m−3
with Rp = 1, Qp = 1, P p = 1. The following derived equations are necessary18 (18) [0, −y, x, 0, y] = 0, βxyp, x(1 + αypm−5 ) + αβ x2 ypm−4 , (19) [−y, 0, x, −y] = 0, bxyp, x(1 + aypm−4 ) + ab x2 ypm−4 , [−y, x, 0, y] = [0, x(1 + dyp), cxypm−4 ].
(20)
From a consideration of (18), (19) and (20) we arrive at the expression for a power of a general operator of G. [z, y, x]s = [sz, sy + Us p, sx + Vs pm−5 ],
(21) where19
s 2
{bxz + βxy + dyz}, n o Vs = 2s αxy + axz + αβ x2 y + cyz + ab x2 z p + α 3s {bxz + βxy + dyz}xp.
Us =
5. Transformation of the groups. All groups of this section are given by equations (15), (16), and (17) with a, b, β, c, d = 0, 1, 2, · · · , p − 1, and α = 0, 1, 2, · · · , p2 − 1, independently. Not all these groups, however, are distinct. Suppose that G and G0 of the above set are simply isomorphic and that the correspondence is given by R, Q, P C= , R10 , Q01 , P10 in which 00
00
00 m−4
R10 = R0z Q0y p P 0x Q01 P10 18 For
m = 6 the term a2
19 When
x xp2 2
0z
0
0y
0
0x p
=R Q P = R0z Q0y P 0x ,
p
0 m−5
,
,
must be added to the exponentnof P in (18). 2
m = 6 the following terms are to be added to Vs :
15
a x 2
s(s−1)(2s−1) 2 y 3!
−
s y 2
o
p.
where x, y 0 and z 00 are prime to p. The operators R10 , Q01 , and P10 must be independent since R, Q, and P are, 2 and that this is true is easily verified. The lowest power of Q01 in {P10 } is Q0 p1 = 1 m−5 0 and the lowest power of R10 in {Q01 , P10 } is R0 p1 = 1. Let Q0s1 = P 0 sp . 1 This in terms of R0 , Q0 , and P 0 is h i 0 0 s0 z 0 , y 0 s0 + d0 s2 z 0 p , s0 x0 pm−5 + c0 s2 y 0 z 0 pm−4 = [0, 0, sxpm−5 ]. From this equation s0 is determined by s0 z 0 ≡ 0 (mod p) y 0 {s0 + d0 2s z 0 p} ≡ 0 (mod p2 ), which give s0 y 0 ≡ 0
(mod p2 ).
Since y 0 is prime to p s0 ≡ 0
(mod p2 ) 2
and the lowest power of Q01 contained in {P10 } is Q0 p1 = 1. s00 Denoting by R0 1 the lowest power of R10 contained in {Q01 , P10 }. s00
s0 p
spm−4
R0 1 = Q0 1 P 0 1
.
This becomes in terms of R0 , Q0 , and P 0 [s00 z 00 , s00 y 00 p, s00 x00 pm−4 ] = [0, s0 y 0 p, {s0 x0 + sx}pm−4 ]. s00 is now determined by s00 z 00 ≡ 0
(mod p)
and since z 00 is prime to p s00 ≡ 0
(mod p). p
The lowest power of R10 contained in {Q01 , P 0 } is therefore R0 1 = 1. Since R, Q, and P satisfy equations (15), (16), and (17) R10 , Q01 , and P10 also satisfy them. Substituting in these equations the values of R10 , Q01 , and P10 and reducing we have in terms of R0 , Q0 , and P 0 (22)
[z, y + θ1 p, x + φ1 pm−5 ] = [z, y + βy 0 p, x(1 + αpm−5 ) + βxpm−4 ],
(23)
[z, y + θ2 p, x + φ2 pm−4 ] = [z, y + by 0 p, x(1 + apm−4 ) + bx0 pm−4 ],
(24) [z 0 , y 0 + θ3 p, (x0 + φ3 p)pm−5 ] = [z 0 , y 0 (1 + dp), x(1 + dp)pm−5 + cxpm−4 ], 16
in which θ1 = d0 (yz 0 − y 0 z) + x(b0 z 0 + β 0 y 0 ), θ2 = d0 yz 00 + b0 xz 00 , θ3 = d0 y 0 z 00 , φ1 = α0 xy 0 + α0 (β 0 y 0 + b0 z 0 ) x2 + a0 xz + c0 (yz 0 − y 0 z) p, φ2 = α0 xy 00 + a0 xz 00 + α0 b0 x2 z 00 + c0 yz 00 , φ3 = c0 yz 00 . A comparison of the members of the above equations give six congruences between the primed and unprimed constants and the nine indeterminates.
(I)
θ1 ≡ βy 0
(II) (III) (IV) (V) (VI)
φ1 θ2 φ2 θ3 φ3
(mod p),
≡ αx + βx0 p (mod p2 ), ≡ by 0 (mod p), ≡ ax + bx0 (mod p), ≡ dy 0 (mod p), ≡ cx + dx0 (mod p).
The necessary and sufficient condition for the simple isomorphism of the two groups G and G0 is, that the above congruences shall be consistent and admit of solution for the nine indeterminates, with the condition that x, y 0 and z 00 be prime to p. For convenience in the discussion of these congruences, the groups are divided into six sets, and each set is subdivided into 16 cases. The group G0 is taken from the simplest case, and we associate with this case all cases, which contain a group G, simply isomorphic with G0 . Then a single group G, in the selected case, simply isomorphic with G0 , is chosen as a type. G0 is then taken from the simplest of the remaining cases and we proceed as above until all the cases are exhausted. Let κ = κ1 pκ2 , and dv1 [κ1 , p] = 1 (κ = a, b, α, β, c, and d). The six sets are given in the table below.
I. A B C
α2 0 0 1
d2 0 1 0
D E F
α2 2 1 2
d2 0 1 1
The subdivision into cases and the results are given in Table II.
17
II. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
a2 1 0 1 1 1 0 0 0 1 1 1 0 0 0 1 0
b2 1 1 0 1 1 0 1 1 0 0 1 0 0 1 0 0
β2 1 1 1 0 1 1 0 1 0 1 0 0 1 0 0 0
c2 1 1 1 1 0 1 1 0 1 0 0 1 0 0 0 0
A
B
A1 A1 A1 A1 A1 A1 A1 A1 A1 A1 A1 A1 A1 A1 A1
B1
B3 B4 B5 B3
B3 B10 B11 B10 B10
C
D
C1 C1 C1 C2 C2 C2 C1 C2 * C2 * C2 C2 *
C2 D1 D1 D1 C2 C2 C2 D1 C2 C1 C2 * C2 C2 *
E
F E2
E3 E5 E3
* E10 E11 E10 E10
E4 E5 F3 E7 E5 F3 E10 E11 E3 E10 E11 E10 E10
The groups marked (*) divide into two or three parts.
Let ad − bc = θ1 pθ2 , α1 d − βc = φ1 pφ2 and α1 b − aβ = χ1 pχ2 with θ1 , φ1 , and χ1 prime to p.
III. * C11 C11 C13 C13 C16 C16 C16
θ2
1 0 1 1 0
φ2 1 0
1 0
χ2
* D13 D13 D16 D16 E12 E12
D1 C1 C1 C2 D1 C1 C2
18
θ2 1 0 1 0
φ2
χ2
1 0
D1 C2 C1 C2 F3 E3
6. Types. The type groups are given by equations (15), (16) and (17) with the values of the constants given in Table IV.
IV. A1 B1 B3 B4 B5 B10 B11 C1 C2 D1
a 0 0 0 0 0 0 0 0 ω 0
b 0 0 1 0 0 1 0 0 0 0
α 1 1 1 1 1 1 1 p p 0
β 0 0 0 1 0 0 1 0 0 0
c 0 0 0 0 1 κ 1 0 0 0
d 1 0 0 0 0 0 0 1 1 1
E1 E2 E3 E4 E5 E7 E10 E11 F1 F3
κ = 1, and a non-residue
a 0 1 0 0 0 1 0 0 0 0
b 0 0 1 0 0 0 1 0 0 1
α p p p p p p p p 0 0
β 0 0 0 1 0 1 0 1 0 0
c 0 0 0 0 1 0 κ 1 0 0
d 0 0 0 0 0 0 0 0 0 0
(mod p),
ω = 1, 2, · · · , p − 1.
The congruences for three of these cases are completely analyzed as illustrations of the methods used. B10 . The congruences for this case have the special forms. (I) (II) (III) (IV) (V) (VI)
b0 xz 0 ≡ βy 0 (mod p), α0 y 0 ≡ α (mod p), 0 b xz 00 ≡ by 0 (mod p), α0 xy 00 + α0 b0 x2 z 00 + c0 yz 00 ≡ ax + bx0 d ≡ 0 (mod p), 0 0 00 c y z ≡ cx (mod p).
(mod p),
Since z 0 is unrestricted (I) gives β ≡ 0 or 6≡ 0 (mod p). From (II) since y 0 6≡ 0, α 6≡ 0 (mod p). From (III) since x, y 0 , z 00 6≡ 0, b 6≡ 0 (mod p). In (IV) b 6≡ 0 and x0 is contained in this congruence alone, and, therefore, a may be taken ≡ 0 or 6≡ 0 (mod p). (V) gives d ≡ 0 (mod p) and (VI), c 6≡ 0 (mod p). Elimination of y 0 between (III) and (VI) gives b0 c0 z 002 ≡ bc (mod p)
19
so that bc is a quadratic residue or non-residue (mod p) according as b0 c0 is a residue or non-residue. The types are given by placing a = 0, b = 1, α = 1, β = 0, c = κ, and d = 0 where κ has the two values, 1 and a representative non-residue of p. C2 . The congruences for this case are (I) (II) (III) (IV) (V) (VI)
d0 (yz 0 − y 0 z) ≡ βy 0 (mod p), α10 xy 0 + a0 xz 0 ≡ α1 x + βx0 (mod p), d0 yz 00 ≡ by 0 (mod p), 0 a xz 00 ≡ ax + bx0 (mod p), d0 z 00 ≡ d (mod p), cx + dx0 ≡ 0 (mod p).
Since z appears in (I) alone, β can be either ≡ 0 or 6≡ 0 (mod p). (II) is linear in z 0 and, therefore, α ≡ 0 or 6≡ 0 (mod p), (III) is linear in y and, therefore, b ≡ 0 or 6≡ 0. Elimination of x0 and z 00 between (IV), (V), and (VI) gives a0 d2 ≡ d0 (ad − bc)
(mod p).
Since z 00 is prime to p, (V) gives d 6≡ 0 (mod p), so that ad − bc 6≡ 0 (mod p). We may place b = 0, α = p, β = 0, c = 0, d = 1, then a will take the values 1, 2, 3, · · · , p − 1 giving p − 1 types. D1 . The congruences for this case are (I) (II) (III) (IV) (V) (VI)
d0 (yz 0 − y 0 z) ≡ βy 0 (mod p), α1 x + βx0 ≡ 0 (mod p), d0 yz 00 ≡ by 0 (mod p), ax + bx0 ≡ 0 (mod p), d0 z 00 ≡ d (mod p), cx + dx0 ≡ 0 (mod p).
z is contained in (I) alone, and therefore β ≡ 0 or 6≡ 0 (mod p). (III) is linear in y, and b ≡ 0 or 6≡ 0 (mod p). (V) gives d 6≡ 0 (mod p). Elimination of x0 between (II) and (VI) gives α1 d − βc ≡ 0 (mod p), and between (IV) and (VI) gives ad − bc ≡ 0 (mod p). The type group is derived by placing a = 0, b = 0, α = 0, β = 0, c = 0 and d = 1.
20
Section 2. 1. Groups with dependent generators. In this section, G is generated by Q1 and P where 2
2
Qp1 = P hp .
(1)
There is in G, a subgroup H1 , of order pm−2 , which contains {P } self-conjugately.20 H1 either contains, or does not contain Qp1 . We will consider the second possibility in the present section, reserving the first for the next section. 2. Determination of H1 . H1 is generated by P and some other operator R1 of G. Rp1 is contained in {P }. Let Rp1 = P lp .
(2)
Since {P } is self-conjugate in H1 ,21 1+kp R−1 1 P R1 = P
(3)
m−4
Denoting Ra1 P b Rc1 P d · · · by the symbol [a, b, c, d, · · · ] we derive from (3) (4)
[−y, x, y] = [0, x(1 + kypm−4 )]
(m > 4),
and h i [y, x]s = sy, x s + k 2s ypm−4
(5)
Placing s = p and y = 1 in (5) we have, from (2) [R1 P x ]p = Rp1 P xp = P (l+x)p . Choosing x so that x+l ≡0
(mod pm−4 ),
R = R1 P x is an operator of order p, which will be used in the place of R1 , and H = {R, P } with Rp = 1. 3. Determination of H2 . We will now use the symbol [a, b, c, d, e, f, · · · ] to denote Qa1 Rb P c Qd1 Re P f · · · . H1 and Q1 generate G and all the operations of G are given by [x, y, z] (z = 0, 1, 2, · · · , p2 − 1; y = 0, 1, 2, · · · , p − 1; x = 0, 1, 2, · · · , pm−3 − 1), since these are pm in number and are all distinct. There is in G a subgroup H2 of order pm−1 which contains H1 self-conjugately. H2 is generated by H1 and 20 Burnside, 21 Burnside,
Theory of Groups, Art. 54, p. 64. Theory of Groups, Art. 56, p. 66.
21
some operator [z, y, x] of G. Qz1 is then in H2 and H2 is the subgroup {Qp1 , H1 }. Hence, p β α1 Q−p , 1 P Q1 = R P
(6)
m−4
p b1 ap Q−p 1 P Q1 = R P
(7)
.
To determine α1 and β we find from (6), (5) and (7) n αp − bp1 βk αp1 − 1 m−4 o 2 2 [−p , 0, 1, p ] = 0, 1 β, αp1 1 + p α1 − b1 2 α1 − 1 n αp−1 αp1 − bp1 o m−4 1 p . + aβ p − α1 − b1 (α1 − b1 )2 By (1) 2
2
Q1−p P Qp1 = P, and, therefore,
αp1 ≡ 1
αp1 − bp1 β ≡ 0 (mod p), α1 − b1 (mod pm−4 ), and α1 ≡ 1 (mod pm−5 )
(m > 5).
Let α1 = 1 + α2 pm−5 and equation (6) is replaced by m−5
p β 1+α2 p Q−p 1 P Q1 = R P
(8)
.
To find a and b1 we obtain from (7), (8) and (5) h bp − 1 m−4 i p . [−p2 , 1, 0, p2 ] = 0, bp1 , a 1 b1 − 1 By (1) and (4) 2
2
2
2
Q−p R Qp1 = P −lp R P lp = R, 1 and, hence, bp1 ≡ 1
(mod p),
a
bp1 − 1 ≡0 b1 − 1
(mod p),
therefore b1 = 1. Substituting b1 = 1 and α1 = 1 + α2 pm−5 in the congruence determining α1 we obtain (1 + α2 pm−5 )p ≡ 1 (mod pm−3 ), which gives α2 ≡ 0 (mod p). Let α2 = αp and equations (8) and (7) are now replaced by (9) (10)
m−4
Qp1 P Qp1 = Rβ P 1+αp m−4
p ap Q−p 1 R Q1 = RP
22
.
,
From these we derive (11)
h i [−yp, 0, x, yp] = 0, βxy, x + αxy + aβx y2 + βk x2 y pm−4 ,
(12)
[−yp, x, 0, yp] = [0, x, axypm−4 ].
A continued use of (4), (11), and (12) yields [zp, y, x]s = [szp, sy + Us , sx + Vs pm−4 ]
(13) where
Us = β Vs =
s 2
s 2
xz, o n αxz + βk 2s z + kxy + ayz + βk 3s x2 z o 1 n1 + aβ s(s − 1)(2s − 1)z 2 − 2s z . 2 3!
4. Determination of G. Since H2 is self-conjugate in G1 we have (14)
γp δ 1 Q−1 1 P Q1 = Q1 R P ,
(15)
d ep Q1−1 R Q1 = Qcp 1 R P
m−4
.
From (14), (15) and (13) [−p, 0, 1, p] = [λp, µ, p1 + vpm−4 ] and by (9) and (1) λp ≡ 0
(mod p2 ),
p1 + νpm−4 + λhp ≡ 1 + αpm−4
(mod pm−3 ),
from which p1 ≡ 1
(mod p2 ),
and
1 ≡ 1
(mod p)
Let 1 = 1 + 2 p and equation (14) is replaced by (16)
γp δ 1+2 p Q−1 . 1 P Q1 = Q1 R P
From (15), (16), and (13) p d −1 p m−4 [−p, 1, 0, p] = c p, d , Kp d−1 where p−1
X dp − 1 dn (dn − 1) K= e+ acd . d−1 2 1 23
(m > 5).
By (10) dp ≡ 1
(mod p),
and
d=1
and by (1) chp2 ≡ apm−4
(mod pm−3 ).
Equation (15) is now replaced by m−4
cp ep Q−1 1 R Q1 = Q1 RP
(17)
.
A combination of (17), (16) and (13) gives i h (1 + p)p − 1 p − 1 2 2 p + cδ . [−p, 0, 1, p] = γ p , 0, (1 + p) 2 2 p2 2 By (9) n (1 + p)p − 1 p − 1o 2 2 γ + cδ hp + (1 + 2 p)p ≡ 1 + αpm−4 2 2 p 2 β ≡ 0 (mod p). A reduction of the first congruence gives 2 n p − 1 o m−4 (1 + 2 p)p − 1 + γh p ≡ α − aδ p 2 p2 2
(mod pm−3 ),
(mod pm−3 )
and, since (1 + 2 p)p − 1 ≡1 2 p2
(2 + γh)p2 ≡ 0
(mod p),
(mod pm−4 )
and (18)
(2 + γh)p2 ≡ α +
aδ m−4 p 2
(mod pm−3 ).
From (17), (16), (13) and (18) h i (19) [−y, x, 0, y] = cxyp, x, exy + ac x2 y pm−4 , h i [−y, 0, x, y] = x γy + cδ y2 p, δxy, x(1 + 2 yp) + θpm−4 (20) where o n aδ x y2 θ = eδx + aδγx + 2 α + 2 o 1 n1 + ac y(y − 1)(2y − 1)δ 2 − y2 δ 2 3! x + αγy + δky + aδxy 2 + (acδ 2 y + acδ) y2 2 . 24
From (19), (20), (4) and (18) 2
2
2
2
{Q1 P x }p = Qp1 P xp = P (h+x)p . If x be so chosen that (mod pm−5 )
h+x≡0
Q = Q1 P x is an operator of order p2 which will be used in place Q1 and 2 Qp = 1. Placing h = 0 in (18) we get 2 p2 ≡ 0
(mod pm−4 ).
Let 2 = pm−6 and equation (16) is replaced by m−5
Q−1 P Q = Qγp Rδ P 1+p
(21) The congruence
apm−4 ≡ chp2
(mod pm−3 )
becomes apm−4 ≡ 0
(mod pm−3 ),
and
a≡0
(mod p).
Equations (19) and (20) are replaced by [−y, x, 0, y] = [cxyp, x, exypm−4 ] h i [−y, 0, x, y] = γy + cδ y2 xp, δxy, x(1 + ypm−5 ) + θpm−4
(22) (23) where
θ = eδx
y 2
x + αγy + δky + αcδ y2 2 .
A formula for any power of an operation of G is derived from (4), (22) and (23) [z, y, x]s = [sz + Us p, sy + Vs , sx + Ws pm−5 ]
(24) where
n1 1 s(s − 1)(2s − 1)z 2 − γxz + cyz + cδx 2 3! Vs = δ 2s xz, n o Ws = 2s xz + (aγ + δk) x2 z + eyz + kxy p Us =
s 2
s 2
o z ,
1 1 γx + y + δkx xzp + cδ (s − 1)z 2 − z 3s xp 2 2 1 1 x s(s − 1)(2s − 1)z 2 − 2s z p. + δex + αcδ 2 2 3! +
s 3
25
5. Transformations of the groups. Placing y = 1 and x = −1 in (22) we obtain (17) in the form m−4
R−1 Q R = Q1−cp P −ep
.
A comparison of the generational equations of the present section with those of Section 1, shows that groups, in which δ ≡ 0 (mod p), are simply isomorphic with those in Section 1, so we need consider only those cases in which δ 6≡ 0 (mod p). All groups of this section are given by m−4 R−1 P R = P 1+kp , m−5 (25), (26), (27) G : Q−1 P Q = Qγp Rδ P 1+p , −1 cp pm−4 Q RQ = Q RP . 2
m−3
Rp = 1, Qp = 1, and P p = 1, (k, γ, c, e = 0, 1, 2, · · · , p − 1; δ = 1, 2, · · · , p − 1; = 0, 1, 2, · · · , p2 − 1). Not all these groups, however, are distinct. Suppose that G and G0 of the above set are simply isomorphic and that the correspondence is given by R, Q, P C= . R10 , Q01 , P10 2
2
m−3
m−3
Since Rp = 1, Qp = 1, and P p = 1, R0 p1 = 1, Q0 p1 = 1 and P 0 p1 The forms of these operators are then
.
P10 = Q0z R0y P 0x , 0
0
0 m−4
00
00
00 m−5
R10 = Q0z p R0y P 0x p Q01 = Q0z R0y P 0x
p
, ,
where dv[x, p] = 1. Since R is not contained in {P }, and Qp is not contained in {R, P } R10 is not contained in {P10 }, and Q0 p1 is not contained in {R10 , P10 }. Let s0
spm−4
R0 1 = P 0 1
.
This becomes in terms of Q0 , R0 and P 0 [s0 z 0 p, s0 y 0 , s0 x0 pm−4 ] = [0, 0, sxpm−4 ], and s0 y 0 ≡ 0
s0 z 0 ≡ 0
(mod p),
26
(mod p).
Either y 0 or z 0 is prime to p or s0 may be taken = 1. Let s00 p
Q0 1
s0
= R0 1 P 0 sp 1
m−4
,
and in terms of Q0 , R0 and P 0 [s00 z 00 p, 0, s00 x00 pm−4 ] = [s0 z 0 p, s0 y 0 , (s0 x0 + sx)pm−4 ], from which s00 z 00 ≡ s0 z 0
(mod p),
and
s0 y 0 ≡ 0
(mod p).
Eliminating s0 we find s00 y 0 z 00 ≡ 0
(mod p),
dv[y 0 z 00 , p] = 1 or s00 may be taken = 1. We have then z 00 , y 0 and x prime to p. Since R, Q and P satisfy equations (25), (26) and (27) R10 , Q01 and P10 do also. These become in terms of R0 , Q0 and P 0 . [z + Φ01 p, y, x + Θ01 pm−4 ] = [z, y, x(1 + kpm−4 )], [z + Φ02 p, y + δ 0 xz 00 , x + Θ02 pm−5 ] = [z + Φ2 p, y + δy 0 , x + Θ2 pm−5 ], [(z 0 + Φ03 )p, y 0 , Θ03 pm−4 ] = [(z 0 + Φ3 )p, y, Θ03 pm−4 ], where Φ01 = −c0 yz 0 , Θ01 = 0 xz 0 + k 0 xy 0 − e0 y 0 z, n o Φ02 = γ 0 z 00 + c0 δ 0 z2 x + c0 (yz 00 − y 00 z), n 00 Θ02 = 0 xz 00 + x2 α0 γ 0 z 00 + α0 c0 δ 0 z2 + δ 0 k 0 z 00 o 00 + δ 0 e0 x z2 + e0 (yz 00 − y 00 z) + k 0 xy 00 p, Φ2 = γz 00 + δz 0 + c0 δy 0 z, Θ2 ≡ x + (γx00 + δx + e0 δy 0 z)p, Φ03 = c0 y 0 z 00 , Θ03 = e0 y 0 z 00 , Φ3 = cz 00 , Θ3 = ex + cx00 . A comparison of the members of these equations give seven congruences (I) (II) (III) (IV) (V) (VI) (VII)
Φ01 Θ01 Φ02 0 δ xz 00
≡ 0 (mod p), ≡ kx (mod p), ≡ Φ2 (mod p), ≡ δy 0 (mod p),
Θ02 ≡ Θ2 (mod p2 ), Φ03 ≡ cz 00 (mod p), Θ03 ≡ Θ3 (mod p). 27
The necessary and sufficient condition for the simple isomorphism of G and G0 is, that these congruences be consistent and admit of solution for the nine indeterminants with x, y 0 , and z 00 prime to p. Let κ = κ1 pκ2 , dv[κ1 , p] = 1 (κ = k, δ, γ, , c, e). The groups are divided into three parts and each part is subdivided into 16 cases. The methods used in discussing the congruences are the same as those used in Section 1. 6. Reduction to types. The three parts are given by
I. A B C
2 0 1 2
δ2 0 0 0
The subdivision into cases and the results of the discussion of the congruences are given in Table II.
II. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
k2 1 0 1 1 1 0 0 0 1 1 1 0 0 0 1 0
γ2 1 1 0 1 1 0 1 1 0 0 1 0 0 1 0 0
c2 1 1 1 0 1 1 0 1 0 1 0 0 1 0 0 0
e2 1 1 1 1 0 1 1 0 1 0 0 1 0 0 0 0
A
B
A2
B1
* A4 A5 A4 A5 A4 A4 A5 A4 A4 A4
B2 B5 B4 B5 B4 B7 B5 B7 B4 B7
C B1 B2 B1 B4 B5 B2 B7 B5 B4 B5 B4 B7 B5 B7 B4 B7
A6 divides into two parts. The groups of A6 in which δk + γ ≡ 0 (mod p) are simply isomorphic with the groups of A1 and those in which δk + γ 6≡ 0 (mod p) are simply isomorphic with the groups of A2 . The types are given by equations (25), (26) and (27) where the constants have the values given in Table III.
28
III. A1 A2 A4 A5 B1 B2 B4 B5 B7
k 0 1 0 0 0 1 0 0 1
δ 1 1 1 1 1 1 1 1 1
γ 0 0 0 0 0 0 0 0 0
1 1 1 1 p p p p p
κ = 1, and a non-residue
c 0 0 1 0 0 0 1 0 ω
e 0 0 0 ω 0 0 0 κ 0
(mod p),
ω = 1, 2, · · · , p − 1.
A detailed analysis of several cases is given below, as a general illustration of the methods used. A1 . The special forms of the congruences for this case are (II) (III) (IV) (V) (VI) (VII)
0 xz 0 ≡ kx (mod p), γz 00 + δz 0 ≡ 0 (mod p), δ 0 xz 00 ≡ δy 0 (mod p), 0 xz 00 ≡ x (mod p), cz 00 ≡ 0 (mod p), ex ≡ 0 (mod p).
Congruence (IV) gives δ 6≡ 0 (mod p), from (II) k can be ≡ 0 or 6≡ 0 (mod p), (III) gives γ ≡ 0 or 6≡ 0, (V) gives 6≡ 0, (VI) and (VII) give c ≡ e ≡ 0 (mod p). Elimination of x, z 0 and z 00 between (II), (III) and (V) gives δk+γ ≡ 0 (mod p). If k ≡ 0, then γ ≡ 0 (mod p) and if k 6≡ 0, then γ 6≡ 0 (mod p). A2 . The congruences for this case are (II) (III) (IV) (V) (VI) (VII)
0 xz 0 + k 0 xy 0 ≡ kx (mod p), γx00 + δz 0 ≡ 0 (mod p), δ 0 xz 00 ≡ δy 0 (mod p), 0 xz 00 ≡ x (mod p), cz 00 ≡ 0 (mod p), ex ≡ 0 (mod p). 29
Congruence (III) gives γ ≡ 0 or 6≡ 0, (IV) gives δ 6≡ 0, (V) 6≡ 0, (VI) and (VII) give c ≡ e ≡ 0 (mod p). Elimination of x, z 0 , and z 00 between (II), (III) and (V) gives δk + γ ≡ k 0 δy 0
(mod p)
from which δk + γ 6≡ 0
(mod p).
If k ≡ 0, then γ 6≡ 0, and if γ ≡ 0 then k 6≡ 0 (mod p). Both γ and k can be 6≡ 0 (mod p) provided the above condition is fulfilled. A5 . The congruences for this case are (II) (III) (IV) (V) (VI) (VII)
0 xz 0 − e0 y 0 z γz 00 + δz 0 δ 0 xz 00 0 xz 00 cz 00 e0 y 0 z 00
≡ kx (mod p), ≡ 0 (mod p), ≡ δy 0 (mod p), ≡ ex (mod p), ≡ 0 (mod p), ≡ ex (mod p).
(II) and (III) are linear in z and z 0 so k and γ are ≡ or 6≡ 0 (mod p) independently, (IV) gives δ 6≡ 0, (V) give 6≡ 0, (VI) c ≡ 0, and (VII) e 6≡ 0. Elimination between (IV), (V), and (VII) gives δ 0 e0 2 ≡ δe02
(mod p).
The types are derived by placing = δ = 1, and e = 1, 2, · · · , p − 1. B5 . The congruences for this case are (II) (III) (IV) (V) (VI) (VII)
−e0 y 0 z γz 00 + δz 0 δ 0 xz 00 00 01 xz 00 + δ 0 e0 x z2 + e0 yz 00 cz 00 0 0 00 eyz
≡ kx (mod p), ≡ 0 (mod p), ≡ δy 0 (mod p), ≡ e1 x + γx00 + δx0 ≡ 0 (mod p), ≡ ex (mod p).
(mod p),
(II), and (III) being linear in z and z 0 give k ≡ 0 or 6≡ 0, and γ ≡ 0 or 6≡ 0 (mod p), (IV) gives δ 6≡ 0, (V) being linear in x0 gives 1 ≡ 0 or ≡ 6 0 (mod p), (VI) gives c ≡ 0 and (VII) e 6≡ 0. 30
Elimination of x and y 0 from (IV) and (VII) gives δ 0 e0 z 002 ≡ δe (mod p). δe is a quadratic residue or non-residue (mod p) according as δ 0 e0 is a residue or non-residue. The two types are given by placing δ = 1, and e = 1 and a non-residue (mod p). Section 3. 1. Groups with dependent generators continued. As in Section 2, G is here generated by Q1 and P , where 2
2
Qp1 = P hp . Qp1 is contained in the subgroup H1 of order pm−2 , H1 = {Qp1 , P }. 2. Determination of H1 . Since {P } is self-conjugate in H1 p 1+kp Q−p 1 P Q1 = P
(1)
m−4
.
Denoting Qa1 P b Qc1 P d · · · by the symbol [a, b, c, d, · · · ], we have from (1) [−yp, x, yp] = [0, x(1 + kypm−4 )]
(2)
(m > 4).
Repeated multiplication with (2) gives h i (3) [yp, x]s = syp, x s + k 2s ypm−4 . 3. Determination of H2 . There is a subgroup H2 of order pm−1 which contains H1 self-conjugately.22 H2 is generated by H1 and some operator R1 of 2 G. Rp1 is contained in H1 , in fact in {P }, since if Rp1 is the first power of R1 in {P }, then H2 = {R1 , P }, which case was treated in Section 1. Rp1 = P lp .
(4)
Since H1 is self-conjugate in H2 (5)
βp α1 R−1 , 1 P R1 = Q1 P
(6)
bp α1 p p R−1 . 1 Q R1 = Q1 P
Using the symbol [a, b, c, d, e, f, · · · ] to denote Ra1 Qb1 P c Rd1 Qe1 P f · · · , we have from (5), (6) and (3) (7) 22 Burnside,
[−p, 0, 1, p] = [0, βN p, αp1 + M p], Theory of Groups, Art. 54, p. 64.
31
and by (4) αp1 ≡ 1
(mod p),
α1 ≡ 1
and
(mod p).
Let α1 = 1 + α2 p and (5) is now replaced by βp 1+α2 p R−1 . 1 P R1 = Q1 P
(8) From (6), (8) and (3)
bp − 1 [−p, p, 0, p] = 0, bp p, a1 p + a1 U p2 , b−1 and by (4) and (2) p p p R−p 1 Q1 R1 = Q1
and therefore bp ≡ 1 (mod p), and b = 1. Placing b = 1 in the above equation the exponent of P takes the form p
a1 p2 (1 + U 0 p) = a1
{1 + (α2 + βh)p} − 1 2 p (α2 + βh)p2
from which a1 p2 (1 + U 0 p) ≡ 0
(mod pm−3 )
or a1 ≡ 0
(mod pm−5 )
(m > 5).
Let a1 = apm−5 and (6) is replaced by m−4
p p ap R−1 1 Q1 R1 = Q1 P
(9)
.
(7) now has the form [−p, 0, 1, p] = [0, βN p, (1 + α2 p)p + M p2 ], where N =p
and
M = βh
(1 + α2 p)p − 1 − 1 , α2 p2
from which (1 + α2 p)p +
(1 + α2 p)p − 1 βhp2 ≡ 1 α2 p2 32
(mod pm−3 )
or (1 + α2 p)p − 1 {α2 + βh}p2 ≡ 0 α2 p2
(mod pm−3 )
and since (1 + α2 p)p − 1 ≡1 α2 p2 (α2 + βh)p2 ≡ 0
(10)
(mod p)
(mod pm−3 ).
From (8), (9), (10) and (3) [−y, 0, x, y] = [0, βxyp, x(1 + α2 yp) + θpm−4 ],
(11)
[−y, xp, 0, y] = [0, xp, axypm−4 ],
(12) where
θ = aβx
y 2
+ βk
x 2
y.
By continued use of (11), (12), (2) and (10) [z, yp, x]s = [sz, (sy + Us )p, xs + Vs p],
(13) where Us = β
s 2
xz n o Vs = 2s α2 xz + ayz + kxy + βk x2 z pm−5 n 1 h1 s(s − 1)(2s − 1)z 2 − + β 3s x2 z + aβ 2 3!
s 2
i o m−5 z x p .
Placing in this y = 0, z = 1 and s = p,23 (R1 P x )p = Rp1 P xp = P (x+l)p , determine x so that x+l ≡0
(mod pm−4 ),
then R = R1 P x is an operator of order p which will be used in the place of R1 , Rp = 1. 4. Determination of G. Since H2 is self-conjugate in G (14)
δp 1 γ Q−1 1 P Q1 = R Q1 P ,
(15)
c dp e1 p Q−1 . 1 R Q1 = R Q1 P
23 Terms of the form (Ax2 + Bx)pm−4 in the exponent of P for p = 3 and m > 5 do not alter the result.
33
From (15) e1 p p (Rc Qdp ) = 1, 1 P
by (13) 2
2
2
e1 p Qdp = P (e1 +dh)p = 1, 1 P
and (e1 + dh)p2 ≡ 0
(16)
(mod pm−3 ).
From (14), (15) and (13) [0, −p, 1, 0, p] = [L, M p, p1 + N p].
(17) By (1) p1 ≡ 1
(mod p),
and
1 ≡ 1
(mod p).
Let 1 = 1 + 2 p and (14) is replaced by δp 1+2 p γ Q−1 . 1 P Q1 = R Q1 P
(18)
From (15), (18), and (13) cp − 1 dp, Kp . [0, −p, 0, 1, p] = cp , c−1 Placing x = 1 and y = −1 in (12) we have [0, −p, 0, 1, p] = [1, 0, −apm−4 ],
(19)
and therefore cp ≡ 1 (mod p), and c = 1. (15) is now replaced by dp e1 p Q−1 . 1 R Q1 = R Q1 P
(20)
Substituting 1 + 2 p for 1 and 1 for c in (17) gives, by (16) [0, −p, 1, p] = [0, M p2 , (1 + 2 p)p + N p2 ], where M = γd
p−1 (1 + 2 p)p − 1 +δ 2 2 p2
and N=
e1 γ (2 + δh)p2
[1 + (2 + δh)p]p − 1 −p . (2 + δh)p 34
By (1) (1 + 2 p)p + (N + M h)p2 ≡ 1 + kpm−4
(mod pm−3 ),
or reducing ψ(2 + δh)p2 ≡ kpm−4
(mod pm−3 ),
where ψ=
(1 + 2 p)p − 1 p−1 + N − e1 γ . 2 2 p 2
Since ψ = 1 (mod p). (2 + δh)p2 ≡ kpm−4
(21)
(mod pm−3 ).
From (18), (20), (13), (16) and (21) [0, −y, x, 0, y] = [γxy, θ1 p, x + φ1 p], [0, −y, 0, x, y] = [x, dxyp, φ2 p],
(22) (23) where
+ δxy + βγ x2 y, φ1 = 2 xy + α2 γ x2 y + e1 γ y2 x + x y2 (2 k + δγ) 1 y 1 1 2 y(y − 1)(2y − 1)γ − γ x + aγ 2 dx y(y + 1)(y − 1) + ad 2 3! 2 3! 1 1 y x 2 2 + e1 γk 3 x + aβ x(x − 1)(2x − 1)γ y − 2 γy 2 3! + x2 (a + k) dy y2 + δy + βγ x3 pm−5 , φ2 = e1 xy + e1 k y2 + ad x2 y pm−5 . θ1 = dγx
y 2
Placing x = 1 and y = p in (23) and by (16) p Q−p 1 R Q1 = R,
and by (19) a≡0
(mod p).
A continued multiplication, with (11), (22), and (23), gives 2
2
2
2
(Q1 P x )p = Qp1 P xp = P (x+l)p . 35
Let x be so chosen that (x + l) ≡ 0
(mod pm−5 ),
then Q = Q1 P x is an operator of order p2 which will be used in place of Q1 , 2 Qp = 1 and h ≡ 0 (mod pm−5 ). From (21), (10) and (16) 2 p2 ≡ kpm−4 ,
α2 p 2 ≡ 0
e1 p2 ≡ 0
and
(mod pm−3 ).
Let 2 = pm−6 , α2 = αpm−5 and e1 = epm−5 . Then equations (18), (20) and (8) are replaced by m−5 Q−1 P Q = Rγ Qδp P 1+p , m−4 G : Q−1 R Q = R Qdp P ep (24), (25), (26) , −1 βp 1+αpm−4 R PR=Q P , Rp = 1,
2
m−3
Qp = 1,
Pp
= 1.
(11), (22) and (23) are replaced by (27)
[−y, 0, x, y] = [0, βxyp, x + φpm−4 ],
(28)
[0, −y, x, 0, y] = [γxy, θ1 p, x + φ1 pm−5 ],
(29)
[0, −y, 0, x, y] = [x, dxyp, φ2 pm−4 ],
where φ = αxy + βk x2 y, θ1 = dγ y2 x + δxy + βγ x2 y, φ1 = exy + eγx y2 + x2 αγy + dγk y2 + δky + βγy x3 p, φ2 = exy. A formula for a general power of any operator of G is derived from (27), (28) and (29) (30)
[0, z, 0, y, 0, z]s = [0, sz + Us p, 0, sy + Vs , 0, sx + Ws pm−5 ],
where δxz + dyz + βxy + βγ x2 z 1 1 s(s − 1)(2s − 1)z 2 − + dx 2 3! Vs = γ 2s xz,
Us =
s 2
36
s 2
z x + βγ
s 2
x2 z,
Ws =
s 2
xz + axy + eyz + (βky + αβγ + δkz) x2 p + 3s αγx2 z + dkxyz + δkx2 z + βkx2 y + 2βγk x2 xz p s 1 1 s 3 2 s(s − 1)(2s − 1)z − z eγx + dγk x2 p + βyk 4 x zp + 2 3! 2 1 1 2 + dγk (s − 1)z − z 3s x2 . 2 2
A comparison of the generational equations of the present section with those of Sections 1 and 2, shows that, γ ≡ 0 (mod p) gives groups simply isomorphic with those of Section 1, while β ≡ 0 (mod p), groups simply isomorphic with those of Section 2 and we need consider only the groups in which β and γ are prime to p. 5. Transformation of the groups. All groups of this section are given by equations (24), (25), and (26), where γ, β = 1, 2, · · · , p − 1; α, δ, d, e = 0, 1, 2, · · · , p − 1; and = 0, 1, 2, · · · , p2 − 1. Not all of these, however are distinct. Suppose that G is simply isomorphic with G0 and that the correspondence is given by R, Q, P C= . R10 , Q01 , P10 An inspection of (30) gives R10 = Q0z
00
Q01 P10
0
0z
p
00
00 m−4
R0y P 0x 0y
0
p
0 m−5
0x p
=Q R P = Q0z R0y P 0x ,
,
,
p
with dv[x, p] = 1. Since Qp is not in {P }, and R is not in {Qp , P }, Q0 1 is not p in {P10 } and R10 is not in {Q0 1 , P10 }. Let s0 p
spm−4
Q0 1 = P 0 1
.
This is in terms of R0 , Q0 , and P 0 , [0, s0 z 0 p, s0 x0 pm−4 ] = [0, 0, sxpm−4 ]. From which s0 z 0 p ≡ 0
(mod p2 ),
and z 0 must be prime to p, since otherwise s0 can = 1. Let s00
s0 p
spm−4
R0 1 = Q0 1 P 0 1
37
,
or in terms of R0 , Q0 , and P 0 , [s00 y 00 , s00 z 00 p, s00 x00 pm−4 ] = [0, s0 z 0 p, (sx + s0 x0 )pm−4 ] and s00 z 00 ≡ s0 z 0
s00 y 00 ≡ 0
(mod p),
(mod p),
and y 00 is prime to p, since otherwise s00 can = 1. Since R, Q, and P satisfy equations (24), (25) and (26), R10 , Q01 , and P10 must also satisfy them. These become when reduced in terms of R0 , Q0 and P 0 [0, z + θ10 p, 0, y + γ 0 xz 0 , 0, x + ψ10 pm−5 ] = [0, z + θ1 p, 0, y + γy 00 , 0, x + ψ1 pm−5 ], [0, (z 00 + θ20 )p, 0, y 00 , 0, (x00 + ψ2 )pm−4 ] = [0, (z 00 + θ2 )p, 0, y 00 , 0, (x00 + ψ2 )pm−4 ], [0, z + θ30 p, 0, y, 0, x + ψ30 pm−4 ] = [0, z + θ3 p, 0, y, 0, x + ψ3 pm−4 ], where n θ10 = d0 (yz 0 − y 0 z) + x d0 γ 0
z0 2
o + δ 0 z 0 + β 0 y 0 + β 0 γ 0 x2 z 0 ,
θ1 = γz 00 + δz 0 + d0 γy 00 z, i h 0 0 ψ10 = 0 xz 0 + e0 γ 0 x z2 + x2 α0 γ 0 z 0 + γ 0 0 d0 k 0 z2 + δ 0 k 0 z 0 + β 0 k 0 y 0 + β 0 γ 0 x3 z 0 + e0 (yz 0 − y 0 z) + α0 xy 0 p, ψ1 = x + {δx0 + γx00 + e0 γy 00 z}p, θ20 = d0 y 00 z 0 , θ2 = dz 0 , ψ20 = e0 y 00 z, ψ2 = dx0 + ex, 0 0 00 0 00 0 θ3 = β xy − d y z, θ3 = βz , 0 00 0 00 0 ψ3 = xz − e y z + α xy 00 + β 0 0 x2 y 00 , ψ3 = αx + βx0 . A comparison of the two sides of these equations give seven congruences (I) (II) (III) (IV) (V) (VI) (VII)
θ10 ≡ θ1 (mod p), γ xz 0 ≡ γy 00 (mod p), 0
ψ10 θ20 ψ20 θ30 ψ30
≡ ψ1 (mod p2 ), ≡ θ2 (mod p), ≡ ψ2 (mod p), ≡ θ3 (mod p), ≡ ψ3 (mod p).
(VI) is linear in z provided d0 6≡ 0 (mod p) and z may be so determined that β ≡ 0 (mod p) and therefore all groups in which d0 6≡ 0 (mod p) are simply isomorphic with groups in Section 2. 38
Consequently we need only consider groups in which d0 ≡ 0 (mod p). As before we take for G0 the simplest case and associate with it all simply isomorphic groups G. We then take as G0 the simplest case left and proceed as above. Let κ = κ1 pκ2 where dv[κ1 , p] = 1, (κ = α, β, γ, δ, , d, e). For convenience the groups are divided into three sets and each set is subdivided into eight cases. The sets are given by A : 2 = 0, β2 = 0, γ2 = 0, B : 2 = 1, β2 = 0, γ2 = 0, C : 2 = 2, β2 = 0, γ2 = 0. The subdivision into cases and results of the discussion are given in Table I.
I. δ2 1 0 1 1 0 0 1 0
1 2 3 4 5 6 7 8
e2 1 1 0 1 0 1 0 0
α2 1 1 1 0 1 0 0 0
A
B
A1
B1
A1 A3 A1 A3 A3
B1 B3 B1 B3 B3
C B1 B1 B3 B1 B3 B1 B3 B3
6. Reduction to types. The types of this section are given by equations (24), (25) and (26) with α = 0, β = 1, λ = 1 or a quadratic non-residue (mod p), δ ≡ 0; = l, e = 0, 1, 2, · · · , p − 1; and = p, e = 0, 1, or a non-residue (mod p), 2p + 6 in all. The special forms of the congruences for these cases are given below. A1 . (I) (II) (III) (IV) (V) (VI) (VII)
β0γ0
z 0 + β 0 xy 0 ≡ γz 00 + δz 0 (mod p), γ 0 xz 0 ≡ γy 00 (mod p), 0 xz 0 ≡ x (mod p), dz 0 ≡ 0 (mod p), ex ≡ 0 (mod p), β 0 xy 00 ≡ βz 0 (mod p), 0 xz 00 + β 0 0 x2 y 0 ≡ αx + βx0 (mod p). x 2
(I) is linear in z 00 and δ ≡ 0 or ≡ 6 0, (II) gives γ 6≡ 0, (III) 6≡ 0, (IV) and (V) d ≡ e ≡ 0, (VI) β 6≡ 0, (VII) is linear in x0 and α ≡ 0 or 6≡ 0 (mod p). 39
Elimination of y 00 and z 0 between (II) and (VI) gives β 0 γ 0 x2 ≡ βγ
(mod p)
and βγ is a residue or non-residue (mod p) according as β 0 γ 0 is a residue or non-residue. A3 . (I) (II) (III) (IV) (V) (VI) (VII)
β0γ0
z 0 + β 0 xy 0 ≡ γz 00 + δz 0 (mod p), γ 0 xz 0 ≡ γy 00 (mod p), 0 z 0 ≡ (mod p), d ≡ 0 (mod p), e0 y 00 z 0 ≡ ex (mod p), β 0 xy 00 ≡ βz 0 (mod p), 0 xz 00 − e0 y 00 z + β 0 0 x2 y 0 ≡ αx + βx0 (mod p). x 2
(I) is linear in z 00 and δ ≡ 0 or 6≡ 0. (II) gives γ 6≡ 0, (III) 6≡ 0, (V) e 6≡ 0 and (VI) β 6≡ 0. (VII) is linear in x0 and α ≡ 0 or 6≡ 0 (mod p). Elimination between (II) and (VI) gives β 0 γ 0 x2 ≡ βγ
(mod p),
and between (II), (III), and (IV) gives 02 γe ≡ 2 γ 0 e0
(mod p).
βγ is a residue, or non-residue, according as β 0 γ 0 is or is not, and if γ and are fixed, e must take the (p − 1) values 1, 2, · · · , p − 1. B1 . (I) (II) (III) (IV) (VI) (VII)
β0γ0
z 0 + β 0 xy 0 ≡ γz 00 + δz 0 (mod p), γ 0 xz 0 ≡ γy 00 (mod p), 01 xz 0 + β 0 xz 0 x3 ≡ 1 x + δx0 + γx00 (mod p), ex ≡ 0 (mod p), 0 β xy 00 ≡ βz 0 (mod p), αx + βx0 ≡ 0 (mod p). x 2
(I) gives δ ≡ 0 or 6≡ 0, (II) γ 6≡ 0, (III) is linear in x00 and gives 1 ≡ 0 or 6≡ 0, (V) e = 0, (VI) β 6≡ 0 and (VII) is linear in x0 and gives α ≡ 0 or 6≡ 0. Elimination between (II) and (VI) gives β 0 γ 0 x2 ≡ βγ
40
(mod p).
B3 . β0γ0
z 0 + β 0 xy 0 ≡ γβ 00 + δz 0 (mod p), γ 0 xz 0 ≡ γy 0 (mod p), 0 01 xz 0 + e0 γ 0 x z2 + β 0 γ 0 x3 + e0 (yz 0 − y 0 z) (III) ≡ 1 x + δx0 + γx00 + e0 γzy 00 (V) e0 y 00 z 0 ≡ ex (mod p). (VI) β 0 xy 00 ≡ βz 0 (mod p), (VII) −e0 y 00 z ≡ αx + βx0 (mod p).
(I) (II)
x 2
(mod p),
(I) gives δ ≡ 0 or 6≡ 0, (II) γ 6≡ 0, (III) is linear in x00 and gives 1 ≡ 0 or 6≡ 0, (V) e 6≡ 0, (VI) β 6≡ 0, (VII) is linear in x0 and gives α ≡ 0 or 6≡ 0 (mod p). Elimination of y 00 and z 0 between (II) and (VI) gives β 0 γ 0 x2 ≡ βγ
(mod p),
and between (V) and (VI) gives β 0 e0 y 002 ≡ βe (mod p) and βγ and βe are residues or non-residues, independently, according as β 0 γ 0 and β 0 e0 are residues or non-residues.
Class III. 1. General relations. In this class, the pth power of every operator of G is contained in {P }. There is in G a subgroup H1 of order pm−2 , which contains {P } self-conjugately.24 2. Determination of H1 . H1 is generated by P and some operator Q1 of G. Qp1 = P hp . Denoting Qa1 P b Qc1 P d · · · by the symbol [a, b, c, d, · · · ], all operators of H1 are included in the set [y, x]; (y = 0, 1, 2, · · · , p − 1, x = 0, 1, 2, · · · , pm−3 − 1). Since {P } is self-conjugate in H1 25 1+kp Q−1 1 P Q1 = P
(1) 24 Burnside, 25 Ibid.,
Theory of Groups, Art. 54, p. 64. Art. 56, p. 66.
41
m−4
.
Hence [−y, x, y] = [0, x(1 + kypm−4 )]
(2)
(m > 4).
and [y, x]s = sy, x s + ky 2s pm−4 .
(3)
Placing y = 1 and s = p in (3), we have, [Q1 P x ]p = Qp1 P xp = P (x+h)p and if x be so chosen that (mod pm−4 ),
(x + h) ≡ 0
Q = Q1 P x will be an operator of order p which will be used in place of Q1 , Qp = 1. 3. Determination of H2 . There is in G a subgroup H2 of order pm−1 , which contains H1 self-conjugately. H2 is generated by H1 , and some operator R1 of G. Rp1 = P lp . We will now use the symbol [a, b, c, d, e, f, · · · ] to denote Ra1 Qb P c Rd1 Qe P ···. The operations of H2 are given by [z, y, x]; (z, y = 0, 1, · · · , p − 1; x = 0, 1, · · · , pm−3 − 1). Since H1 is self-conjugate in H2 f
(4)
β α1 R−1 , 1 P R1 = Q1 P
(5)
b1 αp R−1 1 Q R1 = Q1 P
m−4
.
From (4), (5) and (3) αp − bp1 [−p, 0, 1, p] = 0, 1 β, αp1 + θpm−4 = [0, 0, 1], α1 − b1 where αp βk αp1 − 1 θ= 1 + aβ 2 α1 − 1
αp1 − 1 αp1 − bp1 p− α1 − b1 (α1 − b1 )2
.
Hence (6)
αp1 − bp1 β≡0 α1 − b1
(mod p),
αp1 + θpm−4 ≡ 1
and αp1 ≡ 1 (mod pm−4 ), or α1 ≡ 1 (mod pm−5 ) Equation (4) is replaced by (7)
(m > 5), α1 = 1+α2 pm−5 . m−5
β 1+α2 p R−1 1 P R1 = Q P
42
(mod pm−3 ),
,
From (5), (7) and (3). bp1 − 1 m−4 p [−p, 1, 0, p] = 0, b1 , a p . b−1 Placing x = lp and y = 1 in (2) we have Q−1 P lp Q = P lp , and bp1 ≡ 1
(mod p),
a
bp1 − 1 ≡0 b1 − 1
(mod p).
Therefore, b1 = 1. Substituting 1 for b1 and 1 + α2 pm−5 for α1 in congruence (6) we find (1 + α2 pm−5 )p ≡ 1
(mod pm−3 ),
α2 ≡ 0
or
(mod p).
Let α2 = αp and equations (7) and (5) are replaced by m−4
β 1+αp R−1 1 P R1 = Q P
(8)
m−4
αp R−1 1 Q R1 = QP
(9)
,
.
From (8), (9) and (3) h i (10) [−y, 0, x, y] = 0, βxy, x + αxy + aβx y2 + βky x2 pm−4 , (11)
[−y, x, 0, y] = [0, x, axypm−4 ].
From (2), (10), and (11) [z, y, x]s = [sz, sy + Us , sx + Vs pm−4 ],
(12) where Us = β Vs =
s 2
s 2
xz, αxz + kxy + ayz + βk x2 z 1 1 s 2 s + βk 3 x z + aβ 2 (2s − 1)z − 1 xz. 2 3!
Placing z = 1, y = 0, and s = p in (12)26 [R1 P x ]p = Rp1 P xp = P (x+l)p . If x be so chosen that x+l ≡0
(mod pm−4 )
then R = R1 P x is an operator of order p which will be used in place of R1 , and Rp = 1. 26 The terms of the form (Ax + Bx2 )pm−4 which appear in the exponent of P for p = 3 do not alter the conclusion for m > 5.
43
4. Determination of G. G is generated by H2 and some operation S1 . S p1 = P λp . Denoting S a1 Rb Qc P d · · · by the symbol [a, b, c, d, · · · ] all the operators of G are given by [v, z, y, x]; (v, z, y = 0, 1, · · · , p − 1; x = 0, 1, · · · , pm−3 − 1). Since H2 is self-conjugate in G (13)
γ s 1 S −1 1 P S1 = R Q P ,
(14)
c d ep S −1 1 Q S1 = R Q P
m−4
f
(15)
,
jpm−4
g
S1 R S1 = R Q P
.
From (13), (14), (15), and (12) [−p, 0, 0, 1, p] = [0, L, M, p1 + N pm−4 ] = [0, 0, 0, 1] and p1 ≡ 1
(mod pm−4 )
or
1 ≡ 1
(mod pm−5 )
(m > 5).
Let 1 = 1 + 2 pm−5 . Equation (13) is now replaced by (16)
m−5
γ δ 1+2 p S −1 1 P S1 = R Q P
.
If λ = 0 (mod p) and λ = λ0 p, [1, 0, 0, 1]p = p, 0, 0, p + p2 pm−5 + W pm−4 = [0, 0, 0, p + λ0 p2 + W 0 pm−4 ] and for m > 5 S1 P is of order pm−3 . We will take this in place of S1 and assume dv[λ, p] = 1. m−3 S p1 = 1. There is in G a subgroup H10 of order pm−2 which contains {S1 } self-conjugately. H10 = {S1 , S1v Rz Qy P x } and the operator T = Rz Qy P x is in H10 . There are two cases for discussion. 1◦ . Where x is prime to p. T is an operator of H2 of order pm−3 and will be taken as P . Then H10 = {S1 , P }. Equation (16) becomes m−4
1+p S −1 1 P S1 = P
44
.
There is in G a subgroup H20 of order pm−1 which contains H10 self-conjugately. 0
0
0
0
H20 = {H10 , S v1 Rz Qy P x }. 0
0
T 0 = Rz Qy is in H20 and also in H2 and is taken as Q, since {P, T 0 } is of order pm−2 . H20 = {H10 , Q} = {S1 , H1 } and in this case c may be taken ≡ 0 (mod p). 2◦ . Where x = x1 p. P p is in {S1 } since λ is prime to p. In the present case z y R Q is in H10 and also in H2 . If z 6≡ 0 (mod p) take Rz Qy as R; if z ≡ 0 (mod p) take it as Q. H10 = {S1 , R}
or
{S1 , Q},
and 0 m−4
p R−1 S1 R = S 1+k 1
Q−1 S1 Q = S 1+k 1
or
00 m−4
p
.
On rearranging these take the forms np S −1 1 R S1 = R S 1
m−4
m−4
= R P jp
0 m−4
S1−1 Q S1 = Q S n1 p
or
and either c or g may be taken ≡ 0 (mod p), cg ≡ 0
(17)
(mod p).
From (14), (15), (16), (12) and (17) dp − f p p , d , W pm−4 . [−p, 0, 1, 0, p] = 0, c d−f Place x = λp and y = 1 in (12) Q−1 P λp Q = P λp
S p1 Q S p1 = Q,
or
and dp ≡ 1
(mod p),
d = 1.
Equation (14) is replaced by (18)
m−4
c ep S −1 1 Q S1 = R Q P
.
From (15), (18), (17), (16) and (12) p p p d −f 0 m−4 g, W p . [−p, 1, 0, 0, p] = 0, f , d−f Placing x = λp, y = 1 in (10) R−1 P λp R = P λp , 45
m−4
= Q P ep
,
and f p ≡ 1 (mod p), f = 1. Equation (15) is replaced by m−4
g jp S −1 1 R S1 = R Q P
(19)
.
From (16), (18), (19) and (12) m−4
S 1−p P S p1 = P 1+2 p
=P
and 2 ≡ 0 (mod p). Let 2 = p and (16) is replaced by m−4
S 1−1 P S1 = Rγ Qδ P 1+p
(20)
.
Transforming both sides of (1), (8) and (9) by S1 −1 −1 1+kp −1 S −1 S1 · S −1 1 Q 1 P S1 · S 1 QS1 = S 1 P −1 S −1 S1 1 R
·
S −1 1 P S1
·
S −1 1 RS1
=
β S −1 1 Q S1
m−4
·
S1 ,
1+αpm−4 S −1 S1 , 1 P m−4
−1 −1 −1 ap −1 S −1 S1 · S −1 1 R 1 QS1 · S 1 RS1 = S 1 QS1 · S 1 P
S1 .
Reducing these by (18), (19), (20) and (12) and rearranging 0, γ, δ + βc, 1 + + αc + k + acδ + aβ 2c − aγ pm−4 = [0, γ, δ, 1 + ( + k)pm−4 ]. [0, γ, β + δ, 1 + {kg + + α + aδ − aγg}pm−4 ] h i = 0, γ + βc, β + δ, 1 + + α + βe + α β2 c + aβγ pm−4 , [0, c, 1, (e + a)pm−4 ] = [0, c, 1, (e + a)pm−4 ]. The first gives (21) (22)
βc ≡ 0 (mod p), ac + acδ − aγ ≡ 0 (mod p).
Multiplying this last by g (23)
agγ ≡ 0
(mod p).
From the second equation above (24)
gk + αδ ≡ βe + aβγ
(mod p).
Multiplying by c (25)
acδ ≡ 0
(mod p).
These relations among the constants must be satisfied in order that our equations should define a group. 46
From (20), (19), (18) and (12) (26) [−y, 0, 0, x, y] = [0, γxy + χ1 (x, y), δxy + φ1 (x, y), x + Θ1 (x, y)pm−4 ], (27) [−y, 0, x, 0, y] = [0, cxy, x, Θ2 (x, y)pm−4 ], (28) [−y, x, 0, 0, y] = [0, x, gxy, Θ3 (x, y)pm−4 ], where y 2 , γgx y2 + βγ x2 y, xy + y2 γjx + eδx + aδγ + (αγ + kδ) x2 + y3 [cδj + egγ]x + x2 [αγy + δky + aδγy 2 ] exy + cjx y2 + ac x2 y, jxy + egx y2 + ag x2 y.
χ1 (x, y) = cδx φ1 (x, y) = Θ1 (x, y) = Θ2 (x, y) = Θ3 (x, y) =
+ βγk
x 3
y2 ,
Let a general power of any operator be [v, z, y, x]s = [sv, sz + Us , sy + Vs , sx + Ws pm−4 ].
(29)
Multiplying both sides by [v, z, y, x] and reducing by (2), (10), (11), (26), (27) and (28), we find Us+1 ≡ Us + (cy + γx)sv + cδ sv 2 x (mod p), x Vs+1 ≡ Vs + (gz + δx)sv + γg sv 2 x + βγ 2 sv + β(sz + Us )x (mod p), Ws+1 ≡ Ws + Θ1 (x, sv) + ey + jz + aγxy + ac y2 + ag z2 sv + αx + βk x2 + ay + aδsx + αgsvz sz + ksxy x + sv 2 {cjy + egz} + Us αx + βk 2 + ay + a(δx + gz)sv s + aβ sz+U x + kVs x (mod p). 2 From (29) U1 ≡ 0,
V1 ≡ 0,
W1 ≡ 0
(mod p).
A continued use of the above congruences give 1 1 v + cδxv{ (2s − 1)v − 1} 2s (mod p), 2 3 Vs ≡ {[gz + δx + βγ x2 v + βxz} 2s 1 1 + γgxv{ (2s − 1)v − 1} 2s + βγ 3s x2 v (mod p), 2 3
Us ≡ (cy + γx)
s 2
47
Ws ≡
s 2
n xv + egv + (αγ + δkv + βkz) 2s + βγ k x3 v + ac y2 v o n + jvz + ag z2 v + αxz + kxy + aγxyv + ayz + 3s αcxyv + αγx2 v + 2βγk x2 xv + gkxzv + δkx2 v + βkx2 z + acvy 2 o 2s − 1 n + aγxvy + βkγ 4s x3 v + 2s aδγ x2 v 2 + aδxzv 3 o 1 n 1 on s 2 + agvz + v 2 (2s − 1)v − 1 γjx + eδx + aδγx 2 3 o 1 n x x s 2 + αcδ 2 + γgk 2 + cjy + egz + 2s 2 v − (2s − 1)v 6 o o 1 n 1 + 2 cδjx + egγx v + 3s (s − 1)v − 1 αcδ 2 2 o n 1 2 1 s (2s − 1)z − 1 z + γgk x v + aβx 2 2 3 1 2 s 1 + aδγx v 3 (3s − 1) (mod p) 2 2
Placing v = 1, z = y = s = p in (29)27 [S1 P x ]p = S p1 P xp = P (λ+x)p
(p > 3).
If x be so chosen that x+λ≡0
(mod pm−4 ).
S = S1 P x is an operator of order p and is taken in place of S1 . S p = 1. The substitution of S for S1 leaves congruence (17) invariant. 5. Transformation of the groups. All groups of this class are given by
(30)
with
m−4 Q−1 P Q = P 1+kp , −1 β 1+αpm−4 R PR=Q P , −1 apm−4 R QR = QP , G: −1 γ δ 1+pm−4 S P S = R Q P , m−4 −1 c ep S Q S = R Q P , m−4 −1 S R S = R Qg P jp , m−3
Pp
= 1,
Qp = Rp = S p = 1,
27 For p = 3 and cδ ≡ γg ≡ βγ ≡ 0 (mod p) there are terms of the form (A + Bx + Cx2 + Dx3 )pm−4 in the exponent of P . For m > 5 these do not vitiate our conclusion. For p = 3 and cδ, γg, or βγ prime to p, [S1 P x ]p is not contained in {P } and the groups defined belong to Class II.
48
(k, β, α, a, γ, δ, , c, e, g, j = 0, 1, 2, · · · , p − 1). These constants are however subject to conditions (17), (21), (22), (23), (24) and (25). Not all these groups are distinct. Suppose that G and G0 of the above set are simply isomorphic and that the correspondence is given by S, R, Q, P C= 0 . S1 , R10 , Q01 , P10 Inspection of (29) gives 000
000
000
000 m−4
S10 = S 0v R0z Q0y P 0x R10
=S
0v 00
R
0z 00
0
0y 00
Q
0
P
0
p
0x00 pm−4 0 m−4
Q01 = S 0v R0z Q0y P 0x p P10 = S 0v R0z Q0y P 0x ,
,
,
,
in which x and one out of each of the sets v 0 , z 0 , y 0 , x0 ; v 00 , z 00 , y 00 , x00 ; v 000 , z 000 , y 000 , x000 are prime to p. Since S, R, Q, and P satisfy equations (30), S10 , R10 , Q01 and P10 also satisfy them. Substituting these operators and reducing in terms of S 0 , R0 , Q0 , and P 0 we get the six equations [Vκ0 , Zκ0 , Yκ0 , Xκ0 ] = [Vκ , Zκ , Yκ , Xκ ]
(31)
which give the following twenty-four 0 V ≡ Vκ κ0 Zκ ≡ Zκ (32) Yκ0 ≡ Yκ 0 Xκ ≡ Xκ
(κ = 1, 2, 3, 4, 5, 6),
congruences (mod p), (mod p), (mod p), (mod pm−3 ),
where V10 = v, Z10 Y10
V1 = v, v0 2 , 0 0 γ g x v2 +
= Z + c0 (yv 0 − y 0 v) + γ 0 xv 0 + cδx 0
0
0
0
0
Z1 = z,
= y + g (zv − z v) + δ xv + β 0 xz 0 , Y1 = y, n 0 0 X10 = x + 0 xv 0 + (γ 0 j 0 x + e0 δ 0 x + a0 δ 0 γ 0 x) v2 + c0 δ 0 j 0 v3 + (α0 γ 0 v 0 + δ 0 k 0 v 0 0 + a0 δ 0 γ 0 v 2 + β 0 k 0 z 0 ) x2 + j 0 (zv 0 − z 0 v) + e0 g 0 [z v2 − z 0 v2 ] 0 0 + a0 g 0 [ z2 v 0 + z2 v − zz 0 v] + e0 (yv 0 − y 0 v) + c0 j 0 [y v2 − y 0 v2 ] 0 + a0 c0 [ y2 v 0 + v −y − yy 0 v] + a0 (yz 0 − y 0 z) − a0 β 0 xz 02 + α0 xz 0 2 o 0 + α0 β 0 x z2 + a0 γ 0 x(y − y 0 )v 0 + k 0 xy 0 pm−4 , X1 = x + kxpm−4 ,
49
V20 = v,
V2 = v + βv 0 , v 00 2 , 00 0 0 γ g x v2
Z20 = z + c0 (yv 00 − y 00 v) + γ 0 xv 00 + e0 δ 0
Z2 = z + βz 0 + c0 βy 0 v, + β 0 γ 0 x2 v 00 + β 0 xz 00 ,
Y20 = y + g 0 (zv 00 − z 00 v) + δ 0 xv 00 + Y2 = y + βy 0 + g 0 βz 0 v, n 00 X20 = x + Θ01 (x, v 00 ) + j 0 (zv 00 − z 00 v) + e0 g 0 [z v2 − z 00 v2 ] + a0 g 0 [ x2 v 00 00 00 + −z2 v − zz 00 v] + e0 (yv 00 − y 00 v) + c0 j 0 [y v2 − y 00 v2 ] + a0 c0 [ y2 v 00 00 + −y2 v − yy 00 v 00 ] + a0 g 0 (zv 00 − z 00 v)z 00 + a0 (yz 00 − y 00 z) + a0 δ 0 v 00 z 00 o 00 + a0 γ 0 (y − y 00 )v 00 x + α0 xz 00 + a0 β 0 x z2 + β 0 k 0 x2 z 00 + k 0 xy 00 pm−4 , n X2 = x + αx + βx0 + a0 β2 y 0 z 0 + e0 βvy 0 + (c0 j 0 β + e0 g 0 βz 0 ) v2 o 0 0 + a0 c0 βy2 v + j 0 βvz 0 + a0 g 0 βz2 + a0 β(g 0 z 0 v + y 0 )z pm−4 , V30 = v 0 , V3 = v 0 , Z30 = z 0 + c0 (y 0 v 00 − y 00 v 0 ), Z3 = z 0 , Y30 = y 0 + g 0 (z 0 v 00 − z 00 v 0 ), Y3 = y 0 , n 00 X30 = x0 + j 0 (z 0 v 00 − z 00 v 0 ) + e0 g 0 [ v2 z 0 −
v0 2
0 00 z 00 ] + a0 g 0 [ z2 v 00 + −z2 v 0 00 0 0 00 − z 0 z 00 v 0 ] + e0 (y 0 v 00 − y 00 v 0 ) + c0 j 0 [y 0 v2 − y 00 v2 ] + a0 c0 [ y2 v 00 + −y2 v 0 o − y 00 y 0 v 00 ] + a0 (y 0 z 00 − y 00 z 0 ) pm−4 ,
X4 = (x0 + a0 x)pm−4 , V40 = v, V4 = v + γv 00 + δv 0 , 000 Z40 = z + c0 (yv 000 − y 000 v) + γ 0 xv 000 + c0 δ 0 x v2 , Z4 = z + γz 00 + δz 0 + c0 [ γ2 v 00 y 00 + 2δ v 0 y 0 ] + c0 (γy 00 + δy 0 )v + c0 γδy 00 v, 000 Y40 = y + g 0 (zv 000 − z 000 v) + δ 0 xv 000 + γ 0 g 0 x v2 + β 0 γ 0 x2 v 000 + β 0 xz 000 , Y4 = y + γy 00 + δy 0 + g 0 [ γ2 v 00 z 00 + 2δ v 0 z 0 ] + g 0 (γz 00 + δz 0 )v + g 0 δγv 0 z 00 , n 000 X40 = x + Θ01 (x, v 000 ) + j 0 (zv 000 − z 000 v) + e0 g 0 [ v2 z − v2 z 000 ] + a0 g 0 z2 v 000 i 000 000 + −z2 v − zz 000 v + e0 (yv 000 − y 000 v) + c0 j 0 [y v2 − y 000 v2 ] 000 + a0 c0 [ y2 v 000 + −y2 v − yy 000 v 000 ] + a0 g 0 (v 000 z − vz 000 )z 000 + a0 (yz 000 − y 000 z) + a0 δ 0 xz 000 v 000 + a0 γ 0 x(y − y 000 )v 000 + α0 xz 000 o 000 + a0 β 0 x z2 + β 0 k 0 z 000 x2 + k 0 xy 000 pm−4 .
50
n 00 X4 = x + x + δx0 + γx00 + γ2 [a0 c0 y2 v 00 + a0 y 00 z 00 + e0 v 00 y 00 + j 0 v 00 z 00 00 + a0 g 0 z2 v 00 + (c0 j 0 v 00 y 00 + e0 g 0 v 00 z 00 )(v + δv 0 ) + a0 (z + δz 0 )v 00 z 00 2γ − 1 0 0 00 002 1 1 a g v z + [ (2γ − 1)v 00 − 1](c0 j 0 y 00 + e0 g 0 z 00 )v 00 ] + 3 2 3 0 + γ3 a0 c0 v 00 y 00 + 2δ [a0 c0 y2 v 0 + a0 y 0 z 0 + e0 v 0 y 0 + j 0 v 0 z 0 0 + a0 g 0 z2 v 0 + j 0 c0 vv 0 y 0 + e0 g 0 vv 0 z 0 + a0 g 0 v 0 zz 0 + a0 c0 γy 0 y 00 v 0 2δ − 1 0 0 0 02 1 1 a g v z + { (2δ − 1)v 0 − 1}(c0 j 0 y 0 + e0 g 0 z 0 )] + 3 2 3 00 00 + 3δ a0 c0 v 0 y 02 + (v + δv 0 )[j 0 γz 00 + γz2 a0 g 0 + e0 γy 00 + γy2 a0 c0 0 + a0 g 0 (z + δz 0 )] + v+δv [e0 g 0 γz 00 + c0 j 0 γy 00 ] + δ[(e0 g 0 z 0 2 + c0 j 0 y 0 ) v2 + e0 vy 0 + j 0 z 0 + a0 zy + a0 g 0 vzz 0 + a0 γz 0 y 00 + a0 c0 γvy 0 y 00 ] o 0 0 + a0 g 0 δz2 v + a0 c0 δy2 v + a0 γzy 00 pm−4 , V50 = v 0 , V5 = v 0 + cv 00 , Z50 = z 0 + c0 (y 0 v 000 − y 000 v 0 ), Z5 = z 0 + cz 00 + c0 cy 00 v, Y50 = y 0 + g 0 (z 0 v 000 − z 000 v 0 ), Y5 = y 0 + cy 00 + g 0 cv 0 z 00 , n 000 0 0 X50 = x0 + j 0 (z 0 v 000 − z 000 v 0 ) + e0 g 0 [ v2 z 0 − v2 z 000 ] + a0 g 0 [ z2 v 000 000 000 0 + −z2 v 0 − z 0 z 000 v 0 ] + c0 (y 0 v 000 − y 000 v 0 ) + c0 j 0 [y 0 δ2 − y 000 v2 ] o 0 000 + a0 c0 [ y2 v 000 + −y2 v 0 − y 0 y 000 v 000 ] + a0 (y 0 z 000 − y 000 z 0 ) pm−4 , n 0 X5 = x0 + ex + cx00 + a0 2c y 00 z 00 + j 0 cv 0 z 00 + (e0 g 0 cz 00 + c0 cj 0 y 00 ) v2 + e0 cy 00 v o 00 00 + a0 cy 00 z 0 + a0 g 0 z 0 v 0 + a0 g 0 cz2 + a0 c0 cy2 pm−4 , V60 = v 00 , V6 = v 00 + gv 0 , Z60 = z 00 + c0 (y 00 v 000 − y 000 v 00 ), Z6 = z 00 + gz 0 , Y60 = y 00 + g 0 (z 00 v 000 − z 000 v 00 ), Y6 = y 00 + gy 0 , n h 00 000 00 X60 = x00 + j 0 (z 00 v 000 − z 000 v 00 ) + e0 g 0 [ v2 z 00 − v2 z 000 ] + a0 g 0 z2 v 000 i 000 000 00 + −z2 v 00 − z 00 z 000 v 00 + e0 (y 00 v 000 − y 000 v 00 ) + c0 j 0 [y 00 v2 − y 000 v2 ] o 00 000 + a0 c0 [ y2 v 000 + −y2 v 00 − y 00 y 000 v 000 ] + a0 (y 00 z 000 − y 000 z 00 ) pm−4 , X6 = {x00 + jx + gx0 + a0 gy 00 z 0 }pm−4 . The necessary and sufficient condition for the simple isomorphism of the two groups G and G0 is that congruences (32) shall be consistent and admit of solution subject to conditions derived below. 6. Conditions of transformation. Since Q is not contained in {P }, R is not contained in {Q, P }, and S is not contained in {R, Q, P }, then Q01 is not 51
contained in {P10 }, R10 is not contained in {Q01 , P10 }, and S10 is not contained in {R10 , Q01 , P10 }. Let s0 spm−4 Q0 1 = P 0 1 . This equation becomes in terms of S 0 , R0 , Q0 and P 0 [s0 v 0 , s0 z 0 + c0
s0 2
v 0 y 0 , s0 y 0 + g 0
s0 2
v 0 z 0 , Dpm−4 ] = [0, 0, 0, sxpm−4 ],
and s0 v 0 ≡ s0 z 0 ≡ s0 y 0 ≡ 0
(mod p).
At least one of the three quantities v 0 , z 0 or y 0 is prime to p, since otherwise s may be taken = 1. Let s00 s0 spm−4 R0 1 = Q0 1 P 0 1 , 0
or in terms of S 0 , R0 , Q0 and P 0 [s00 v 00 , s00 z 00 + c0
s00 2
v 00 y 00 , s00 y 00 + g 0
s00 2
v 00 z 00 , Epm−4 ] 0 = [s0 v 0 , s0 z 0 + c0 s2 v 0 y 0 , s0 y 0 + g 0
s0 2
v 0 z 0 , E1 pm−4 ],
and s00 v 00 ≡ s0 v 0 00 00
00
s z + c0 s2 v 00 y 00 00 s00 y 00 + g 0 s2 v 00 z 00
(mod p), 0 ≡ s z + c0 s2 v 0 y 0 (mod p), 0 ≡ s0 y 0 + g 0 s2 v 0 z 0 (mod p). 0 0
Since c0 g 0 ≡ 0 (mod p), suppose g 0 ≡ 0 (mod p). Elimination of s0 between the last two give by means of the congruence Z30 ≡ Z3 (mod p), s00 {2(y 0 z 00 − y 00 z 0 ) + c0 y 0 y 00 (v 0 − v 00 )} ≡ 0
(mod p),
between the first two s00 {2(v 0 z 00 − v 00 z 0 ) + c0 v 0 v 00 (y 0 − y 00 )} ≡ 0
(mod p),
and between the first and last s00 (y 0 v 00 − y 00 v 0 ) ≡ 0
(mod p).
At least one of the three above coefficients of s00 is prime to p, since otherwise s may be taken = 1. Let s000 s00 s0 spm−4 S 0 1 = R0 1 Q0 1 P 0 1 00
52
or, in terms of S 0 , R0 , Q0 , and P 0 [s000 v 000 , s000 z 000 + c0
s000 000 000 000 000 2 v y ,s y + 0 0 00 00 0 0 0
g0
s000 2
v 000 z 000 , E2 pm−4 ] 0 = [s00 v 00 + s v , s z + s z + c { s2 v 00 y 00 + s2 v 0 y 0 + s0 s00 y 00 v 0 }, 00 0 s00 y 00 + s0 y 0 + g 0 { s2 v 00 z 00 + s2 v 0 z 0 + s0 s00 v 0 z 00 }, E3 pm−4 ]
00
and s000 v 000 ≡ s00 v 00 + s0 v 0 (mod p), 000 s000 z 000 + c0 s2 v 000 y 000 ≡ s00 z 00 + s0 z 0 + c0 { 000 s000 y 000 + g 0 s2 v 000 z 000
s00 2
≡ s00 y 00 + s0 y 0 + g 0 {
s00 2
v 00 y 00 +
s0 2
v 00 z 00 +
s0 2
v 0 y 0 + s0 s00 y 00 v 0 } (mod p),
v 0 z 0 + s0 s00 z 00 v 0 } (mod p).
If g 0 ≡ 0 and c0 6≡ 0 (mod p) the congruence Z30 ≡ Z3 (mod p) gives (y 0 v 00 − y 00 v 0 ) ≡ 0
(mod p).
Elimination in this case of s00 between the first and last congruences gives s000 (y 00 v 000 − y 000 v 00 ) ≡ 0
(mod p).
Elimination of s00 between the first and second, and between the second and third, followed by elimination of s0 between the two results, gives c02 00 00 000 002 0 00 00 0 y v (y 0 v 000 − y 000 v 0 ) ≡ 0 (mod p). s z −cy z v + 4 Either (y 00 v 000 − y 000 v 00 ), or (y 0 v 000 − y 000 v 0 ) is prime to p, since otherwise s000 may be taken = 1. A similar set of conditions holds for c0 ≡ 0 and g 0 6≡ 0 (mod p). When c0 ≡ g 0 ≡ 0 (mod p) elimination of s0 and s00 between the three congruences gives 0 v v 00 v 000 s000 ∆ ≡ s000 y 0 y 00 y 000 ≡ 0 (mod p) z 0 z 00 z 000 and ∆ is prime to p, since otherwise s000 may be taken = 1.
53
7. Reduction to types. In the discussion of congruences (32), the group G0 is taken from the simplest case and we associate with it all simply isomorphic groups G.
I. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
a2 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
β2 1 1 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1
A. c2 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1
g2 1 1 1 1 0 0 1 1 0 0 1 0 1 1 0 0 1 0 1 1 1 1 0 1 1 0 0 1 0
γ2 1 1 1 1 1 1 1 0 1 1 1 1 0 1 0 1 0 0 1 0 1 0 1 0 1 0 1 0 0
δ2 1 1 1 0 1 0 1 1 1 0 1 1 1 0 1 0 0 0 1 1 0 0 1 1 0 1 0 0 0
54
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
k2 1 0 1 1 1 1 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 1 0
B. α2 1 1 0 1 1 1 0 1 1 1 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0 1 0 0 0 1 0 0
2 1 1 1 0 1 1 1 0 1 1 0 1 1 0 0 1 0 1 1 0 0 1 0 0 1 0 0 0 1 0 0 0
e2 1 1 1 1 0 1 1 1 0 1 1 0 1 0 1 0 1 0 1 0 1 0 0 1 0 0 0 1 0 0 0 0
j2 1 1 1 1 1 0 1 1 1 0 1 1 0 1 0 0 1 1 0 1 0 0 1 0 0 0 1 0 0 0 0 0
II. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 B. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
1 × × 12 12 21 21 12 12 21 24 12 24 21 21 21 21 12 24 24 21 24 24 24 21 24 21 24 24 * 24 24 *
2 × 21 21 × 21 21 21 24 21 24 24 24 21 24 24 21 24 24 24 24 * 24 * 24 24 24 * * * * * *
3 × 31 31 ×
4
A. 5 196
6 196
196 196 *
31 31 34
196 196
* 34 34
196
196 196 196 196
7 196 196
31
196 196
* 34 *
196
*
196
* *
196
34 *
34 * * *
196
* * * * *
196 196 196
55
196 196 196 196 196 196 196 196 196 196 196 196 196 196 196 196 196
196 196 196 196
10 196
196
196 196 196
196
34
196
196 196 196 196
196 196 196
196
*
9 196
196 196 196
196 *
8
196 196 196
196 196 196 196 196 196
196 196 196
196 196 196 196 196 196
196 196 196 196 196 196
196 196 196 196
196 196 196
196 196 196 196
196 196 196
196
196 196
II. (continued) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 B. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
11 × 252 111 242
12 191
× 252 252
192
192 192
13 × × 131 131
A. 14 111 252 242 242
× 132 132
116 252 252
196
15 191
16 191
212 191
192 192
212
192
216
196
17 131 132 131 131
18 191
136 132 132
212
212 191
216
2510 242
192
× 131
2510 *
212
192
1310 131
212
116
*
136
116
*
192
136
*
116
192
136
116
212
192
136
212
132
252 216
196
252 196 2510
1310
2510
1310
2510
196 2510
196
216 216
192 196 196
136
*
1310
116
*
216
196
196 196
216 216
196 196
216 216
196
216
196
216
136
192 196 196
216
196
192
192 192 196
1310
216
116
192 191 ×
132
216
196
19 × ×
*
196 196 2510
1310
251 0
56
216 1310 196
II. (concluded) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 B. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
20 191 192
21 191 ×
22 191 212
192 192 196
191 191 ×
191 191 216
192 192 196
212 212 216
212 212 216
* 196 196
191 216 216
191 216 216
192 196 196
212 216 216
212 216 216
196
216 216
216 216
196
216
216
23 191 192 192 196 191
192 196 192 196 192 196
196 192 196 196
196
A. 24 111 × 252 111 116 31 252 242 116 34 252 2510 31 116 31 31 252 2510 34 116 34 34 2510 31 34 31 2510 34 * 34 34 *
25 111 × 242 111 31 116 252 252 31 × 242 34 116 31 116 31 252 34 2510 31 2510 34 34 116 34 31 34 2510 * 34 34 *
26 191
27 191
212 191 216 191
192 192 196 192
212 216 212 216 191 216
192 196 192 196 * 196
216 212 216 216
196 192 196 196
216
196
28 111 252 252 111 31 31 * 252 31 34 252 34 31 31 31 * * 34 34 31 34 * 34 31 * * 34 34 * * * *
29 191 212 191 216 191
212 216 212 216 191 216
216 212 216 216
216
For convenience the groups are divided into cases. The double Table I gives all cases consistent with congruences (17), (21), (23) and (25). The results of the discussion are given in Table II. The cases in Table II left blank are inconsistent with congruences (22) and (24), and therefore have no groups corresponding to them. Let κ = κ1 pk2 where dv[κ1 , p] = 1 (κ = a, β, c, g, γ, d, k, α, , e, j). In explanation of Table II the groups in cases marked rs are simply isomorphic with groups in Ar Bs .
57
The group G0 is taken from the cases marked × . The types are also selected from these cases. The cases marked ∗ divide into two or more parts. Let a − αe + jk = I1 , aδ(a − e) + 2I1 = I3 , αδ − β = I5 , c − eγ = I7 , δe + γj = I9 ,
a − jk αg − βj g − δj αe − jk αγ + δk
= I2 , = I4 , = I6 , = I8 , = I10 .
The parts into which these groups divide, and the cases with which they are simply isomorphic, are given in Table III.
III. A1,2 B ∗ A3 B ∗ A4 B ∗ A12 B13 A14 B11 A15,18 B ∗ A16 B24 A20 B14 A24,25 B ∗ A27 B15 A29 B7,17 A29 B16,26 A29 B22,25,30,31 A29 B29,32 A29 B29,32
dv[I1 , p] = p dv[I2 , p] = p dv[I3 , p] = p dv[I4 , p] = p dv[I5 , p] = p dv[I4 , p] = p dv[I6 , I5 , p] = p dv[I7 , p] = p dv[I8 , p] = p dv[I6 , p] = p dv[I10 , p] = p dv[I9 , p] = p dv[I9 , p] = p dv[I8 , I9 , p] = p [I8 , p] = 1, [I9 , p] = p
58
21 31 31 191 111 191 191 191 31 191 242 116 2510 116 2510
dv[I1 , p] = 1 dv[I2 , p] = 1 dv[I3 , p] = 1 dv[I4 , p] = 1 dv[I5 , p] = 1 dv[I4 , p] = 1 dv[I6 , I5 , p] = 1 dv[I7 , p] = 1 dv[I8 , p] = 1 dv[I6 , p] = 1 dv[I10 , p] = 1 dv[I9 , p] = 1 dv[I9 , p] = 1 [I8 , p] = p, [I9 , p] = 1 [I8 , p] = 1, [I9 , p] = 1
24 34 34 192 242 212 192 192 34 192 252 31 34 31 34
8. Types. The types for this class are given by equations (30) where the constants have the values given in Table IV.
IV. 11 21 31 111 ∗131 191 12 ∗132 192 ∗212 242 252 24 34 116 ∗136
a 0 1 κ 0 0 0 0 0 0 0 0 0 1 κ 0 0
β 0 0 1 1 1 0 0 1 0 0 0 0 0 1 1 1
c 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0
g 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
γ 0 0 0 0 1 0 0 1 0 0 1 0 0 0 0 1
δ 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0
k 0 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0
α 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0
e 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
j 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1
κ = 1, and a non-residue (mod p). ∗For p = 3 these groups are isomorphic in Class II.
A detailed analysis of congruences (32) for several cases is given below as a general illustration of the methods used. A3 B1 . The special forms of the congruences for this case are β 0 xz 0 ≡ 0 (mod p), a0 (yz 0 − y 0 z) ≡ kx (mod p), βv 0 ≡ 0, βz 0 ≡ 0, βy 0 ≡ β 0 xz 00 (mod p), 00 a0 (yz 00 − y 00 z) + a0 β 0 x z2 ≡ αx + βx0 + a0 βy 0 z (mod p), a0 (y 0 z 00 − y 00 z 0 ) ≡ ax (mod p), γv 00 + δv 0 ≡ 0 (mod p), γz 00 + δz 0 ≡ 0 (mod p), γy 00 + δy 0 ≡ β 0 xz 00 (mod p), 000 a0 (yz 000 − y 000 z) + a0 β 0 x z2 ≡ x + γx00 + δx + a0 δy 0 z + a0 γy 00 z + a0 γ2 y 00 z 00 (mod p),
(II) (III) (IV),(V),(VI) (VII) (X) (XI) (XII) (XIII) (XIV)
(XV),(XVI),(XVII) (XVIII)
cv 00 ≡ 0, cz 00 ≡ 0, cy 00 ≡ 0 (mod p), a0 (y 0 z 000 − y 000 z 0 ) ≡ ex (mod p), 59
(XIX),(XX),(XXI) (XXII)
gv 0 ≡ 0, gz 0 ≡ 0, gy 0 ≡ 0 a0 (y 00 z 000 − y 000 z 00 ) ≡ jx (mod p),
From (II) z 0 ≡ 0 (mod p). The conditions of isomorphism give 0 v v 00 v 000 0 ∆ ≡ y y 00 y 000 6≡ 0 z 0 z 00 z 000
(mod p),
(mod p).
Multiply (IV), (V), (VI) by γ and reduce by (XII), βγv 0 ≡ 0, βγz 0 ≡ 0, βγy 0 ≡ 0 (mod p). Since ∆ 6≡ 0 (mod p), one at least of the quantities, v 0 , z 0 or y 0 is 6≡ 0 (mod p) and βγ ≡ 0 (mod p). From (XV), (XVI) and (XVII) c ≡ 0 (mod p), and from (XIX), (XX) and (XXI) g ≡ 0 (mod p). From (IV), (V), (VI) and (X) if a ≡ 0, then β ≡ 0 and if a 6≡ 0, then β 6≡ 0 (mod p). At least one of the three quantities β, γ or δ is 6≡ 0 (mod p) and one, at least, of a, e or j is 6≡ 0 (mod p). A3 : Since z 000 ≡ 0 (mod p), (XVIII) gives e ≡ 0. Elimination between (III), (X), (XIV) and (XXII) gives a − kj ≡ 0 (mod p). Elimination between (VI) 2 and (X) gives a0 β 0 z 00 ≡ aβ (mod p) and aβ is a quadratic residue or non-residue 0 0 according as a β is or is not, and there are two types for this case. A4 : Since y 0 and z 00 are 6≡ 0 (mod p), e 6≡ 0 (mod p). Elimination between (VI), (X), (XIII) and (XVIII) gives aδ − βe ≡ 0 (mod p). This is a special form of (24). Elimination between (III), (VII), (X), (XIII), (XIV), (XVIII) and (XXII) gives 2jk + aδ(a − e) + 2(a − αe) ≡ 0 (mod p). A24 : Since from (XI), (XII) and (XIII) y 00 and z 000 6≡ 0 (mod p), and z 00 ≡ v ≡ 0 (mod p), (xxii) gives j 6≡ 0 (mod p). Elimination between (III), (X), (XVIII) and (XXII) gives 00
αe − jk ≡ 0
(mod p).
A25 : (XI), (XII) and (XIII) give v 0 ≡ z 0 ≡ 0 and y 0 , z 000 6≡ 0 (mod p) and this with (XVIII) gives e 6≡ 0. Elimination between (III), (VII), (XVIII) and (XXII) gives αe − jk ≡ 0
(mod p).
A28 : Since a ≡ 0 then e or j 6≡ 0 (mod p). Elimination between (III), (VII), (XVIII) and (XXII) gives αe − jk ≡ 0 60
(mod p).
Multiply (XIII) by a0 z 000 and reduce 2
δe + γj ≡ a0 β 0 z 000 6≡ 0
(mod p).
A11 B1 . The special forms of the congruences for this case are (II) (III) (IV),(V),(VI) (VII) (X) (XI) (XII) (XIII) (XIV) (XV),(XVI),(XVII) (XVIII) (XIX),(XX),(XXI) (XXII)
β 0 xz 0 ≡ 0 (mod p), kx ≡ 0 (mod p), βv 0 ≡ βz 0 ≡ 0, βy 0 ≡ β 0 xz 00 , αx + βx0 ≡ 0 (mod p), ax ≡ 0 (mod p), 00 γv + δv ≡ 0 (mod p), γz 00 + δz ≡ 0 (mod p), γy 00 + δy ≡ β 0 xz 000 (mod p), x + γx00 + δx0 ≡ 0 (mod p), cv 00 ≡ cz 00 ≡ cy 00 ≡ 0 (mod p), ex ≡ 0 (mod p), gv 0 ≡ gz 0 ≡ gy 0 ≡ 0 (mod p), jx ≡ 0 (mod p),
(II) gives z 0 = 0, (III) gives k ≡ 0, (X) gives a ≡ 0, (XV), (XVI), (XVII) give c ≡ 0(∆ 6≡ 0), (XVIII) gives e ≡ 0, (XIX), (XX), (XXI) give g ≡ 0, (XXII) gives j ≡ 0. One of the two quantities z 00 or z 000 6≡ 0 (mod p), and by (VI) and (XIII) one of the three quantities β, γ or δ is 6≡ 0. A11 : (XIV) gives ≡ 0 (mod p). Multiplying (IV), (V), (VI) by γ gives, by (XII), βγv 0 ≡ βγz 0 ≡ βγy 0 ≡ 0 (mod p), and βγ ≡ 0 (mod p). A14 : Elimination between (VII) and (XIV) gives αδ − β ≡ 0 (mod p). A24 : (VII) gives α ≡ 0 (mod p), (XIV) ≡ 0 or 6≡ 0 (mod p). A25 : (VII) gives α ≡ 0 (mod p), (XIV) ≡ or 6≡ 0 (mod p). A28 : (VII) gives α ≡ 0 (mod p), (XIV) ≡ or 6≡ 0 (mod p).
61
A19 B1 . The special forms of the congruences for this case are (I) (III) (IV),(V),(VI) (VII) (VIII) (X) (XI)
c0 (yv 0 − y 0 v) ≡ 0 (mod p), kx ≡ 0 (mod p), βv ≡ 0, βz ≡ c0 (yv 00 − y 00 v), βy 0 ≡ 0 αx + βx0 ≡ 0 (mod p), c0 (y 0 v 00 − y 00 v 0 ) ≡ 0 (mod p), ax ≡ 0 (mod p), 00 γv + δv 0 ≡ 0 (mod p),
(mod p),
(XII) γz 00 + δz 0 + c0 γδy 00 v + c0 2δ v 0 y 0 + c0 γ2 v 00 y 00 ≡ c0 (yv 000 − y 000 v) (mod p), (XIII) γy 00 + δy 0 ≡ 0 (mod p), (XIV) x + γx00 + δx0 ≡ 0 (mod p), (XV),(XVI),(XVII) cv 00 ≡ 0, cz 00 ≡ c0 (y 0 v 000 − y 000 v 0 ), cy 00 ≡ 0 (mod p), (XVIII) ex + cx00 ≡ 0 (mod p), (XIX),(XX),(XXI) gv 0 ≡ 0, gz 0 ≡ c0 (y 00 v 000 − y 000 v 00 ), gy 0 ≡ 0 (mod p), (XXII) jx + gx0 ≡ 0 (mod p). (III) gives k ≡ 0, (X) gives a ≡ 0. Since dv[(y 0 v 000 − y 000 v 0 ), (y 00 v 000 − y 000 v 00 ), p] = 1 then dv[c, g, p] = 1. If c 6≡ 0, v 00 ≡ y 00 ≡ 0 (mod p) and therefore g ≡ 0 (mod p) and if g 6≡ 0, then c ≡ 0 (mod p). A12 : (XVIII) gives e ≡ 0 (mod p). Elimination between (VII) and (XXII) gives αg − βj ≡ 0 (mod p), (XIV) gives ≡ 0 (mod p). A15 : (XVIII) gives e ≡ 0 (mod p). Elimination between (VII) and (XXII) gives αg − βj ≡ 0 (mod p), (XIV) gives ≡ 0 or 6≡ 0 (mod p). A16 : (XVIII) gives e ≡ 0. Elimination between (XIV) and (XXII) gives g − δj ≡ 0 (mod p), between (VII) and (XIV) gives αδ − β ≡ 0. A18 : (XVIII) gives e ≡ 0 (mod p). Elimination between (VII) and (XXII) gives αg − βj ≡ 0 (mod p), (XIV) gives ≡ 0 or 6≡ 0 (mod p). A19 : (VII) gives α ≡ 0 (mod p), (XIV) gives ≡ 0, (XXII) gives j ≡ 0 (mod p), (XVIII) gives e ≡ 0 or 6≡ 0 (mod p). A20 : (VII) gives α ≡ 0, (XXII) gives j ≡ 0. Elimination between (XIV) and (XVIII) gives c − eγ ≡ 0 (mod p). A21 : (VII) gives α ≡ 0, (XIV) gives ≡ 0 or 6≡ 0 (mod p), (XVIII) gives e ≡ 0, or 6≡ 0, and (XXII) gives j ≡ 0 (mod p). A22 : (VII) gives α ≡ 0, (XIV) gives ≡ 0 or 6≡ 0, (XVIII) gives epsilon ≡ 0 or 6≡ 0, (XXII) gives j ≡ 0 (mod p). 62
A23 : (VII) gives α ≡ 0, (XIV) gives ≡ 0, (XVIII) gives ≡ 0, (XXII) gives j ≡ 0 or 6≡ 0 (mod p). A26 : (VII) α ≡ 0, (XIV) ≡ 0 or 6≡ 0, (XVIII) ≡ 0, (XXII) j ≡ 0 or 6≡ 0 (mod p). A27 : (VII) α ≡ 0, (XIV) ≡ 0 or 6≡ 0, (XVIII) ≡ 0, (XXII) j ≡ 0 or 6≡ 0 (mod p). Elimination between (XIV) and (XXII) gives g − δj ≡ 0 (mod p). A29 : (VII) α ≡ 0, (XIV) ≡ 0 or 6≡ 0, (XVIII) ≡ 0, (XXII) j ≡ 0 or 6≡ 0 (mod p).
63
End of the Project Gutenberg EBook of Groups of Order p^m Which Contain Cyclic Subgroups of Order p^(m-3), by Lewis Irving Neikirk *** END OF THE PROJECT GUTENBERG EBOOK GROUPS OF ORDER P^M *** This file should be named grdpm10p.pdf or grdpm10p.zip Corrected EDITIONS of our eBooks get a new NUMBER, grdpm11p.pdf VERSIONS based on separate sources get new LETTER, grdpm10ap.pdf Produced by Cornell University, Joshua Hutchinson, Lee Chew-Hung, John Hagerson, and the Online Distributed Proofreading Team. Project Gutenberg eBooks are often created from several printed editions, all of which are confirmed as Public Domain in the US unless a copyright notice is included. Thus, we usually do not keep eBooks in compliance with any particular paper edition. We are now trying to release all our eBooks one year in advance of the official release dates, leaving time for better editing. Please be encouraged to tell us about any error or corrections, even years after the official publication date. Please note neither this listing nor its contents are final til midnight of the last day of the month of any such announcement. The official release date of all Project Gutenberg eBooks is at Midnight, Central Time, of the last day of the stated month. A preliminary version may often be posted for suggestion, comment and editing by those who wish to do so. Most people start at our Web sites at: http://gutenberg.net or http://promo.net/pg These Web sites include award-winning information about Project Gutenberg, including how to donate, how to help produce our new eBooks, and how to subscribe to our email newsletter (free!).
Those of you who want to download any eBook before announcement can get to them as follows, and just download by date. This is also a good way to get them instantly upon announcement, as the indexes our cataloguers produce obviously take a while after an announcement goes out in the Project Gutenberg Newsletter. http://www.ibiblio.org/gutenberg/etext03 or ftp://ftp.ibiblio.org/pub/docs/books/gutenberg/etext03
Or /etext02, 01, 00, 99, 98, 97, 96, 95, 94, 93, 92, 92, 91 or 90 Just search by the first five letters of the filename you want, as it appears in our Newsletters.
Information about Project Gutenberg (one page) We produce about two million dollars for each hour we work. The time it takes us, a rather conservative estimate, is fifty hours to get any eBook selected, entered, proofread, edited, copyright searched and analyzed, the copyright letters written, etc. Our projected audience is one hundred million readers. If the value per text is nominally estimated at one dollar then we produce $2 million dollars per hour in 2002 as we release over 100 new text files per month: 1240 more eBooks in 2001 for a total of 4000+ We are already on our way to trying for 2000 more eBooks in 2002 If they reach just 1-2% of the world’s population then the total will reach over half a trillion eBooks given away by year’s end. The Goal of Project Gutenberg is to Give Away 1 Trillion eBooks! This is ten thousand titles each to one hundred million readers, which is only about 4% of the present number of computer users. Here is the briefest record of our progress (* means estimated): eBooks Year Month 1 10 100 1000 1500 2000 2500 3000 4000 6000 9000 10000
1971 1991 1994 1997 1998 1999 2000 2001 2001 2002 2003 2004
July January January August October December December November October/November December* November* January*
The Project Gutenberg Literary Archive Foundation has been created to secure a future for Project Gutenberg into the next millennium. We need your donations more than ever! As of February, 2002, contributions are being solicited from people and organizations in: Alabama, Alaska, Arkansas, Connecticut, Delaware, District of Columbia, Florida, Georgia, Hawaii, Illinois,
Indiana, Iowa, Kansas, Kentucky, Louisiana, Maine, Massachusetts, Michigan, Mississippi, Missouri, Montana, Nebraska, Nevada, New Hampshire, New Jersey, New Mexico, New York, North Carolina, Ohio, Oklahoma, Oregon, Pennsylvania, Rhode Island, South Carolina, South Dakota, Tennessee, Texas, Utah, Vermont, Virginia, Washington, West Virginia, Wisconsin, and Wyoming. We have filed in all 50 states now, but these are the only ones that have responded. As the requirements for other states are met, additions to this list will be made and fund raising will begin in the additional states. Please feel free to ask to check the status of your state. In answer to various questions we have received on this: We are constantly working on finishing the paperwork to legally request donations in all 50 states. If your state is not listed and you would like to know if we have added it since the list you have, just ask. While we cannot solicit donations from people in states where we are not yet registered, we know of no prohibition against accepting donations from donors in these states who approach us with an offer to donate. International donations are accepted, but we don’t know ANYTHING about how to make them tax-deductible, or even if they CAN be made deductible, and don’t have the staff to handle it even if there are ways. Donations by check or money order may be sent to: Project Gutenberg Literary Archive Foundation PMB 113 1739 University Ave. Oxford, MS 38655-4109 Contact us if you want to arrange for a wire transfer or payment method other than by check or money order. The Project Gutenberg Literary Archive Foundation has been approved by the US Internal Revenue Service as a 501(c)(3) organization with EIN [Employee Identification Number] 64-622154. Donations are tax-deductible to the maximum extent permitted by law. As fund-raising requirements for other states are met, additions to this list will be made and fund-raising will begin in the additional states. We need your donations more than ever!
You can get up to date donation information online at: http://www.gutenberg.net/donation.html
*** If you can’t reach Project Gutenberg, you can always email directly to: Michael S. Hart
Prof. Hart will answer or forward your message. We would prefer to send you information by email.
**The Legal Small Print**
(Three Pages) ***START**THE SMALL PRINT!**FOR PUBLIC DOMAIN EBOOKS**START*** Why is this "Small Print!" statement here? You know: lawyers. They tell us you might sue us if there is something wrong with your copy of this eBook, even if you got it for free from someone other than us, and even if what’s wrong is not our fault. So, among other things, this "Small Print!" statement disclaims most of our liability to you. It also tells you how you may distribute copies of this eBook if you want to. *BEFORE!* YOU USE OR READ THIS EBOOK By using or reading any part of this PROJECT GUTENBERG-tm eBook, you indicate that you understand, agree to and accept this "Small Print!" statement. If you do not, you can receive a refund of the money (if any) you paid for this eBook by sending a request within 30 days of receiving it to the person you got it from. If you received this eBook on a physical medium (such as a disk), you must return it with your request. ABOUT PROJECT GUTENBERG-TM EBOOKS This PROJECT GUTENBERG-tm eBook, like most PROJECT GUTENBERG-tm eBooks, is a "public domain" work distributed by Professor Michael S. Hart through the Project Gutenberg Association (the "Project"). Among other things, this means that no one owns a United States copyright on or for this work, so the Project (and you!) can copy and distribute it in the United States without permission and without paying copyright royalties. Special rules, set forth below, apply if you wish to copy and distribute this eBook under the "PROJECT GUTENBERG" trademark.
Please do not use the "PROJECT GUTENBERG" trademark to market any commercial products without permission. To create these eBooks, the Project expends considerable efforts to identify, transcribe and proofread public domain works. Despite these efforts, the Project’s eBooks and any medium they may be on may contain "Defects". Among other things, Defects may take the form of incomplete, inaccurate or corrupt data, transcription errors, a copyright or other intellectual property infringement, a defective or damaged disk or other eBook medium, a computer virus, or computer codes that damage or cannot be read by your equipment. LIMITED WARRANTY; DISCLAIMER OF DAMAGES But for the "Right of Replacement or Refund" described below, [1] Michael Hart and the Foundation (and any other party you may receive this eBook from as a PROJECT GUTENBERG-tm eBook) disclaims all liability to you for damages, costs and expenses, including legal fees, and [2] YOU HAVE NO REMEDIES FOR NEGLIGENCE OR UNDER STRICT LIABILITY, OR FOR BREACH OF WARRANTY OR CONTRACT, INCLUDING BUT NOT LIMITED TO INDIRECT, CONSEQUENTIAL, PUNITIVE OR INCIDENTAL DAMAGES, EVEN IF YOU GIVE NOTICE OF THE POSSIBILITY OF SUCH DAMAGES. If you discover a Defect in this eBook within 90 days of receiving it, you can receive a refund of the money (if any) you paid for it by sending an explanatory note within that time to the person you received it from. If you received it on a physical medium, you must return it with your note, and such person may choose to alternatively give you a replacement copy. If you received it electronically, such person may choose to alternatively give you a second opportunity to receive it electronically. THIS EBOOK IS OTHERWISE PROVIDED TO YOU "AS-IS". NO OTHER WARRANTIES OF ANY KIND, EXPRESS OR IMPLIED, ARE MADE TO YOU AS TO THE EBOOK OR ANY MEDIUM IT MAY BE ON, INCLUDING BUT NOT LIMITED TO WARRANTIES OF MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE. Some states do not allow disclaimers of implied warranties or the exclusion or limitation of consequential damages, so the above disclaimers and exclusions may not apply to you, and you may have other legal rights. INDEMNITY You will indemnify and hold Michael Hart, the Foundation, and its trustees and agents, and any volunteers associated with the production and distribution of Project Gutenberg-tm
texts harmless, from all liability, cost and expense, including legal fees, that arise directly or indirectly from any of the following that you do or cause: [1] distribution of this eBook, [2] alteration, modification, or addition to the eBook, or [3] any Defect. DISTRIBUTION UNDER "PROJECT GUTENBERG-tm" You may distribute copies of this eBook electronically, or by disk, book or any other medium if you either delete this "Small Print!" and all other references to Project Gutenberg, or: [1]
Only give exact copies of it. Among other things, this requires that you do not remove, alter or modify the eBook or this "small print!" statement. You may however, if you wish, distribute this eBook in machine readable binary, compressed, mark-up, or proprietary form, including any form resulting from conversion by word processing or hypertext software, but only so long as *EITHER*: [*]
The eBook, when displayed, is clearly readable, and does *not* contain characters other than those intended by the author of the work, although tilde (~), asterisk (*) and underline (_) characters may be used to convey punctuation intended by the author, and additional characters may be used to indicate hypertext links; OR
[*]
The eBook may be readily converted by the reader at no expense into plain ASCII, EBCDIC or equivalent form by the program that displays the eBook (as is the case, for instance, with most word processors); OR
[*]
You provide, or agree to also provide on request at no additional cost, fee or expense, a copy of the eBook in its original plain ASCII form (or in EBCDIC or other equivalent proprietary form).
[2]
Honor the eBook refund and replacement provisions of this "Small Print!" statement.
[3]
Pay a trademark license fee to the Foundation of 20% of the gross profits you derive calculated using the method you already use to calculate your applicable taxes. If you don’t derive profits, no royalty is due. Royalties are payable to "Project Gutenberg Literary Archive Foundation" the 60 days following each date you prepare (or were legally required to prepare) your annual (or equivalent
periodic) tax return. Please contact us beforehand to let us know your plans and to work out the details. WHAT IF YOU *WANT* TO SEND MONEY EVEN IF YOU DON’T HAVE TO? Project Gutenberg is dedicated to increasing the number of public domain and licensed works that can be freely distributed in machine readable form. The Project gratefully accepts contributions of money, time, public domain materials, or royalty free copyright licenses. Money should be paid to the: "Project Gutenberg Literary Archive Foundation." If you are interested in contributing scanning equipment or software or other items, please contact Michael Hart at: [email protected] [Portions of this eBook’s header and trailer may be reprinted only when distributed free of all fees. Copyright (C) 2001, 2002 by Michael S. Hart. Project Gutenberg is a TradeMark and may not be used in any sales of Project Gutenberg eBooks or other materials be they hardware or software or any other related product without express permission.] *END THE SMALL PRINT! FOR PUBLIC DOMAIN EBOOKS*Ver.02/11/02*END*
