The generating set for some maximal soluble subgroups of GL(4,Pk)

(1)

The generating set for some maximal soluble

subgroups of GL(4,P

k

)

Behnam Razzaghmaneshi

Assistant professor of Department of Mathematics Talesh Branch, Islamic Azad University, Talesh, Iran

Corresponding author:

Behnam Razzaghmaneshi

ABSTRACT: In this paper we show that the number of conjugacy classes of irreducible soluble GL(4,3) subgroup of is 108, whose :65 consist of imprimitive groups and 43 of primitive groups.

Keywords: generating set, soluable, maximal

INTRODUCTION

In determining all permutation groups of a deferent degree The first work is probably due to Ruffini(1799)

,

who gives The possible orders of the groups of degree 5.This them is taken up-again By Cauchy (1845)

,

who determines the possible orders of the groups of degree Up to 6. Mathieu(1858)continus cauchy’s work

,

determining the Possible Orders of the groups of degrees 7 and 8. Miller(1896 a)Suggests that Ruffini’s work is in Complate

,

and citesan omission in Cauchy’s list for degree 6 .Serret(1850)determines all subgroups of

s

4_and

s

5_{.Miller(1896 a)says that}

thise work is correct.

kirkman(1862–3)Lists the transitive groups of degrees 3 to 10.Miller(1896 a) Says that This list is correct up to degree

7

,

has missing six groups of Degree 8

,

another six of degree 9

,

and is highly inaccurate for degree 10.

Jordan(1871a)gives a table containing the numbers of conjugacy classes of primitive maximal soluble groups of

degree lees then

10

6. He claims there are five such classes of degree 81

,

but there are only four

,

(see to Chapter

6).This error is Likely to lead to errors for larger degrees.Also

,

the Second And third entries in the last row of this table should be Swapped.In The same paper

,

Jordan also gives a table containing the numbers of conjugacy classes of transitive maximale soluble groups of degree Up to 10000.(This is an a stounding achievment if for no other reason Than the amount of counting required to prepare it.for example

,

if

p

is a prime greater than 3

,

Jordan claims that the number of conjugacy classes of transitive maximal soluble groups of degree

p

2 6

3

2

_{is 8306).Again}

,

the error in the first table is Likely to errors in this on too.Jordan (1872)caunts all the primitive groups of degrees 4 to 17.His count Matches that of sims except that Jordan has one les for degrees 9

,

12 and 15

,

eight less for degree 16 and two less For degree 17.These errors are Pointed out by Miller(1894

,

4b

,

1895c

,

1897a

,

1897b and 1900a).Jordan(1874)states that every transitive group of degree 19 is either alternating

,

symetric or affine.This agrees With sim’s list.Veronese(1885)determines all groups of degree up to 6.

Miller(1983)cites several errors for degree 6.Ask with(1890) and Cayley(1891) determine the groups of degree

6 and get the same answers.Their count of the intransitive ones is correct but undercount the transitive groups by three

,

as cole(1893a) points out

,

as kirkman(1862–3) had already (success fully

,

according to Miller)determined the transitinves

,

degree 6 was then complete.Askwith(1890a) and Cayley(1891) also arrive at the same determination of the groups of degree 7.They undercount both the transitive and intransitives by one

,

as cole(1893a) Points out.With kirkman’s transitives and col’s additional intransitive

,

Degree 7 was then complete .

2. Periliminares:

The first we determine the JS-imprimitives of

(

4 ,

)

k

p

GL

_.

Recall that the number JS-maximals of

(

2 ,

)

k

p

(2)

1200

(

2 ,

)

:

(

)

(

2 ,

2 ),

1 (mod

4 ).

),

4 (mod

3 ),

2 ,

2 (

)

(

:

)

,

2 (

,

:

)

,

2 (

,

2 ,

)

,

1 (

:

)

,

2 (

8 1 4 8 1 3 2 1 2 2 1 2

















    k p k k p k p k k k k

p

Sp

N

Q

C

p

M

p

O

N

Q

C

p

M

C

p

M

p

S

wr

p

GL

p

M

k k k



Therefore the Js-imprimitives of

(

4 ,

)

k

p

GL

_{are listed below.}

).

4 (mod

1 ,

)

,

2 (

:

)

,

4 (

),

4 (mod

3 ,

)

,

2 (

:

)

,

4 (

,

)

,

2 (

:

)

,

4 (

,

2 ,

)

,

1 (

:

)

,

4 (

2 4 4 2 3 3 2 2 2 4 1













k k k k k k k k k k k

p

S

wr

p

M

p

M

p

S

wr

p

M

p

M

S

wr

p

M

p

M

p

S

wr

p

GL

p

M

Now we listed the JS-primitives of

(

4 ,

)

k

p

GL

_.

6.1-The JS-primitives of

(

4 ,

)

k

p

GL

_.

There is one JS-primitive of

(

4 ,

)

k

p

GL

_{whose unique maximal abelian normal subgroup has order}

p

4k



1 ,

namely.

M

5

(

4 ,

p

)

:

C

_P4K 1

C

4

,

k

_





the normaliser of a singer cycle.

There is one JS-primitive of

(

4 ,

)

k

p

GL

_{whose unique maximal abelian normal subgroup has order}

p

2k



1 ,

namely

(

4 ,

)

:

(

2 ,

)

2

,

2 .

2

4

6

p



M

p

C

p



M

k k



We think the filed of k

p

2 _{elements here as the filed of}₂_by₂_{matrices that is the linear span of our fixed singer cycle} of

(

2 ,

)

k

p

GL

_.

Now we come to the JS-primitives of

GL

(4

,

k

p

_{) whose unique maximal abelian normal subgroup is the scalar}

group.Let M be such a group. There are three cases examine.

6.1.1-Cas1:

Assume that k

p



3 (mod

4 )

_{and that the Fitting subgroup} F _of M _is

C

k ₁

D

8

D

8

.

p 



_Then

F

M

is

isomorphic to a completely reducible maximal soluble subgroup of

(

4 ,

2 )



O

_{that dosenot fix any non-zero isotropic}

subspace of the natural module for

(

4 ,

2 ).



O

_However

,

O



(

4 ,

2 )

_{is isomorphic to}

S

3

wr

S

2

,

_{so inparticular it is}

soluble. Therefore

F

M

).

2 ,

4 (





O

_So _in _this _case _we _find _exactly _one _JS_-primitive

,

).

2 ,

4 (

)

(

:

)

,

4 (

₁ ₈ ₈

7

 







C

D

O

p

M

_pk

(3)

1201

6.1.2-Case2:

Assume that



3 (mod

4 )

k

p

_{and that the Fitting subgroup}

F

_of

M

_is

C

k ₁

D

8

Q

8

.

p 



_Then

F

M

is

isomorphic to a completely reducible maximal soluble subgroup of

(

4 ,

2 )



O

_{that dose not fix any non-zero isotropic}

subspace of the natural module for

(

4 ,

2 )



O

_.

Threre is a unique

(

4 ,

2 )



O

_{-conjugacy class of such groups and that each group in this class is isomorphic to}

Hol

(

C

5_).Therefore

F

M

=

Hol

(

C

5

)

_.So _in _this _case _we _find _exactly _on _JS_-primitive

,

).

(

)

(

:

)

,

4 (

₁ ₈ ₈ ₅

8

p

C

D

Q

Hol

C

M

_pk

k









6.1.3-Case3:

Assume that



1 (mod

4 ).

k

p

_{Therefore we can write the Fitting subgroups}

F

_of

M

_as

C

k ₁

D

8

D

8

.

p 



Then

F

M

is isomorphic to a completely reducible maximal soluble subgroup of

Sp

(

4 ,

2 )

that dosenot fix any non-zero isotropic subgroup of the natural module for

Sp

(

4 ,

2 )

.

Every such subgroup is

Sp

(

4 ,

2 )

-conjugacy to either

O

(

4 ,

2 )

o r

Ho l

(

C

5

).



So in this case we find exactly two JS -primitives

).

(

)

(

:

)

,

4 (

),

2 ,

4 (

)

(

:

)

,

4 (

5 8

8 1 1 0

8 8 1 9

C

Ho l

D

C

p

M

O

D

C

p

M

k k

p k















 

Although 9

(

4 ,

)

k

p

M

and 1 0

(

4 ,

)

k

p

M

admit the same description as 7

(

4 ,

)

k

p

M

and 8

(

4 ,

),

k

p

M

respectively

,

they arise in different way and so are given different numbers. For example the JS-maximals of

GL

(

4 ,

3 )

are

.

,

2 3 5 6 8

1

M

a n d

M

Of these

M

3

,

M

5

,

M

7

a n d

M

8 are maximal soluble

,

while

M

1

a n d

M

6 are conjugate to subgroups of

M

7

,

a n d

M

2 is a conjugate to a subgroup of

M

3

.

_{The following table show the}_JS

-maximals of

(

4 ,

),

k

p

GL

_for

p

k₌₁

,

_…

,

(4)

1202

Table 1. The JS-maximals of

(

4 ,

),

k

p

GL

_for

p

k



1 ,...

30

_.

10

M

9

M

6

M

5

M

4

M

2

M

1

M

)

1 ,

4 (

GL

5

M

2

M

)

2 ,

4 (

GL

8

M

7

M

6

M

5

M

3

M

2

M

1

M

)

3 ,

4 (

GL

5

M

2

M

1

M

)

4 ,

4 (

GL

10

M

9

M

6

M

5

M

4

M

2

M

1

M

)

5 ,

4 (

GL

5

M

2

M

1

M

)

6 ,

4 (

GL

8

M

7

M

6

M

5

M

3

M

2

M

1

M

)

7 ,

4 (

GL

5

M

2

M

1

M

)

8 ,

4 (

GL

10

M

9

M

6

M

5

M

4

M

2

M

1

M

)

9 ,

4 (

GL

5

M

2

M

1

M

)

10 ,

4 (

GL

8

M

7

M

6

M

5

M

3

M

2

M

1

M

)

11 ,

4 (

GL

5

M

2

M

1

M

)

12 ,

4 (

GL

10

M

9

M

6

M

5

M

4

M

2

M

1

M

)

13 ,

4 (

GL

5

M

2

M

1

M

)

14 ,

4 (

GL

8

M

7

M

6

M

5

M

3

M

2

M

1

M

)

15 ,

4 (

GL

5

M

2

M

1

M

)

16 ,

4 (

GL

10

M

9

M

6

M

5

M

4

M

2

M

1

M

)

17 ,

4 (

GL

5

M

2

M

1

M

)

18 ,

4 (

GL

8

M

6

M

5

M

3

M

2

M

1

M

)

19 ,

4 (

GL

7

M

5

M

2

M

1

M

)

20 ,

4 (

GL

10

M

9

M

6

M

5

M

4

M

2

M

1

M

)

21 ,

4 (

GL

5

M

2

M

1

M

)

22 ,

4 (

GL

8

M

7

M

6

M

5

M

3

M

2

M

1

M

)

23 ,

4 (

GL

5

M

2

M

1

M

)

24 ,

4 (

GL

10

M

9

M

6

M

5

M

4

M

2

M

1

M

)

25 ,

4 (

GL

5

M

2

M

1

M

)

26 ,

4 (

GL

8

M

7

M

6

M

5

M

3

M

2

M

1

M

)

27 ,

4 (

GL

5

M

2

M

1

M

)

28 ,

4 (

GL

10

M

9

M

6

M

5

M

4

M

2

M

1

M

)

29 ,

4 (

GL

5

M

2

M

1

M

)

30 ,

4 (

GL

(5)

1203

6.2-The imprimitive soluble subgroups of

GL

(

4 ,

2 )

.

By above table there is one JS-imprimitives of

GL

(

4 ,

2 )

,

namely

M

2

:



GL

(

2 ,

2 )

wr

S

2_or

.

)

2 ,

2 (

:

₂ ₂

2

M

wr

S

M



A polycyclic presentation for

M

2_is:

,

1 ,

,

{

2



a

e

d

c

b

a

,

1 }.

,

1 ,

,

1 ,

,

1 ,

3 2 2 3

2 2



e

c

e

d

b

d

c

e

c

b

d

b

d c b a

c b

a b a a

6.2.1-Theorem:

A complete and irredundant list of

GL

(

4 ,

2 )

-jugacy class representatives of the irreducible subgroups of

M

2_is:



a

,

b

,

c

,

d

,

e



,



ab

,

c

,

e



,



a

,

bd

,

c

,

e



,



a

,

c

,

e



.

proof:

See short (1992

,

7.2

,

p.85).

By use from above table

,

obviously that the JS-imprimitives of

GL

(

4 ,

3 )

are

M

1

(

4 ,

3 ),

M

2

(

4 ,

3 )

_and

M

3

(

4 ,

3 ).

There are exactly 18

GL

(

4 ,

3 )

-cojugacy classes of irreducible soluble subgroups whose gurdion is

M

1،₃₃

GL

(

4 ,

3 )

-conjugacy classes of irreducible soluble subgroups whose guardian is

M

2_and₁₄

GL

(

4 ,

3 )

_{-conjugacy classes of}

irreducible soluble subgroups whose gardian is

M

3_{.Therefore there are 65}

GL

(

4 ,

3 )

_{-conjagacy classes of}

imprimitives soluble subgroups. We have picked exactly one representative from each of these classes. The following table details how many groups of each order there are in this set of representatives.(see short(1992

,

7.3

;

p.86-93)).

Table 1. The imprimitive soluble subgroups of

GL

(

4 ,

3 )

.

Order 16 32 48 64 96 128 192 256 384 512 768 1152 2304 4608 Number 5 12 2 12 5 10 4 6 3 1 1 1 2 1

The JS-primitives of

GL

(

4 ,

)

k

p

,

for



3

k

p

_are

M

5

,

M

6

,

M

7

and

M

8

and for

p

2 is

M

5

.

k



Now we determine the primitive subgroups of

M

5

,

M

6

,

M

7

and

M

8.

6.4-The primitive subgroups of

M

5

(

4 ,

3 )

_.

Recall that

M

5

(

4 ,

p

)

C

p4k 1

C

4

k

_





_,

the normaliser of a singer cycle. A polycyclic presentation for

M

5_{is :}

,

1 |

,

4





a

b

a

,

1 .

1

4





_pk _p k

a

(6)

1204

6.4.1-Theorem:

A complete and irredundant list of

GL

(

4 ,

2 )

-conjugacy class representatives of the primitive subgroups of

)

2 ,

4 (

5

M

_is:

,

.

3 3

2 3

2

























a

b

a

b

a

b

a

b

proof:

See short (1997

,

8.5

,

pp.95-100).

Thus the six groups listed in the theorem are pairwise non-isomorphic

,

and so no two of them can be conjugate in

)

2 ,

4 (

GL

_.

Therefore there are 10

GL

(

4 ,

3 )

-conjugacy classes of irreducible soluble subgroups:4 consist of imprimitive groups and 6 of primitives ones.

6.4.2-Theorem(short

,

1992

,

8.5

,

pp.100):

There are exactly 21

GL

(

4 ,

3 )

-conjugacy classes of primitive soluble subgroups that have

M

5

(

4 ,

3 )

_{as their}

guardian.

Thus by use above theorem we have obtained a complete and irredundant set of representatives of these classes.

6.5-A generating set for 6

(

4 ,

)

k

p

M

.

Recall that 6

(

4 ,

)

k

p

M

₍

₂

_,

2

₎

₂

_.

4

p

C

M

k





_{We already saw a polycyclic presentation for} 4

(

2 ,

2

)

k

p

M

_in

chapter 2.From the proof of theorem 2.17

,

we see that a extra element of order 2 needed to generate 6

(

4 ,

)

k

p

M

can be chosen so that it acts

P-th poweringly on a generator of the center of

(

2 ,

),

2 4

k

p

M

_{and trivially on each of the other members of our}

canonical generating set for

(

2 ,

).

2 4

k

p

M

_{Therefore we have.}

6.5.1-Proposition: 6

(

4 ,

)

2

(

2 ,

)

(

2 ,

),

k i k k

p

M

p

M

p

M





where

i

is 3 or 4 according as k

p

_{is congruent to}₃_or

1 modulo 4

,

respectively.

Proof:

Set

:

2

(

2 ,

)

(

2 ,

).

k i k

p

M

p

M

G





Then

|

48 (

1 ).

2





k

p

G

_Let

A

_{be our fixd singer cycle of}

GL

(

2 ,

p

k

).

_{It is}

not difficult to see that

A



I

2_{is the unique maximal abelian normal subgroup of both}

G

_and 6

(

4 ,

)

k

p

M

.

Therefore (4, )

(

)

6

(

4 ,

)

k p

GL

A

M

p

N

G



k



(this last equality because

Sp

(

2 ,

2 )

is soluble).Since

G



M

6

,

the result follows.

It is now easly to write down a polycyclic presentation for 6

(

4 ,

).

k

p

M

6.6-The generating set for 7

(

4 ,

)

k

p

M

.

Recall that 7

(

1 8 8

)

(

4 ,

2 ).

 







C

D

O

M

k

p _Let

F

_{be the Fitting subgroup of}

M

7_{. We could find a}

generating set by the methods given in chapter 2. The following structure theorem yields a much simpler way.

6.6.1-Theorm:

7

M

_{is conjugate to}

(

M

3

(

2 ,

p

)

M

3

(

2 ,

p

))

S

2

,

k





where the non-trivial element of the

S

2_{is the permutation}

(7)

1205

Proof:

Let

G

be the group

(

M

3

(

2 ,

p

)

M

3

(

2 ,

p

))

S

2

,

k k





as defined in the statement of the theorem. Them

Fit

(

G

)

is the tensor square of

(

3

(

2 ,

)),

k

p

M

Fit

and so

Fit

(

G

)



C

pk1



Q

8



Q

8

.

Since

Q

8_{is absolutely irreducible in}

),

,

2 (

p

k

GL

_{it follows from the outer Tensor product Theorem that}

Fit

(

G

)

_{is irreducible.Then by Theorem}_4.3

)

(

G

Fit

_{is conjugat to}

F

_.

So

G

is conjugate to a subgroup of the normaliser in

(

4 ,

)

k

p

GL

_of

F

,

_namely

M

7

.

_Since

|

G

|



|

M

7

|,

_the

result follows.

We now redefine 7

(

4 ,

)

k

p

M

to be

(

M

3

(

2 ,

p

)

M

3

(

2 ,

p

))

S

2

.

k k





It is then easly to write down a polycyclic presentation for

M

7

.

Note that



















 

),

8 (mod

7 ),

(

),

8 (mod

3 ),

)

3 ,

2 (

(

2 1

7

k p

p

if

S

cr

BO

C

p

if

S

cr

GL

C

M

k k

where by

G

cr

T

we mean the crown product of the group

G

and the transitive permutation groups T.In our case the crown product can be defined (abstractly) as the quotient of the wreath prooduct by the unigue diagonal central subgroup of the base group.

Obviously it is important to find the primitive subgroups of

GL

(

2 ,

3 )

cr

S

2 and

BO

cr

S

2

.

_Let

N

_{be either of}

these groups

,

let a be the centre of

N

(the scalar subgroup of order2) and let

E

be the Fitting subgroups of

N

.

Then

A

E

is symplectic space on which

E

N

acts in the natural way as

(

4 ,

2 ).



O

_{The following theorem identifies}

those primitive subgroups of

N

which contain

E

.

6.6.2-Theorem:

Let

G

be a subgroup of

N

that contains

E

. Then

G

is imprimitive if and only if

E

G

normalises amaximal

isotropic subspace of

A

E

.

Proof: see short (1992

,

8.3

,

pp.95-97)

6.7-The generating set for 8

(

4 ,

).

k

p

M

Let

F

be the field of k

p

_elements

,

_when

p

k



3 (mod

4 ).

_{we construct a generating set for}

M

8_{by the following}

methods.

Let

z

be a generator for the scalar group

,

and define

u

1

,

v

1

,

u

2

and

v

2 by

,

1

0

1 :

,

0

1

0 :

,

0

1

0 :

2 2 2 1 2 2 2

1

I

u

I

v

I

v

I

u

_











































































Where

,



and



belong to the prime subfield of

F

and

1 .

(8)

1206

Then

{

u

1

,

v

1

,

u

2

,

v

2

,

z

}

_{generates the Fitting subgroup}

F

_of

M

8

.

_{To extend this set to}_{a generating set for}

M

8

,

_{we first require a generating set for}

Hol

(

C

₅

).

_Actully

,

_{it is more convenient to choose a generating set for}

).

2 ,

4 (



O

_{we choose this set to consist of the three matrices}

f



,

g



and

h



defined below











































1

0

1

0

1

0

1

0 :

1

0

1

0

1

0

1

0 :

1

0

1

0

1

0

1

0

1 :



h

g

f

setting

b



:



h



f



and

a



g



f



b



g



2

)

(

:



,

we find that

a



has order 4,

b



has order 5

,

and

,

)

(

)

(

b



a



b



2 _{so that}



a



,

b





Hol

(

C

₅

).

There are tow

resons we work with

f



,

g



and

h



.One resons is that each is an involution

;

this means that the solutions to the liner equations we must solve are less complicated .The other reson is that each of these matrices has mainly

zeros in its last colum

;

experience shows that calculations not involving

v

2_{are much simpler.It is also useful that the}

first three rows of

g



and

h



are the same . In this case

,

there exists a matrix

f

of

GL

(

4 ,

F

)

such that

,

1 1

1

u

f





,

2 1 1 1 1

u

v

f





u



f

2



2_u1_u2

,

v



f

2



2_v2_.

Where the



i_.And



j

are scalars.Setting



1₌



2₌



1_{=1 and}



2_{=-1 .}

We find that one such matrix is

f

_:=21

.

1

1 





















































(9)

1207

Then

f

has determinat 1

,

and its square is



I

4_.

Let



be an element of

F

such that

2



=











;

)

8 (mod

7

2 ,

)

8 (mod

3

2

k k

p

if

p

if

Then

,

g

of

GL

(

4 ,

F

)

such that

,

1 1

1

v

u

g





,

2 2 2

1 1 1

u

v

g g







v

2 2

v

2

.

g





Where the



i_and



j_{are scalars.Seting}



1





1





1

_and



2





2



1 ,

we find that one such matrix is

.

1

0

1

0

1

0

1

0

1

0

1

0

1 :

1

















_



g

Then

g

has determinat 1

,

and its square is



I

4

or

I

4

,

_{acording as} k

p

_{is congruent to}₃_or₇_modulo₈_.Then

,

h

of

GL

(

4 ,

F

)

such that

.

,

2 2 2 2

2 2 2

1 1 1

v

u

v

u

v

u

h h h h











Where the



i_and



j

are scalars.Setting



2





1 and



1





1





2



1 ,

we find that one such matrix is



















































































































1

2 :

h

.

Then

h

has determinat 1

,

and its square is



I

4_.

(10)

1208



























1

2

1































b

.

Also

,

b

has determinant 1 and order 5.The action of

b

on

F

is given by the matrix















1

0

1

0

1

0

1

0

1

0

1

1 

b

.

Now set

a

gf

bg

2

)

(

:



_.Then

































































































































1

3

a

.

Also

,

a

has determinate 1 and its fourth power is -

I

4

.

_Furthermore

,

b

a



b

2

.

_{The action of}

a

_on

F

_{is given by the}

matrix

.

1

0

1

0

1

0

1

0

1

0 

















a

Finally we have that

M

8



a

,

b

,

u

1

,

v

1

,

u

2

,

v

2

,

z



.

Set

N

:



a

,

b

,

u

1

,

v

1

,

u

2

,

v

2



.

_{It is clear that}

N



M

8



SL

(

4 ,

F

)

_{and that}

M

8



N

Y



z



.

_{The given}

element

,

whitout

z

,

from a polycyclic generating sequence for

N

,

and the corresponding relations are:

(11)

1209

Note that each of these relations is independent of the value of k

p

_modulo₈_. It can be checked via CAYLEY that

N

dosenot split over

N



F

.

6.8-Some primitive subgroups of 6

(

4 ,

)

k

p

M

Recall that 6

(

4 ,

)

2

(

2 ,

)

(

2 ,

),

k i k k

p

M

p

M

p

M





where i is 3 or 4

,

acording as

k

p

_{is congruent to}₃_or₁_modulo₄

,

respectively

,

Also recall that





















_

.

)

8 (mod

5 ,

)

8 (mod

3 )

3 ,

2 (

,

)

8 (mod

1 )

,

4 (

₁

k k k

p k i

p

if

NS

p

if

GL

p

if

BO

C

p

M

k

Therefore

,

since the scalar group is the subgroup that is amalgamated in the tensor product

,

we may write





















_

).

8 (mod

5 ),

8 (mod

3 )

3 ,

2 (

),

8 (mod

1 )

(

)

,

4 (

₂

1 6

k k k

p k

p

if

NS

p

if

GL

p

if

BO

C

p

M

k



Then subgroup amalgamated in this central product of order 2

,

except when



5 (mod

8 ),

k

p

_{in which case it}

has order 4. In this central decomposition of

M

6

,

_{denote the first central factor by}

X

_{and the second by}

Y

_.

In this section we determine some primitive subgroups of 6

(

4 ,

)

k

p

M

by using our knowledge of

X

,

Y

and of the subgroups of central products.

6.8.1-Proposition:

Every primitive subgroup of 6

(

4 ,

)

k

p

M

contains the scalar matrix



I

4_.

Proof:

The proof when





1 (mod

8 )

k

p

_{is sufficient to indicate the method. Let}

G

_{be a subgroup of} 6

(

4 ,

)

k

p

M

that

dosenot contain



I

4_.

By examining the Burnside inclusion diagram of

BO

,

we see that

G



Y

must be of order 1 or 3.Therefore

6

C

X

G





_{. Observe that}

X



C

6



X



C

3_.If

p



3 ,

_{then this group is reducible}

,

_{and if}

p



3 ,

_{then it is not}

primitive

,

because it has a

non–cyclic abelian normal 3-subgroup.

We now specialise to the case



3 (mod

8 ).

k

p

_Then

Y



GL

(

2 ,

3 ).

Let

G

be a primitive subgroup of 6

(

4 ,

)

k

p

M

(12)

1210

Where

X

1



X



GY

,

X

0



X



G

,

Y

1



XG



Y

,

Y

0



G



Y

_and



_{is an isomorphism from}

X

1

X

0_to 0

1

Y

_.Since

_G

_{is primitive}

,

_{it is clear(from elementary propertise of the tensor product)that both}

X

1_and

Y

1_must

be primitive.The primitive subgroups of

X

were discussed in section 2

,

and the primitive subgroups of

Y

were discussed in section 3.

6.8.2-Lemma:

The only possibilities for the pair

(

Y

1

,

Y

0

)

_are

)

),

3 ,

2 (

(

))

3 ,

2 (

),

3 ,

2 (

(

),

3 ,

2 (

(

)),

3 ,

2 (

),

3 ,

2 (

(

)),

3 ,

2 (

),

3 ,

2 (

(

GL

SL

GL

Q

₈

SL

and

SL

Q

₈ _.

Proof:By use from the burnside inclusion diagram of

GL

(

2 ,

3 )

yeilds that

,

if

Y

1

did not contain

SL

(

2 ,

3 )

,

Then

G

would be a subgroup of

X



SD

1 6

or

X



D

1 2

.

The first of these obviously has a non-cyclic abelian normal 2-subgroup and so cannot be primitive.If

p



3 ,

then

1 2

D

X



is reducible

,

and if

p



3 ,

then

31 (

1 )

2k



p

_{and so}

X



D

1 2_{can not be primitive}

,

_{because it has a}

non-cyclic abelian normal 3-subgroup.Hence

Y

1



SL

(

2 ,

3 ).

_Since

X

_{is metacyclic}

,

_{it follows that}

X

1

X

0_is

metacyclic.Therefor

Y

1

Y

0_{must be metacyclic too.}

6.8.3-Teorem:

Let

G

be a subgroup of

M

6_{such that}

G



Y



SL

(

2 ,

3 ).

_If

B

_{is an abelain normal subgroup of}

G

,

_then 0

X

B



_.

Proof:

Without loss of generality we can assume that

B





I

4



.

Clearly

B



Y

is an abelian normal subgroup of

G



Y

,

which is

SL

(

2 ,

3 )

or

GL

(

2 ,

3 ).

Therefore

B



Y





I

4



.

_Also

G



Y

_{must normalise}

XB



Y

.

_{The only subgroups of}

GL

(

2 ,

3 )

_{that are}

normalised by

SL

(

2 ,

3 )

are the normal subgroups of

GL

(

2 ,

3 )

.Since

B

is abeline

,

we conclude that

XB



Y

is







I

4 _or

Q

8_.

Since

B

is normal in

G

,

we have from by Theorem 3.1.5 that

(

X



BY

)

/(

X



B

)





(

XB



Y

)

/(

B



Y

),

Where



_{is the set of outomorphism of}

X



Y

induced by the elements of G acting by conjugation.

Since the 3-elements of

SL

(

2 ,

3 )

act trivially on

X

but non-trivially on

Q

8

/





I

4



,

_{we conclude that}

B

_{cannot be}

a diagonal subgroup of

X



Y

,

and thus

B



X

.

6.8.4-Theorem:

(13)

1211

Let

G

:



X

1



SL

(

2 ,

3 )

and

H



K



SL

(

2 ,

3 ),

_Where





I

2



K



X

1

.

_{Then by proof of above theorem}

,

_we

have the Figure following

1

6.8.5-Corollary 1:

If

X

1_{is a primitive subgroup of}

X

,

_then

X

1



GL

(

2 ,

3 )

_{is a primitive subgroup of}

M

6_. 6.8.6-Theorem K:

If

X

1_{is a primitive subgroup of}

X

,

_{then the diagonal between}

X

0



SL

(

2 ,

3 )

and

X

1



GL

(

2 ,

3 )

_is

primitive.

Proof:

See short(1992

,

theorem 8.6.5

,

pp.103-104) Now we specialise to the case



3 .

k

p

_{We have}

).

3 ,

2 (

)

3 ,

2 (

)

3 ,

2 (

)

3 ,

4 (

₂ ₃ ₁₆

6

M

SD

GL

M









The primitive subgroups of

SD

1 6_are

SD

1 6

,

Q

8_and

C

8_.When

X

1



X

0

,

_{we get six group}

,

_{all of which are primitive}

by theorem

K

and it corollary when

X

1

X

0_has

Order 2

,

we get five groups

,

all of which are primitive by theorem

K

.

ISOTEST shows that among these eleven groups there are ten distinct isomorphism types of sylow 2-subgroups .The two groups whose sylow 2-subgroups are isomorphic have derived groups of deffirent orders.Therefore these eleven groups are pairwise non-isomorphic.None of these groups is isomorphic to a subgroup of

M

5

(

4 ,

3 )

_because

that group is metacyclic.So there are exactly

11 GL

(

4 ,

3 )

-conjugacy classes of primitive soluble subgroups whose gurdian is

M

6_.

6.9-The primitive subgroups of 7

(

4 ,

)

k

p

M

when



3 (mod

8 ).

k

p

we assume that



3 (mod

8 )

k

p

_and

F

_{is the field of}

p

k



_{elements.Also}

,

_{We denote}

GL

(

4 ,

F

)

_by

L

_.We established in section 3 that

).

)

3 ,

2 (

(

)

,

4 (

₍ ₁₎_/₂ 2

7

p

C

GL

cr

S

M

_pk

k







Denote by

N

the second direct factor in the decomposition.We must find the primitive subgroups of

N

.For this work

,

we have: