Groups of Order p^m Which Contain Cyclic Subgroups of Order p^(m-3), doc

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Title: Groups of Order p^m Which Contain Cyclic Subgroups of Order p^(m-3)Author: Lewis Irving Neikirk

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GROUPS OF ORDER pm WHICH CONTAIN

byLEWIS IRVING NEIKIRK

sometime harrison research fellow in mathematics

1905

INTRODUCTORY NOTE.

This monograph was begun in 1902-3 Class I, Class II, Part I, and the conjugate groups of Class III, which contain all the groups with independentgenerators, formed the thesis which I presented to the Faculty of Philosophy

self-of the University self-of Pennsylvania in June, 1903, in partial fulfillment self-of therequirements for the degree of Doctor of Philosophy

The entire paper was rewritten and the other groups added while the authorwas Research Fellow in Mathematics at the University

I wish to express here my appreciation of the opportunity for scientific search afforded by the Fellowships on the George Leib Harrison Foundation atthe University of Pennsylvania

re-I also wish to express my gratitude to Professor George H Hallett for hiskind assistance and advice in the preparation of this paper, and especially toexpress my indebtedness to Professor Edwin S Crawley for his support andencouragement, without which this paper would have been impossible

Lewis I Neikirk

University Of Pennsylvania, May, 1905.

GROUPS OF ORDER pm, WHICH CONTAIN CYCLIC

bylewis irving neikirkIntroduction

The groups of order pm, which contain self-conjugate cyclic subgroups oforders pm−1, and pm−2

respectively, have been determined by Burnside,2andthe number of groups of order pm, which contain cyclic non-self-conjugate sub-groups of order pm−2

has been given by Miller.3Although in the present state of the theory, the actual tabulation of allgroups 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 oddprime) which contain cyclic subgroups of order pm−3 and none of higher orderare determined The method of treatment used is entirely abstract in characterand, in virtue of its nature, it is possible in each case to give explicitly thegenerational 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 oforder 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 iscontained 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:

I There is in G at least one operator Q1, such that Qp12 is not contained in{P }

II The p2 power of every operator in G is contained in {P }, and there is atleast 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.

2 Theory of Groups of a Finite Order, pp 75-81.

3 Transactions, vol 2 (1901), p 259, and vol 3 (1902), p 383.

The number of groups for Class I, Class II, and Class III, together with thetotal number, are given in the table below:

We have the relation

There is in G, a subgroup H1of order pm−2, which contains {P }

self-conjugate-ly.5 The subgroup H1is generated by P and some operator Qy1Pxof G; it thencontains Qy1 and is therefore generated by P and Qp12; 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

Q−p1 2P Qp12= P1+kpm−4,(2)

Q−p1 P Qp1= Qβp1 2Pα1,(3)

Q−11 P Q1= Qbp1 Pa1.(4)

4 With J W Young, On a certain group of isomorphisms, American Journal of matics, vol 25 (1903), p 206.

Mathe-5

Burnside: Theory of Groups, Art 54, p 64.

6 Ibid., Art 56, p 66.

2 Determination of H1 Derivation of a formula for [yp , x] —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,

(5) [yp2, x]s=syp2, x{s + k s2
ypm−4}

3 Determination of H2 Derivation of a formula for [yp, x]s.—It followsfrom (3) and (5) that

α1= 1 + α2p2,the congruence

(1 + α2p2)p− 1

α2p3 (α2+ hβ)p3≡ kpm−4 (mod pm−3);

and so

(α2+ hβ)p3≡ 0 (mod pm−4),since

(1 + α2p2)p− 1

α2p3 ≡ 1 (mod p2)

From the last congruences

(α2+ hβ)p3≡ kpm−4 (mod pm−3)

(6)

Equation (3) is now replaced by

Q−p1 P Q−p1 = Qβp1 2P1+α2 p2.(7)

From (7), (5), and (6)

[−yp, x, yp] =βxyp2, x{1 + α2yp2} + βk x

2
ypm−4

A continued use of this equation gives

(8) [yp, x]s= [syp + β s2
xyp2,

and from (1) and (2)

These defining relations may be presented in simpler form by a suitablechoice of the second generator Q1 From (9), (6), (8) and (10)

[1, x]p3 = [p3, xp3] = [0, (x + h)p3] (m > 6),and, if x be so chosen that

x + h ≡ 0 (mod pm−6),

Q1Px is an operator of order p3 whose p2 power is not contained in {P } Let

Q1Px= Q The group G is generated by Q and P , where

corre-[−y, 1, y] = [byp, 1 + aypm−6] (m > 8)

It is important to notice that by placing y = p and p2 in the precedingequation we find that8

b ≡ β (mod p), a ≡ α ≡ k (mod p3) (m > 7)

A combination of the last equation with (8) yields9

(12) [−y, x, y] = [bxyp + b2 x2
yp2,

x(1 + aypm−6) + ab x2
ypm−5+ ab2 x3
ypm−4] (m > 8)

7 For m = 8 it is necessary to add a 2 y

2
p 4 to the exponent of P and for m = 7 the terms a(a +abp2 ) y2
p 2 + a 3 y

3
p 3 to the exponent of P , and the term ab y2
p 2 to the exponent of Q The extra term 27ab 2 k y3
is to be added to the exponent of P for m = 7 and p = 3.

(m > 8)

5 Transformation of the Groups.—The general group G of Class I is ified, 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

spec-a = spec-a1pλ, b = b1pµ,where

where

Q01= Q0y0P0x0pm−6, and P10 = Q0yP0x,

10 For m = 8 it is necessary to add the term 1axy s2
[ 1 y(2s − 1) − 1]p 4 to the exponent of

P , and for m = 7 the terms

3s − 1

3 +a

2 b 2

s(s − 1) 2 (s − 4)

s

3

yp 3 o with the extra terms

27abxy

n bk 3!

 s

2
y 2 − (2s − 1)y + 2  s

3
+ x(b 2 k + a2)(2y2+ 1) s3
o

, for p = 3, to the exponent of P , and the termsab2s − 1 y − 1 s
xyp 2 to the exponent of Q.

with y and x prime to p.

Since

Q−1P Q = QbpP1+apm−6,then

Q0−11 P10Q01= Q0bp1 P01+ap1 m−6,

or in terms of Q0, and P0

y + b0xy0p + b02 x2
y0p2, x(1 + a0y0pm−6) + a0b0 x2
y0pm−5

+ a0b02 x3
y0pm−4 = [y + by0p, x + (ax + bx0p)pm−6] (m > 8)and

by0 ≡ b0xy0+ b02 x2
y0p (mod p2),(14)

ax + bx0p ≡ a0y0x + a0b0 x2
y0p + a0b02 x3
y0p2 (mod p3)

(15)

The necessary and sufficient condition for the simple isomorphism of these twogroups G(a, b) and G(a0, b0) is, that the above congruences shall be consistentand admit of solution for x, y, x0 and y0 The congruences may be written

b1pµ≡ b01xpµ0+ b021 x2
p2µ0+1 (mod p2),

a1xpλ+ b1x0pµ+1≡

y0{a01xpλ0+ a01b01 x2
pλ0+µ0+1+ a01b021 x3
pλ0+2µ0+2} (mod p3).Since dv[x, p] = 1 the first congruence gives µ = µ0 and x may always be sochosen that b1= 1

We may choose y0 in the second congruence so that λ = λ0and a1= 1 exceptfor the cases λ0≥ µ + 1 = µ0+ 1 when we will so choose x0 that λ = 3

The type groups of Class I for m > 811are then given by

(I) G(pλ, pµ) : Q−1P Q = Qp1+µP1+pm−6+λ, Qp3 = 1, Ppm−3 = 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 group {P } self-conjugate, while the group G(p3, p2) is the abelian group oftype (m − 3, 3)

sub-11 For m = 8 the additional term ayp appears on the left side of the congruence (14) and G(1, p 2 ) 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, p 2 ) become simply isomorphic, also G(p, p) and G(p, p 2 ).

2 Groups with independent generators.

Consider the first possibility in the above paragraph There is in H2, a group H1of order pm−2, which contains {P } self-conjugately.12 H1is generated

sub-by Qp1 and P H2contains H1 self-conjugately and is itself self-conjugate in G.From these considerations13

Q−p1 P Qp1 = P1+kpm−4,(2)

Q−11 P Q = Qβp1 Pα1.(3)

[−p, 1, p] = αp1− 1

α1− 1βp, α

p 1



1 + βk2

Equation (3) now becomes

2αk

 13!s(s − 1)(2s − 1)y

Q1Pxwill be the required Q of order p2

Placing h = 0 in congruence (5) we find

α2p2≡ kpm−4 (mod pm−3)

Let α2= αpm−6 H2 is then generated by

Qp2 = 1, Ppm−3 = 1

Two of the preceding formulæ now become

[−y, x, y] =βxyp, x(1 + αypm−5) + βk x2
ypm−4,

(8)

[y, x]s= [sy + Usp, xs + Wspm−5],(9)

where

Us= β s2
xyand14

R−11 Q R1= Qd1Pc1 p m−5

.(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)

Rp1Q Rp1= P−µpQ Pµp = Q P−aµpm−4.From these equations we obtain

dp1 ≡ 1 (mod p) and d1≡ 1 (mod p)

Let d1= 1 + dp Equation (12) is replaced by

and from the last two equations

ap1≡ 1 (mod pm−5)and

a1≡ 1 (mod pm−6) (m > 6); a1≡ 1 (mod p) (m = 6).Placing a1= 1 + a2pm−6 (m > 6); a1= 1 + a2p (m = 6)

K ≡ 0 (mod p),and16

The preceding relations will be simplified by taking for R1an operator of order

p This will be effected by two transformations

and if y be so chosen that

λ + y ≡ 0 (mod p),

R2= R1Qy is an operator such that Rp2 is in {P }

Let

Rp2 = Plp.Using R2 in the place of R1, from (15), (9) and (14)

[1, 0, x]p=hp, 0, xp +ax

2 p

m−4i=h0, 0, (x + l)p +ax

2 pm−4i,

16 K has an extra term for m = 6 and p = 3, which reduces to 3b 1 c 1 This does not affect the reasoning except for c 1 = 2 In this case change P2to P and c 1 becomes 1.

17 The extra terms appearing in the exponent of P for m = 6 do not alter the result.

and if x be so chosen that

x + l +ax

2 pm−5≡ 0 (mod pm−4),

then R = R2Pxis the required operator of order p

Rp= 1 is permutable with both Q and P Preceding equations now assumethe final forms

Q−1P Q = QβpP1+apm−5,(15)

R−1P R = QbpP1+apm−4,(16)

R−1Q R = Q1+dpPcpm−4,(17)

with Rp= 1, Qp2= 1, Ppm−3 = 1

The following derived equations are necessary18

[0, −y, x, 0, y] =0, βxyp, x(1 + αypm−5) + αβ x2
ypm−4,

5 Transformation of the groups All groups of this section are given byequations (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 thecorrespondence is given by

R01, Q01, P10

,

in which

R01= R0z00Q0y00pP0x00pm−4,

Q01= R0z0Q0y0P0x0pm−5,

P10 = R0zQ0yP0x,

18 For m = 6 the term a 2 x

2
xp 2 must be added to the exponent of P in (18).

19 When m = 6 the following terms are to be added to V s : a22x

n

s(s−1)(2s−1) 3! y2− s2
yop.

where x, y and z are prime to p.

The operators R10, Q01, and P10 must be independent since R, Q, and P are,and that this is true is easily verified The lowest power of Q0

1in {P0

1} is Q0p2

1 = 1and the lowest power of R01 in {Q01, P10} is R0p

1= 1 Let Q0s10 = P0sp1m−5.This in terms of R0, Q0, and P0 is

h

s0z0, y0s0+ d0 s20
z0p , s0x0pm−5+ c0 s20
y0z0pm−4i= [0, 0, sxpm−5].From this equation s0 is determined by

s0z0 ≡ 0 (mod p)

y0{s0+ d0 s2
z0p} ≡ 0 (mod p2),which give

s0y0≡ 0 (mod p2)

Since y0 is prime to p

s0≡ 0 (mod p2)and the lowest power of Q01 contained in {P10} is Q0p2

[z, y + θ1p, x + φ1pm−5] = [z, y + βy0p, x(1 + αpm−5) + βxpm−4],(22)

[z, y + θ2p, x + φ2pm−4] = [z, y + by0p, x(1 + apm−4) + bx0pm−4],(23)

[z0, y0+ θ3p, (x0+ φ3p)pm−5] = [z0, y0(1 + dp), x(1 + dp)pm−5+ cxpm−4],(24)

φ1≡ αx + βx0p (mod p2),(II)

θ2≡ by0 (mod p),(III)

φ2≡ ax + bx0 (mod p),(IV)

θ3≡ dy0 (mod p),(V)

G0 is then taken from the simplest of the remaining cases and we proceed asabove until all the cases are exhausted

Let κ = κ1pκ2, and dv1[κ1, p] = 1 (κ = a, b, α, β, c, and d)

The six sets are given in the table below

The groups marked (*) divide into two or three parts.

Let ad − bc = θ1pθ2, α1d − βc = φ1pφ2 and α1b − aβ = χ1pχ2 with θ1, φ1,and χ1 prime to p

6 Types.

The type groups are given by equations (15), (16) and (17) with the values

of the constants given in Table IV

b0xz0 ≡ βy0 (mod p),(I)

α0y0 ≡ α (mod p),(II)

b0xz00≡ by0 (mod p),(III)

α0xy00+ α0b0 x2
z00+ c0yz00≡ ax + bx0 (mod p),(IV)

d ≡ 0 (mod p),(V)

c0y0z00≡ cx (mod p)

(VI)

Since z0 is unrestricted (I) gives β ≡ 0 or 6≡ 0 (mod p)

From (II) since y06≡ 0, α 6≡ 0 (mod p)

From (III) since x, y0, z006≡ 0, b 6≡ 0 (mod p)

In (IV) b 6≡ 0 and x0 is contained in this congruence alone, and, therefore, amay be taken ≡ 0 or 6≡ 0 (mod p)

(V) gives d ≡ 0 (mod p) and (VI), c 6≡ 0 (mod p)

Elimination of y0 between (III) and (VI) gives

b0c0z002≡ bc (mod p)

so that bc is a quadratic residue or non-residue (mod p) according as bc is aresidue or non-residue.

The types are given by placing a = 0, b = 1, α = 1, β = 0, c = κ, and d = 0where κ has the two values, 1 and a representative non-residue of p

C2.The congruences for this case are

d0(yz0− y0z) ≡ βy0 (mod p),(I)

α01xy0+ a0xz0≡ α1x + βx0 (mod p),(II)

d0yz00≡ by0 (mod p),(III)

a0xz00≡ ax + bx0 (mod p),(IV)

d0z00≡ d (mod p),(V)

cx + dx0≡ 0 (mod p)

(VI)

Since z appears in (I) alone, β can be either ≡ 0 or 6≡ 0 (mod p) (II)

is linear in z0 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 z00 between (IV), (V), and (VI) gives

a0d2≡ d0(ad − bc) (mod p)

Since z00 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

d0(yz0− y0z) ≡ βy0 (mod p),(I)

α1x + βx0 ≡ 0 (mod p),(II)

d0yz00≡ by0 (mod p),(III)

ax + bx0 ≡ 0 (mod p),(IV)

d0z00≡ d (mod p),(V)

cx + dx0 ≡ 0 (mod p)

(VI)

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 α1d − βc ≡ 0 (mod p), andbetween (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

Section 2.

1 Groups with dependent generators In this section, G is generated by Q1and P where

There is in G, a subgroup H1, of order pm−2, which contains {P }

self-conjugate-ly.20 H1 either contains, or does not contain Qp1 We will consider the secondpossibility 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

and

[y, x]s=hsy, xs + k s

2
ypm−4 i(5)

Placing s = p and y = 1 in (5) we have, from (2)

[R1Px]p= Rp1Pxp= P(l+x)p.Choosing x so that

of order pm−1 which contains H1 self-conjugately H2 is generated by H1 and

some operator [z, y, x] of G Q1is then in H2and H2is the subgroup {Q1, H1}.Hence,

Q−p1 P Qp1= RβPα1,(6)

Q−p1 P Qp1 = Rb1Papm−4.(7)

To determine α1 and β we find from (6), (5) and (7)

α1− b1

p

1− bp1(α1− b1)2

o

pm−4



By (1)

Q−p1 2P Qp12 = P,and, therefore,

By (1) and (4)

Q−p1 2R Qp12= P−lp2R Plp2 = R,and, hence,

bp1 ≡ 1 (mod p), ab

p

1− 1

b1− 1 ≡ 0 (mod p),therefore b1= 1

Substituting b1= 1 and α1= 1 + α2pm−5in the congruence determining α1

we obtain (1 + α2pm−5)p≡ 1 (mod pm−3), which gives α2≡ 0 (mod p).Let α2= αp and equations (8) and (7) are now replaced by

Qp1P Qp1= RβP1+αpm−4,(9)

Q−p1 R Qp1= RPapm−4.(10)

From these we derive

[−yp, 0, x, yp] =h0, βxy, x +αxy + aβx y

2
+ βk x

2
y pm−4i,(11)

[−yp, x, 0, yp] = [0, x, axypm−4]

2aβ

n13!s(s − 1)(2s − 1)z

2− 2s
zo

4 Determination of G

Since H2is self-conjugate in G1 we have

Q−11 P Q1= Qγp1 RδP1,(14)

Q−11 R Q1= Qcp1 RdPepm−4.(15)

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) (m > 5).Let 1= 1 + 2p and equation (14) is replaced by

1acdd

n(dn− 1)

(1 + 2p)p− 1

2p2 ≡ 1 (mod p), (2+ γh)p2≡ 0 (mod pm−4)and

[−y, 0, x, y] =hxγy + cδ y

2
p, δxy, x(1 + 2yp) + θpm−4i(20)

where

θ =neδx + aδγx + 2



α +aδ2



xo y2
+1

2ac

n13!y(y − 1)(2y − 1)δ

2− y

2
δo+αγy + δky + aδxy2+ (acδ2y + acδ) y2
x

2

Equations (19) and (20) are replaced by

[−y, x, 0, y] = [cxyp, x, exypm−4]

(22)

[−y, 0, x, y] =hγy + cδ y

2
xp, δxy, x(1 + ypm−5) + θpm−4i(23)

2δex + αcδ x

2

 13!s(s − 1)(2s − 1)z

2

− 2s
z p

5 Transformations of the groups Placing y = 1 and x = −1 in (22) weobtain (17) in the form

R−1Q R = Q1−cpP−epm−4

A comparison of the generational equations of the present section with those ofSection 1, shows that groups, in which δ ≡ 0 (mod p), are simply isomorphicwith 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

Since Rp= 1, Qp2= 1, and Ppm−3 = 1, R0p1= 1, Q0p12 = 1 and P0p1m−3

The forms of these operators are then

P10 = Q0zR0yP0x,

R10 = Q0z0pR0y0P0x0pm−4,

Q01= Q0z00R0y00P0x00pm−5,where dv[x, p] = 1

Since R is not contained in {P }, and Qp is not contained in {R, P } R01 isnot contained in {P10}, and Q0p

[s0z0p, s0y0, s0x0pm−4] = [0, 0, sxpm−4],and

s0y0≡ 0 (mod p), s0z0≡ 0 (mod p)

Either y or z is prime to p or s may be taken = 1.

Θ01≡ kx (mod p),(II)

Φ02≡ Φ2 (mod p),(III)

δ0xz00≡ δy0 (mod p),(IV)

Θ02≡ Θ2 (mod p2),(V)

Φ03≡ cz00 (mod p),(VI)

Θ03≡ Θ3 (mod p)

(VII)

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 nineindeterminants with x, y0, and z00prime to p

A6 divides into two parts

The groups of A6in which δk + γ ≡ 0 (mod p) are simply isomorphic withthe groups of A1and those in which δk + γ 6≡ 0 (mod p) are simply isomorphicwith the groups of A2 The types are given by equations (25), (26) and (27)where the constants have the values given in Table III

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

0xz0 ≡ kx (mod p),(II)

γz00+ δz0 ≡ 0 (mod p),(III)

δ0xz00≡ δy0 (mod p),(IV)

0xz00≡ x (mod p),(V)

cz00≡ 0 (mod p),(VI)

ex ≡ 0 (mod p)

(VII)

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, z0and z00between (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

0xz0+ k0xy0≡ kx (mod p),(II)

γx00+ δz0≡ 0 (mod p),(III)

δ0xz00≡ δy0 (mod p),(IV)

0xz00≡ x (mod p),(V)

cz00≡ 0 (mod p),(VI)

ex ≡ 0 (mod p)

(VII)

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, z0, and z00 between (II), (III) and(V) gives

δk + γ ≡ k0δy0 (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

0xz0− e0y0z ≡ kx (mod p),(II)

γz00+ δz0≡ 0 (mod p),(III)

δ0xz00≡ δy0 (mod p),(IV)

0xz00≡ ex (mod p),(V)

cz00≡ 0 (mod p),(VI)

−e0y0z ≡ kx (mod p),(II)

γz00+ δz0 ≡ 0 (mod p),(III)

δ0xz00≡ δy0 (mod p),(IV)

01xz00+ δ0e0x z200
+ e0yz00≡ e1x + γx00+ δx0 (mod p),

(V)

cz00≡ 0 (mod p),(VI)

e0y0z00≡ ex (mod p)

(VII)

(II), and (III) being linear in z and z0 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

Elimination of x and y from (IV) and (VII) gives

Qp1 is contained in the subgroup H1of order pm−2, H1= {Qp1, P }

2 Determination of H1 Since {P } is self-conjugate in H1

R−11 QpR1= Qbp1 Pα1 p.(6)

Using the symbol [a, b, c, d, e, f, · · · ] to denote RaQb

1PcRdQePf· · · , wehave from (5), (6) and (3)

22

Burnside, Theory of Groups, Art 54, p 64.

and by (4)

αp1≡ 1 (mod p), and α1≡ 1 (mod p)

Let α1= 1 + α2p and (5) is now replaced by

a1p2(1 + U0p) = a1{1 + (α2+ βh)p}p− 1

(α2+ βh)p2 p2from which

a1p2(1 + U0p) ≡ 0 (mod pm−3)or

N = p and M = βh (1 + α2p)p− 1

α2p2 − 1

,

from which

(1 + α2p)p+(1 + α2p)

p− 1

α2p2 βhp2≡ 1 (mod pm−3)

(1 + α2p)p− 1

α2p2 {α2+ βh}p2≡ 0 (mod pm−3)and since

2aβ

h13!s(s − 1)(2s − 1)z

2− s

2
zixopm−5.Placing in this y = 0, z = 1 and s = p,23

(R1Px)p= Rp1Pxp= P(x+l)p,determine x so that

x + l ≡ 0 (mod pm−4),then R = R1Pxis 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

Q−11 P Q1= RγQδp1 P1,(14)

Q−11 R Q1= RcQdp1 Pe1 p

.(15)

23 Terms of the form (Ax2+ Bx)pm−4in the exponent of P for p = 3 and m > 5 do not alter the result.

M = γdp − 1

(1 + 2p)p− 1

2p2and

(2+ δh)p2

 [1 + (2+ δh)p]p− 1(2+ δh)p − p



By (1)

(1 + 2p)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 + 2p)

p− 1

2p2 + N − e1γp − 1

2 .Since

ψ = 1 (mod p)

From (18), (20), (13), (16) and (21)

[0, −y, x, 0, y] = [γxy, θ1p, x + φ1p],(22)

[0, −y, 0, x, y] = [x, dxyp, φ2p],(23)

2ad

 13!y(y − 1)(2y − 1)γ

2aβ

 13!x(x − 1)(2x − 1)γ

Q−p1 R Qp1= R,and by (19)

a ≡ 0 (mod p)

A continued multiplication, with (11), (22), and (23), gives

(Q1Px)p2 = Qp12Pxp2 = P(x+l)p2