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THE EULERIAN FUNCTIONS OF CYCLIC

GROUPS, DIHEDRAL GROUPS,

AND P- GROUPS

THESIS

Presented to the Graduate Council of the

University of North Texas in Partial

Fulfillment of the Requirements

For the Degree of

MASTER OF ARTS

by

Cynthia M. Sewell, B.S.

Denton, Texas

August, 1992

Z3~:A

Sewell, Cynthia M., The Eulerian Functions of Cyclic

Groups, Dihedral Groups, and P-Groups. Master of Arts

(Mathematics), August, 1992, 50 pp., 13 illustrations,

bibliography, 5 titles.

In 1935, Philip Hall developed a formula for finding

the number of ways of generating the group of symmetries of

the icosahedron from a given number of its elements. In

doing so, he defined a generalized Eulerian function.

This thesis uses Hall's generalized Eulerian function

to calculate generalized Eulerian functions for specific

groups, namely: cyclic groups, dihedral groups, and

p- groups.

TABLE OF CONTENTS

Page

LIST OF ILLUSTRATIONS ... ............. iv

Chapter

I. INTRODUCTION .. ............. 1

Historical significanceDefinition of the Eulerian Function

II. THE MOBIUS FUNCTION..... .. ...... 4

Definition and calculation of the MObius functionThe Eulerian function of cyclic groupsFrattini subgroup

III. THE EULERIAN FUNCTION OF DIHEDRAL GROUPS... ... 14

Classification of subgroupsCalculation of Mt*bius function

IV. THE EULERIAN FUNCTION OF P-GROUPS.... ... . 35

Weisners Theorem

BIBLIOGRAPHY. . ..... ............... 50

iii

LIST OF ILLUSTRATIONS

Figure

1. Mbuis function example....... .....

2. MObius function example.. .........

3. Subgroup structure of the integers modulo 12.

4. Subgroup structure of the integers modulo p .

5. Subgroup structure of the integers modulo pq.

6. Subgroup structure of the Quaternion group. .

7. Subgroup structure of D2p

8. Subgroup structure of D2p n

9. Subgroup structure of D2pq... ..........

10. Subgroup structure of D2pqs. .........

11. Subgroup structure of the elementary abeliangroup of order 4.............. ..........

12. Subgroup structure of the elementary abeliangroup of order 9............ . ...........

13. Subgroup structure of the elementary abeliangroup of order 8............. ...........

Page

. . . .18

. . . .20

. . . .21

. . . .25

. . . .39

. . . 4.

. . . 18

. 0 20

. 0 21

. . . 25

.. .39

.. .40

.. .44

iv

CHAPTER I

INTRODUCTION

In 1935, Philip Hall developed a formula for finding the

number of ways of generating the group of symmetries of the

icosahedron from a given number of its elements. In doing

so, he showed that this problem can be solved for any finite

group whose subgroups are sufficiently known. Given any

finite group G, Hall defined the function $n (G) to be the

number of ways of generating G with n of its elements.

This function is known as the generalized Eulerian function

and is defined as follows:

#n(G) = p,(H,G) JH|H: H < G [In

In order to find the Eulerian function of a finite group

G, we must first calculate, for each subgroup H of G, the

integer valued function p(H,G), known as the M6bius function.

Only those subgroups whose Mtbius functional values differ

from zero need to be considered. Chapter II provides a

theorem for determining these subgroups. The MObius function

is. defined and discussed in Chapter II. The second chapter

also provides examples of the calculation of both the M*bius

and Eulerian functions and is concluded with the generalized

Eulerian function for cyclic groups.

1

2

The third chapter is devoted to dihedral groups. This

chapter addresses the problem of finding all the subgroups

of a given finite dihedral group and calculating the Mbius

functional value for each subgroup. Again a Eulerian

function is calculated for finite dihedral groups.

In the final chapter, the Eulerian function of p-groups

is determined. The calculation of the Mobius function on

the subgroups of a given p-group is simplified, through the

use of several theorems and lemmas, to the calculation of

the Mbius function on an elementary abelian group. It is

here that Weisner's Theorem is employed. The chapter

concludes with the generalized Eulerian function for

p-groups.

CHAPTER BIBLIOGRAPHY

P. Hall, The Eulerian Functions of a Croup, Quart. J.Math., 7 (1936), 134-151.

3

CHAPTER II

THE MOBIUS FUNCTION

The Mbius function , p , is the integer-valued function

defined inductively on the set S of all subgroups of a finite

group G by:

(*) tp(G,G) = 1

and for each subgroup H of G

(**) p(K,G) = 0.K:H < K

Consider any group G such that all the subgroups of G are

known. Since the elements of S are subgroups, we can

arrange them according to their order; that is, the elements

of S are arranged in the following way:

G = Si, S 2 , S3 ..., Sn = {e}, where IGI 1S21 > --- > > Sn.-

The M5bius function assigns one and only one numerical

value to each element of S. The initial condition (*)

stipulates that p(G,G) = 1. To assign a value to S2 one must

first find all the members of S that contain S2 . The sum of

the functional values of these members of S containing S2

along with the functional value of S2 must be zero. S2 is

4

5

S2S3 -S2i

SE

Figure 1 Figure 2

of a subgroup of G and only G; therefore, by (**):

p(S 2 ,G) + p(G,G) = 0.

Solving for p(S2 ,G) we find that the functional value of S2

is -1. Now let us consider S3 . Suppose that S3 is not a

subgroup of S2 . Then by (**):

g(S3 ,G) + p(G,G) = 0

and hence, A(S53 ,G) = -1. (See Figure 1)

However, if S3 is a subgroup of S2 (then again by (**)):

p(Sa,G) + p(S 2 ,G) + p(G,G) = 0

and hence,

1(4S 3 ,G) = - ( p(52 ,G) + IL(G,G) ) = - ( -1 + 1 ) = 0

(See Figure 2)

One can continue in this manner to assign a functional value

to each subgroup.

Example 2.1. Let us now demonstrate this procedure on

the abelian group (112 = { ,iHTHH,7,H,9,ThK} of the

6

3iZ1-2

< 22 > < 3

I<I< "<

FT < 0 >

Figure 3

integers modulo 12 under addition. The subgroups of E12

are as follows (Figure 3):

E1 2

< > ={,,1,i, ,~}

<H > = {5,,,9},

< > =

<K > = {tH}.

< > = {}.

The M6bius functional values can now be calculated for each

of the above subgroups. (See figure 3)

p1(112 ,112) = 1

S >2112) = -1

p( 3>,iE12)

7

p Dz p q = < X > [ I I

-1< X P> < -

{e} E-] {e} LIII|Figure 4 Figure 5

The subgroup <~4 > is contained in 112 and < > ; and,

therefore, by (**):

Pu(112 ,F1 2) + A(<~ >,1!> 2) + pt(< 24 >,12) = 0.

Hence

A(< 4 >,112) = 0.

The subgroup < ~B > is contained in 912, < ~ >, and < ~3 >;

and, again by (**):

(1 2 ,Z112) + p(< ~2 >,12) + (< ~3 >,L12) + A(<6 >,12) = 0.

Hence

p( 6>,2712) = -

The subgroup < ~ > is contained in every subgroup of E12.

Once again using (**) we find that:

A(< b ~ >, 112) = 0.

Once the M5bius function values are determined for every

subgroup of 112, the Eulerian Function for 112 can be

calculated as follows:

On (E2) = p(HA12) I

Hn

_ 12 [ 1_ ]+ 2n IJ J [J + [

We now have a function for determining the number of n-bases

that generate 112. For example:

8

01 (112) = 4.

This reveals that E12 can be generated by four distinct

elements of the group.

Example 2.2. The Eulerian Function for HP, for any

prime p, is as follows (See Figure 4):

#n (21P)=n n

Example 2.3. Consider the group Epq such that p and q

are distinct primes. (See Figure 5) The Mbbius Function is

easily computed and the Eulerian Function for "pq is given

by the following formula:

On(11pq) = [pq [] [n]+[]Example 2.4. Consider the Quaternion Group,

Q = { 1, + i, + j, 2=k } where i2 = j2 = k2 = -1; ij = k;

jk = i; ki = j; ji = -k; kj = -i; ik = -j; and the usual

rules apply for multiplying by -- 1. The subgroups of G are:

G,

< i > = { i, -i, -1, 1 },

< j > = { j, -j, -1, 1 },

< k > = { k, -k, -1, 1 },

< -1 > = { -1, 1 },

< 1 > = { I }.

See figure 6 on the following page.

The Eulerian Function of the Quaternion Group

is #n (G) = []- 3 [ ] 2 [J.

The following theorem is useful in calculating

Eulerian functions on larger groups.

9

G

i> < <j> < k>

< -1>

{e}

Figure 6

THEOREM 2.1. Let G be a group and H be any subgroup of

G. The Mdbius function p(H,G) can differ from zero if and

only if H can be written as the intersection of a certain

number of maximal subgroups of G.

Proof. The proof of the theorem is by induction.

Let us suppose that Theorem 2.1 holds for all Ki > H,

where H < G and Ki < G for all i. Suppose further that H is

not the meet of any set of maximal subgroups of G. In other

words, there does not exist a set {M1 ,M2 ,...,M} of maximal

subgroups of G such that H = M1 f M2 f ... fl M.

Let M represent the intersection of all the maximal subgroups

of G that contain H. Hence, M = Mi where H < Mi for i=11

to n. Using the definition of the Mbbius function we know:

K:HK< K p(KG) = 0 or

10

[ pK(K,G) ] + p(H,G) = 0.K:H < K

Therefore,

( K:HK (K,G) = - pa(H,G).

For all K such that H < K but K does not contain M, we know

that K is not the meet of any set of maximal subgroups of G;

hence, for all such K, p(K,G) = 0 (by hypothesis). Thus, (*)

can be rewritten as follows:

I(KG) = - u(H,G).

K:K MBut M is, itself, a subgroup of G, and therefore

pL(KG) = 0.K:K M

Hence pL(H,G) = 0.

COROLLARY 2.1. If p(H,G) 0 then H1 (G), where 4(G)

is the subgroup of G formed by the intersection of all the

maximal subgroups of G.

Theorem 2.1 enables us to calculate the Eulerian

function on the group in the following example.

Example 2.4. Consider the cyclic groups Ela1 ... ar ofI r

integers modulo pla, P a2 ... Pr ar under addition where each pi

is a distinct prime raised to the power ai. There are r

maximal subgroups of the form Ela, ... p aj- 1... a for1 j r

1 < j < r. Each of these subgroups has a M~bius value of -1.

The maximal subgroups are referred to as Type 1 subgroups.

There are subgroups of the form 2Z a. .. paj-1... ak-1... ar,1 jPk r

11

for j < k ; 1 < j K (r-1) and 2 < k < r. Each of these

subgroups is contained in two maximal subgroups, namely

Ll a...aj -1...par and 21 pa1...pak-1...par. Hence, theI j r 1 k r

M*bius value for subgroups of this type is computed to be 1.

These subgroups are henceforth referred to as Type 2

subgroups. Continuing in this manner, the next type of

subgroup is of the form Epa, ...pa-1...pak-1...pai-I..par,1I k 1 r

where 1 < j < k < 1 < r. Each of the subgroups of this type

is contained in [3] Type 2 subgroups and [3] Type 1 subgroups.

Thus, subgroups of this form, which are denoted Type 3

subgroups, have a Mbbius functional value of -1. Type 4

subgroups are found to have a Mtbius value of 1. The

functional values of the subgroups continue to oscillate

in this same manner. By Theorem 2.1 we know that the Mbius

function of a subgroup differs from zero if and only if the

subgroup is the intersection of some set of maximal

subgroups of the group. The subgroup formed by intersecting

all the maximal subgroups, f, of a group is known as the

Frattini subgroup, i(G) = nl { MI M E if }. This is the

smallest subgroup with a nonzero Mbius functional value.

The Frattini subgroup of E a,...a ,ris p a-1... ar-1.p pp p

1 r 1 r

The Eulerian function for Ip a, ... par is written as follows:1 r

Ira, ar a, aj -1 a.

#n(R a,...ar) Pi= * ..'Pr Pi .n.'p j ... r +

p a.

j k

+ [ a1-1

i=1

aj-1 ak-.&P. .. p knk

al ar-... p. ... p1 rn

1 parr -

p ,- -1 ar-fl.

nrWhen each ai is equal to 1, we find that the Eulerian

function

forllr p ..Pris:

[ Pi0 -- Pr]nOn (p .p 1 - ) =

jj = 1

k Pi *Pj -iPj +1.*.Pk-1Pk+1 -Prn

r

i= 1

C&nPi~ Ln

12

+

P i - -j 1 Pj+1 - Prn

CHAPTER BIBLIOGRAPHY

1. R. P. Burns, Groups, a Path to Geometry, CambridgeUniversity Press, Cambridge, 1985.

2. P. Hall, The Eulerian Functions of a Group, Quart. J.Math., 7 (1936), 134-151.

3. I. N. Herstein, Topics in Algebra, John Wiley & Sons,Toronto, 1975.

4. J. S. Rose, A Course on Group Theory, CambridgeUniversity Press, Cambridge, 1978, 266-271.

13

CHAPTER III

THE MOBIUS FUNCTION OF DIHEDRAL GROUPS

The dihedral group D211 of order 2n is the symmetry

group of a regular n-gon. The elements of this non-abelian

group consist of rotations, ri for 1 ( i ( n; and reflections,

ris also for 1 ( i < n. In order to compute the M6bius

function on D2 n, we must first find all its subgroups.

THEOREM 3.1. Subgroups of D2 n are of the following form:

1. For every integer m, such that m divides n, there exists

a cyclic subgroup of order m.

2. For every integer m, such that m divides n, there exist

n/m dihedral subgroups isomorphic to D2 m.

Proof. The proof of Theorem 3.1 is accomplished by the

use of the following lemmas:

LEMMA 3.1. Every subgroup of the dihedral group D2 n is

either cyclic or dihedral.

Proof. Let D2 n = {e,r,...,rn-1, srs,....,rn-s}.

Since the subgroup generated by <r> is a normal subgroup of

D2, and D2 n is isomorphic to the group generated by r and s,

let us consider the following ways to generate subgroups:

a. Subgroups can be generated by a single

rotation, ri, for i = 1 to n-1. The

resulting subgroup will consist only of

14

15

rotations, hence it is cyclic. The order

of <ri> is k where ik = n.

b. Subgroups can also be generated by a single

reflection, ris, for i = 1 to n-1.

(ris) (ris) = rirn-iss = rns2 = e. Since

reflections have order 2, these subgroups

are also cyclic.

c. Now consider the subgroup H generated by a

reflection and a rotation, such that

H * D2n. Let H = <rk, r1 s> for some k and

some 1 from 1 to n-1. Let R = <ri> and let

S = <rls>, hence H is isomorphic to R x S.

Given that the order of <ri> is m, there

exists a minimal j such that <rj> = <ri>.

H = <rj , r1s> = {e,ri , r2 j,....,rmi-i

r's,ri 1 s,....,rmi-i 1 s}.

Let rYs represent the minimal reflection in

H such that H = <rj, rY>= <rj> <rYs>. Hence

H is isomorphic to the dihedral group of

order 2m.

LEMMA 3.2. For every integer k such that k divides the

order of the dihedral group, D2 n, there exists a subgroup, H,

of D2n such that the order of H is k.

Proof. Let D2 n = {e,r,...,rn-1,s,rs,...,rn-Is} be

the dihedral group of order 2n. Since every reflection has

order two, there exists a subgroup of order two. The subgroup

16

generated by <r> has order n, hence providing a subgroup of

order n. For any integer, k, such that k divides n, there

is an integer, 1, such that ki = n. Consider <rl>. This is

the rotational subgroup of order k. Finally consider any

integer w, such that w divides 2n but w does not divide n.

The integer w can be represented as 2s, and s must divide n.

Therefore, there exists a minimal t such that st=n and, hence,

the order of <rt> is s. Consider H = <rt ,s>. H has order w. o

LEMMA 3.3. If H and K are subgroups of the dihedral

group, D, and the order of H is 2m and the order of K is 2n;

then H n K is isomorphic to either the rotational subgroup of

order t, where t is the greatest common divisor of m and n;

or H 0 K is isomorphic to the dihedral subgroup of order 2t.

Proof. Consider the dihedral group, D, with subgroups

H isomorphic to D2 m and K isomorphic to D21. Let F = H A K.

Consider the following cases:

a. Suppose that m=n but H t K. It is true,

however, that H is isomorphic to K. Since

the order of the rotational subgroups

generated by r in H and K are m and n,

respectively, F = <r> and the order of F

is the gcd(m,n) = m = n.

b. Suppose m j n and H A K contains no

reflections. There exists an integer 1

such that the order of <r1> is equal to n,

and similarly, there exists an integer s

17

such that the order of <rs> is equal to m.

<rl> l <rs> = <rsl> and the order of <rsl>

is equal to the gcd(m,n).

c. Now suppose that m 0 n and H f K contains a

reflection. Since F must be either cyclic or

dihedral and F contains a reflection, it can

be concluded that F is the dihedral group of

order 2 (the order of the rotational

subgroup of H f K) = 2 gcd(m,n). 0

Thus, Theorem 3.1 is verified. c

Example 3.1. Let us now consider the Mobius function

on D2 Pn where p is a prime. From Lemma 3.2, it is known

that there exists a subgroup of order 2pi for all i from 1

to n-1. Moreover:

LEMMA 3.4. There are exactly pk dihedral subgroups of

D2 pn having order 2pi (where ik=n) for all i from 1 to n-1.

nProof. Consider the dihedral group of order 2p with

the dihedral subgroup, H, of order 2p

H = {e,rp,r2p,... .,r(p" ~ )Ps,rps,....,r(pn ~1-ps

The rotations form a normal subgroup and will remain

consistent under conjugation. Let us consider the

reflections of H:

i) s(rps)s~1 = srp = rn - Ps = r(pn s1)p5 E H

ii) (rbs)(rps)(rbs)~ 1= (rbs)(rps)(srn - P)

rbrp - Prbs = rPn - p + 2 bs 0 H.

18

D2

| 1 |< r > 4o

s > < rs >

-1i -1

2S>-r .. r -

e

Figure 7

Using ii), the elements of H can be conjugated for all b

from 1 to p-1 producing p-1 new subgroups isomorphic to H.

Hence, including H, there are p subgroups of order 2pn - I

It follows directly from Lemma 3.4 that every reflection

is contained in one and only one dihedral subgroup of order

2pi; and the intersection of any two dihedral subgroups of

the form D2 p is the rotational subgroup of order pi.

We begin by calculating the Eulerian function on

the dihedral group D2 pn, where p is a prime and n=1.

2P= {e,r,r2 ,...,r , s,rs,r2 s,...,r s}. The subgroups of

D2p are as follows (Figure 7):

Si = 12P ,

S2 = {r,r2,...,r 1

S3 = {e,s},

S4 = {e,rs},

01

19

S5 = {e,r2 s}, ... , ... , .

P -Sp+2 {e,r },

Sp+3 {e}-

Hence, the Mbius functional values can be assigned as

follows:

p(D2PD2p) = 1,

u(< r >,D2 P) = p(c s >,D2 P) = rs >,D2P) =

= p(< r 1 >,D2P) = -1,

u({e},D2 P)= P-

The Eulerian function for D2 P is defined as follows:

#m(D2 p) =14(HD2 P) [ I I,3]= [ 2 - [ 3 - p[ 3 + p .

Calculating #m (D2 p) for m = 1, the desired result is obtained;

that is, the dihedral group can not be generated by a single

element. Let us now verify the Eulerian function for D2P

when m = 2. There are 2p-J 31subgroups of order two not

containing the identity; of which, all generate D2 p except

those subgroups consisting of two rotations. The number of

subgroups of order two containing two rotations is [ PHence, the number of subgroups of order two that generate D2 P

is 2p- 1 p- ; .Simplifying, we have:

-(2p-1)(2p-2) - -(p-1)(p-2) = -2p - -p.

Calculating #2 (D2P) we find:

=2 (D2 = p3 - - p[2 = -(2p) (2p- 1) - -(p) (p- 1) - p

= .2 -

Thus verifying the formula for m = 2.

20

D2pn

< r >

D2Pn -1 D2 Pn -1 .. . ... . .. D n -

< rp >

Figure 8

Consider the dihedral group of order 2pn for n > 1. The

Frattini subgroup of D2 pn, as defined in chapter two, is the

subgroup formed by the intersection of all its maximal

subgroups. Since there are p maximal dihedral subgroups of

the form D 2 pn-1 and one maximal cyclic subgroup, namely, <r>,

the Frattini subgroup of D2 Pn is found to be the rotational

subgroup of order pn-1, namely, <rP>.

Since, by Theorem 2.1, the M~bius function takes on a

value of zero for all subgroups of D2pn which are not the

meets of maximal subgroups, it can be concluded that the

only significant subgroups in the calculation of the

Eulerian function are those of order greater than or equal

to pn-1, the order of the Frattini subgroup of D2Pn.

Therefore, we are concerned only with the subgroups of the

form D2pn, <r>, <rP>, and D2pn-i. The Mbbius values are

calculated as follows (See figure 8):

p(D 2 pnD2 p") = 1,

21

D2pq

D p of these

D q of these

< s > < rs > < r 2 s ><......< rP'- 1 >

{e}

Figure 9

/z(<r>,D2 p") = -1,

(<rp>D2 pn) =P

u(H,D2 P) = -1 where H Lv D2p-ln

The Eulerian Function for D2Pn can now be written as follows:

Om(D2 p") = [2p"] p2p"-1+]- +.

Our next goal is to calculate the Eulerian function for

D 2p, ... Pk ,where the pi are distinct prime. In order

to do this let us begin with a few simple cases.

Example 3.2. Consider the dihedral group D2pq, where p

and q are distinct primes. Maximal subgroups of D2pq are of

the form D2 p, D2q, and <r>. See figure 9.

r rq

LEMMA 3.5. The Frattini subgroup of D2pq is the

identity subgroup, {e}.

Proof. From Lemma 3.4, we know that there are exactly

p subgroups of the form D2q. The intersection of these p

subgroups is the rotational subgroup < rp > of order q.

We also know, by Lemma 3.4, that there are exactly q

subgroups of the form D2P. The intersection of these q

subgroups is the rotational subgroup < rq > of order p.

Hence, by -intersecting the p subgroups of the form D2q with

the q subgroups of the form D2 P, and the maximal subgroup

< r >, we have: < rp > L < rq > L < r > = {e}.

Thus, the intersection of all the maximal subgroups of

D2pq is the identity subgroup.

Hence, by Lemma 3.5, it is necessary to calculate the

M~bius values of all the subgroups of D2pq. Figure 9

represents the subgroup structure of this dihedral group.

We can now calculate the Mbius function for each subgroup.

p(D2pq,D2pq) = 1,

p( H , D2pq) = -1 where H N D2q,

IL( K , D2pq) = -1 where K L D2P-

Since < rp > is contained in < r > and each of the p dihedr;

subgroups of order 2q, as well as D2pq, we find that:

4(< rp >, D2pq) = P-

Since < rq > is contained in < r > and each of the q dihedr;

subgroups of order 2q, as well as D2pq, we find that:

pi(< r(1 >, D2pq) = q.

0

al

al

22

23

From a previous result, we know that each reflection is

contained in exactly one of the dihedral subgroups of each

type. Hence, < ris > is contained in exactly one of the

D2q subgroups and in exactly one of the D2P subgroups.

Hence:

/(< ris >, D2pq) = 1 for i = 0 to (pq - 1).

The identity is contained in each of the subgroups.

Calculating the Mbius function for {e}, we find that:

p(HD2 pq) = 0e < H

p(D2pqD2pq) + p(< r >,D2pq) + p (D2q,D2pq) + q pL(D 2 pD 2 pq)

+ p/,(< rp >,D2pq) + 4(< rq >,D2pq)

+ pq p(< ris >,D2pq) + u({e},D2pq) = 0

1 + (-1) + p(-l) + q(-1) + p + q +pq(1) + k({e}) = 0

Hence:

/p({e},D2pq) = -pq.

The order of each of the subgroups is as follows:

jD2pqg = 2pq,

JD2q| = 2q,

ID2pI = 2 p,

j< r >| =pq,

< rp>I = q,

< rq >| = P,

< ris >1 = 2,

f{e}I = 1.

24

The Eulerian Function for D2pq can be written as follows:

#n(D2pq)= [2p] -[2q q [ 2p pqI] + p [q+ q[P] + pg - pq .

Example 33. Consider the dihedral group of order

2pqs where p, q, and s are distrinct primes.

The following page provides a detailed diagram of the

subgroup structure of this group along with the calculation

of the Mbius functional values of each of the many

subgroups. Notice that the Frattini subgroup is the

identity subgroup subgroup and, therefore, every subgroup

has a nonzero Mbius functional value.

The Eulerian function for D2pqs is:

n (D2 p q s 2pqs s 2pq 2qs] q 2ps pqs

+s[p + p qs + q - qs - ]PSpqs + qs [p]+ ps 2q]+ pq[2s

pqs [n] + pqs[n'

Example 3.4. We can now begin to define the Eulerian

function for G = D2p . .pk , where each pi is a distinct prime.

We must first find A the subgroups of this dihedral group.

The maximal subgroups are of the form:

Maximal Subgroup order number

< r > P1P2...Pk

H D2 ...- I, ---)...p 2 PIP2 -.- Pj -1 ,Pj +1 ,-...,Pk Pj

25

F 1 D2 p qs

order pqs < r > D2pq s of these

< rs >e K LIII ZE D2qs p of theseorder

q

< rp > F7-1 D2 ps -- q oforder q these

< r > q'order s

D2P D q of these

D2q .ps of these

D2Su . ps of these 1

r rqs > -sorder p

< r ps > -sorder q

< r pq > I-porder s

{e,s} { rs ... {e,r pqs-ls

- 1..... ........

Figure 10

26

Each of the above subgroups has a Mbius functional value of

-1. The Frattini subgroup of this group is the identity

subgroup, hence, all the subgroups of this dihedral group

have a Mt*bius functional value not equal to zero. Let us

begin by calculating the M~bius function on each of the

dihedral subgroups. The maximal dihedral subgroups are

referred to as Type 1 dihedral subgroups. Type 2 dihedral

subgroups will be of the form:

Hj1 j1 j1D2rPn-,+ . M-1 m+1' P k

where 1 < j,m < k. There are pjpm subgroups of the form

Hjm for all 1 < j,m < k according to Theorem 3.1. Each of

the Type 2 dihedral subgroups is contained in exactly two

Type 1 dihedral subgroups, namely:

some Hj rD2.P-- ,..+ p , and

some Hm D2P -P P , .P

We can now calculate the M~bius function on Type 2 dihedral

subgroups to be:

p (Hjm,G) = -[ p (Hj ,G) + p(HmG) + Ip(G,G) 3

= - [ (-1) + (-1) + 1 3 = 1.

Hence, the Type 2 dihedral subgroups have a M~bius functional

value of 1. Type 3 dihedral subgroups are of the form:

Hj ms r D2P -- P. ,P. 7,...P ,p ),...P ,P ,... P1 j-1 j+1 M-1 M+1 s-1 s+1 k

for all 1 < j,ms < k. There are pjpmps subgroups of the

form Hj.s for each I < j,m,s ( k according to Theorem 3.1.

Each subgroup of the form Hj ms is contained in [ 3 ] Type 2

27

dihedral subgroups, namely, one of the form Hjm ,one of the

form Hj ,, and one of the form Hs. Each of the Type 3

subgroups is also contained in[ [ 3]Type 1dihedralsubgroups, namely, one of the form Hj, one of the form Hm,

and one of the form Hs. Calculating the Mbius function on

the Type 3 dihedral subgroups we find:

p(Hj mns ,G) = - [ p(Hj .,G) + ap(Hj s ,G) + p(Hms ,G) + pU(Hj ,G)

+ pA(H,G) + p(Hs,G) +p(G,G)]

=- [1 + 1 + 1 + (-1) + (-1) + (-1) +1=-i

The Mbius functional value of the Type 3 dihedral subgroups

is found to be -1. Each Type 4 dihedral subgroup has a

M~bius functional value of 1 since each is contained in [ 3Type 3 dihedral subgroups; [ ]3Type 2 dihedral subgroups;

and [ 4 3 Type 1 dihedral subgroups. In general, the Mbius

functional values of the dihedral subgroups continue to

alternate from 1 to -1 until the last dihedral subgroups

are reached, namely, those subgroups of the form D2Pi for

1 ( i < k. For each pi there are Pi...Pi-1Pi+I...Pk of

the form D2Pi. The Mdbius functional values of these

subgroups are again -- 1. There are P1P2...Pk cyclic

subgroups of the form {e,ris} where 0 < i < PiP2.-..Pk -1

These subgroups, although cyclic, are included in this

calculation since their Mbius functional value is 1 or - 1

depending on the Mbius functional value of the last group

of dihedral subgroups. At this point, we can calculate the

first half of the Eulerian function for D with the

28

dihedral subgroups and those cyclic subgroups of order two as

mentioned above.

Subgroup Type

G

Hi for 1 < i < k

Hij for 1 K<i j k

D2 p' for 1 K i K k

{e ,rs}for0 i -PIP2 - -Pk

Hence, we have:

. (G)= (G,G) [II

order

2 pP2 - - -Pk

2pi. Pi-iPi+1...P2k

2p i..Pi-ipi+1 *..pj-1pj+1*...Pk

pP... Pi-iPi+i .. Pk

12

1

Pi+ (HG1=n

+ p(H ,G) Hij ] Pipj +

i ,j =1i * j

+~~~ p(er }G) |e,r s}|

] k-[

2 p121---.Pk [ 2 P ... Pi-xPi+1...Pk Dp11=1

k

+ [ 2p ipn Jpk Psn 4 PI P2 'Pi-l Pi + i a aPk ns [2-3

number

I

Pi

Pi Pi

2pi

.. +

29

Next, we calculate the Mbius function on the cyclic

subgroups of D2p - . We know that the maximal cyclic

subgroup < r > has order P1P2...Pr and a Mbius functional

value of -1. Cyclic subgroups of the form < r'i > have

order PIP2---Pi-1Pi+1...-Pk. These cyclic subgroups are

referred to as Type 1 cyclic subgroups. Each of these is

contained in < r >, and in pi dihedral subgroups of the

form Hi D2 P, *-1- ,... . Dihedral subgroups of

this form have a M5bius functional value of -1. We can

now calculate the M6bius function on Type 1 cyclic

subgroups to be:

p(< ri >,G) = - [ A(< r >,G) + pi p(Hi,G) + p(GG) ]

- [(-1) + pi (-1) + 1I] =-pi.

Hence the Mtbius functional value of a Type 1 cyclic subgroup

< rpi > is found to be pi. Type 2 cyclic subgroups are of

the form < rPIPJ > for 1 < i,j K k. Type 2 cyclic subgroups

have order P1P2...Pi-IPi+1...Pj-1Pj+1...-Pk. Each Type 2

cyclic subgroups is contained in pipj dihedral subgroups of

the form Hij. Dihedral subgroups of this form have a

Mbius functional value of 1. Each Type 2 cyclic subgroup

is also contained in pi dihedral subgroups of the form Hi

and pj dihedral subgroups of the from Hj. These dihedral

subgroups have a M~bius functional value of -1. Type 2

cyclic subgroups are also contained in two Type 1 cyclic

subgroups, namely, < rPi > and < rP >. Recall that the

30

Mbbius functional values of these Type 1 cyclic subgroups

are pi and pj, respectively. Hence we find that:

j(< rPiP >,G) = - [ (pipj) u(HijG) + pi z(HiG) + pj (HjG)

+ p< rpi >,G) + u(< rPj >,G)

+ p(< r >,G) + A(G,G) 3

- [pipJ - pi - pj + pi + pj - 1+1]

= - PiPj.

The Mbbius functional value of a Type 2 cyclic subgroup,

< rP'Pi >, is -p1pj. Continuing in this same fashion,

Type 3 cyclic subgroups are of the form < rpipipm > and

will have a M~hius functional value of pipjpm. This

process can be repeated until {e} is reached. If we

consider {e} = < rP1P2 ---Pk >, we see that the Mdbius

functional value of {e} is +PIP2 -- Pk-

The Eulerian function for G = pjP2 --.Pk is:

k

On(G) =p(G,G)[ I +I 3I + pa(HiG)[I'I3 pij=1

k

+ p(Hyj ,G) [IHii I]pipj + ... +

Ij =1i * j

+ j({e,r s},G) I{erS}I

k

+ pa(< r >,G) [<r>1 + k(< rP' >,G) [< rPi >1n=1

+ k (< rP'P >,G) < n rP + ... + ... +

+ p({e,G)j.

31

k_ 2p, P2 --- Pk _ 2p, ... .pi -1 pi +1 . . . Pk P

+ [2Pp2... p . [ 2 P... Pi -P+1.. Pk PD pii=1k

+ 2pi P.2..Pi..Pi+1...Pjj.. P21...pk] P2

i ,j =1.i #jk

-I 2 . .] P P2--- Pi- .i+-1Pk - ] P P2---Pk=1

kP1 ....Pi -IPi+1 ... Pj.-1Pj+1 ... Pk pi +...-

i ,j =1.1 Ji j

+ n1] PI P2*.Pk.+ 111

Example 3.5. We can now begin calculate the generalized

Eulerian function for G L D2 a1 , ak where each pi is a

1 k

distinct prime and each ai > 1. The maximal subgroups are

of the form:

Maximal Subgroup order number

< r > Pi a ... Pkak

Hj Da 1 aj - 1 ak 2pa' p aj- - *a2p- ...p ...p 1 j k1 j k

for 1 < j < k.

Each of these subgroups has a M~bius functional value of -1.

The Frattini subgroup of this group is the rotational

subgroup generated by rP r Pk > of order pa,-1 ak1 k

32

The calculation of the Mtbius function for Dai ak2p 1... Pk

1 k

can be done in the same manner as Example 3.4. Type 1

dihedral subgroups are the maximal dihedral subgroups as

shown above. Type 2 dihedral subgroups are subgroups

of the form:

Hjm N D a, aj--1 a. ak for 1 < j < m K k.2p 1 ... P. ... p ... p1 j m k

Each of these subgroups is contained in 2 maximal subgroups

and, as in the previous example, Type 2 dihedral subgroups

have M6bius functional values of 1. Type 3 dihedral

subgroups have Mbius functional values of -1, and it is

again found that the functional values alternate between

1 and -1.

Next, we calculate the M~bius function on the cyclic

subgroups of Dpa, yak. The functional values are found in

the same manner as the previous example.

M5biusfunction

Subgroup Type order value

< r > p a, pak_

Pi a, ai-1 ak< r > p .p . ... p kPifor 1< <i K k 1 k

'~rpi> a1 ai-1 ~ aj-I ak< rpi > p ... ... .. a-p1... . P ak -Pi Pjfor1 <:i,j < k k i k

PI --- Pk a1-1 ak-i +<>IPk1 k

The Eulerian function for D a, ak can now be written as2p 1... Pk

a, ak -2p 1 ... k1 k

I J

[ a12p ...

1( i ,j < kka-

~ 2 a1-1 - ap+2pp ... pin

p ... p a- 1 k +

n J

I ij I k

+ p a,

a1p ..1

1

k at ai ak]-2p ... p a ... p a

1i kn

pi

aj- Ip.in

-1 akppk PiPj

ak-S k

- a 1 ai-IP['*2Pi1 n

ai- aj-1. p. ... P. ..

n

ai ak-ln kI

n

ak- a op k

ak I.p 3p1

pi pi

Pi ..-PI-lpi+1 ---Pk-

f ollows:

-[

33

aj

Pi ...-Pi-1Pi +1 - - -Pk

CHAPTER BIBLIOGRAPHY

1. R. P. Burns, Groups, a Path to Geometry, CambridgeUniversity Press, Cambridge, 1985.

2. P. Hall, The Eulerian Functions of a Group, Quart. J.Math., 7 (1936), 134-151.

3. I. N. Herstein, Topics in Algebra, John Wiley & Sons,Toronto, 1975.

4. J. S. Rose, A Course on Group Theory, CambridgeUniversity Press, Cambridge, 1978, 266-271.

34

CHAPTER IV

THE MOBIUS FUNCTION OF P-GROUPS

In order to calculate a generalized Eulerian function

for p-groups it is necessary to begin with a few basic

definitions. A group G is a p-group if the order of G is

nfop. for some prime p and some positive integer n. A group

G is an elementary abelian group if the following three

conditions hold:

1. G is a p-group,

2. G is abelian, and

2. there exists a prime p such that for all x c G

XP = 1.

Elementary abelian groups of order pn can be represented as

the product of n cyclic groups of order p. For example, the

elementary abelian group of order 125 = 53 can be expressed

as Z15 X 7J5 x215. In this chapter we show that the subgroup

structure of elementary abelian groups can be generalized.

Since the subgroup structure is known the Mbius function

can be calculated for these groups. In order to define the

Mbius function of any given p-group the following theorem

is necessary.

THEOREM 4.1. If G is a p-group then G/(G) is

elementary abelian.

35

36

Proof. The proof of this theorem is accomplished with

the use of the following lemmas.

LEMMA 4.1. G/(G) is a p-group.

Proof. G is a p-group, therefore IGI = pn for some

prime p and some positive integer n. Since 4(G) is a

subgroup of G, we know that |I(G)I = pk for some k < n.

IG/ (G)I = IGI / j4(G)j = pn1pk = pn-k

Hence G/ (G) is a p-group. o

LEMMA 4.2. G/P(G) has trivial Frattini subgroup.

Proof. Let G be a p-group and K be a normal subgroup

of G. Consider the group G/K, G modulo K. Let ff be the set

of maximal subgroups of G. We know that for all M E ff, such

that K < M we have M/K < G/K with M/K a maximal subgroup of

G/K. Suppose that K < (G). This means that for all M E ff

we have K < M and hence K <lff. We have

n ff/ K =n ff /K]

that is,

(G) / K = [ G / K ].

Let K = f(G).

Then we have 1[ G / f(G) ] = (G)/ (G) =1.0n

LEMMA 4.3. If G is a p-group of order p then the

maximal subgroups of order pn-1 are normal subgroups of G.

Proof. The proof of this lemma is by induction.

When n = 1 the lemma is obviously true. Assume that there

exists a k > 0 such that for any group G of order pk'

where k' < k it can be concluded that the maximal subgroup

37

H of order pk is a normal subgroup of G. Let us now

consider the group G of order pk+1. The center of the

group G, Z(G), is nontrivial, therefore there exists an

element x c Z(G) such that the order of x is equal to p.

Consider the subgroup H = < x >. The order of H is p and

H is a normal subgroup of Z(G). Now consider the quotient

group G/H. The order of this quotient group is p . By

the induction assumption we know that G/H has a normal

subgroup N/H such that the order of N/H is pk-1 From

this information it can be concluded that N is a normal

subgroup of G and the order of N is pK.

LEMMA 4.4. If G is a p-group and 4(G) = 1 then G is

abelian.

Proof. The commutator subgroup, G', of the p-group G

gnerated by the elements in G of the form [x,y] = x~1yt1xy

is the unique smallest normal subgroup of G such that G/G'

is abelian. By Lemma 4.3, every maximal subgroup of G is

normal in G. Let M be an arbitrary maximal subgroup of

G. Using Lemma 4.3, we find that G/MI = p and thus the

group G/M is cyclic.

G/M is abelian <==> (xM)(yM) = (yM)(xM) for all x,y c G

<==> xyM = yxM for all x,y c G

<==> x~1yt1xyM = M for all x,y e G

<==> G' K M.

If G' K M for every maximal subgroup, M, of G, then

G' (G). Recall that f(G) = 1, therefore G' = 1.

38

Hence G/G' N G is abelian. o

LEMMA 4.5. If G is a p-group and I(G) = 1, then there

exists a prime p such that for all x E G, xP = 1.

Proof. Let ff be the set of maximal subgroups of G.

Let M E !. For all x e G, 1 = x1G -=xIMI ' = (xp)

Therefore, xp e M for all x e G and for all ME Af.

Therefore xp E (G) = 1 for all x E G. Hence, for all

x e G, xP = 1.

Theorem 4.1 is now proved. E

The problem of calculating the M5bius function for a

given p-group, G, has now been reduced to the problem of

calculating the M6bius function for the elementary abelian

group G/f(G). Therefore, our next task is to define the

subgroup structure of elementary abelian groups.

The elementary abelian group G 2 Ep x E x ... x Ep

of order pn is an n-dimensional vector space over the field

E . The elements of this group can be thought of as vectorsp

and the subgroups as subspaces. The total number of

subgroups of order pk for 1 < k < n in the group G is

equivalent to the total number of subspaces of dimension

k in the vector space. Let us consider the following

examples.

Example 4.1. The elements of the elementary abelian

group E2 X 112 are:

E X 2 = { (0,0), (0,1), (1,0), (1,1) }.

39

E2 xI2

{ (0,0), (0,1) } { (0,0), (1,1) }{(0,0) 7(1,0)}

{ (0,0) }

Figure 11

The subgroups of E2 x 112 are:

< (0,0) > = { (0,0) },

< (0,1) > = { (0,0), (0,1) },

< (1,0) > = { (0,0), (1,0) },

< (1,1) > = { (0,0), (1,1) }.

See figure 11.

The vector space E2 x 112 has one subspace of dimension

zero, namely (0,0) which corresponds to the one subgroup of

order 2 . This space also has three subspaces of dimension 1

which correspond to the three subgroups of order 21.

Example 4.2. The elements of the elementary abelian

group 113 x 113 are:

113 x3 = { (0,0), (0,1), (0,2), (1,0), (1,1),

(1,2), (2,0), (2,1), (2,2) }.

40

3 x F 3

< (0,1) > < (1,0) > < (1,c) > < (1,2) >

{ (M,) }

Figure 12

The subgroups of 113 x 113 are:

< (0,0) > = { (0,0) },

< (0,1) > = { (0,0), (0,1), (0,2) },

< (1,0) > = { (0,0), (1,0), (2,0) },

< (1,1) > = { (0,0), (1,1), (2,2) },

< (1,2) > = { (0,0), (1,2), (2,1) }.

See figure 12.

The vector space 113 x 113 has one subspace of dimension

zero , in other words, the group 113 x 113 has one subgroup of

order 30. There are four subspaces with dimension 1 which

correspond to the four subgroups of order 3I.

The number of subspaces of a given dimension k for the

elementary abelian group G 11 x 11 x ... x 11 of order pn-p p p

is found by calculating the total number of sets of k

linearly independent vectors in the space and dividing by

41

the total number of bases in each subspace of dimension k.

If there are m subspaces of dimension k we can conclude

that G has m subgroups of order pk. The following chart

is representation of the calculation of the number of

subgroups.

Number of subgroups

dimension Number of subspaces Order of subgroups

0 1 p 0 =

1 n 1p-I

2 (n - 1)(Pn - P) 2

(p2 1)(P2 P)

3 (pn 1)(n _p)(Pn P2) p3

(p3 - 1)(p3 - p)(p3 - p)

In general

k (n 1(Pn P .(n -Pk-1)

kk 1 )(kk-1

_ (pn 1)(Pn-1 1 )(Pn- k

(Pk k-1 - 1)

The total number of subspaces of dimension k for an

n .elementary abelian group G of order p is denotes as

follows:[n

The task of calculating the number of subgroups of the

group G is now complete.

The next step in the calculation is to determine the

M6bius functional values for each of the subgroups. The

following theorem is a special case of Weisner's Theorem

for subgroup lattices of elementary abelian groups.

42

THEOREM 4.2. (Weisner's Theorem) Let G be an elementary

abelian group and let A be any subgroup of G not equal to G,

Then: p(HG) = 0.H: H l A = {1}

H < G

Proof. By definition we know:

< (H,G) = 0 for all A < G.H: A ( H < G

Let A be a subgroup of the elementary abelian group G.

Partition the subgroups of G by placing each subgroup H of

G in an equivalence class according to the relationship that

exists between A and H. If H1 and H2 are both subgroups of

G such that H, f A = H2 n A then H, and H2 are in the same

partition of G. Let U1 , U2 , ... U be the representative

subgroup from each of then equivalence classes such that

Ui n A = Uj and |Uji <Ujif1. Obviously A = U1 and {1} = U .

Since each subgroup of G is represented in one and only one

equivalence class we can write:

n

0= p(H,G) = ap(H,G)H: H ( G 1i 1 H: H A = Ui

Let A < G such that A G. Suppose that jGj = p. Then A

must be the identity subgroup and,

n

0 = p(H,G) = A(HG)=0.i= 1 H: H n A = Ui H: 1 A = {1}

H< G

Hence the theorem is true when IGI = p.

Now suppose that |Gj = p2 and A = {1} or A is a subgroup

of order p. The first case when A = {1} is trivial and the

43

theorem obviously holds. Let us consider the second case

when JAI = p. In this case the partition is A = U1, U2 = {1}-

n

Therefore, 0 = p(H,G)i=1 H: Hfl A = Ui

p(H,G) + 1 (H,G).H:HIn A = A H: HfnA={i}

Since the collection of subgroups of G that contain A is

isomorphic to the quotient group G/A we know that:

p(HG) = 0,H: n A = A

from which we conclude that:

a(H,G) = 0.H: H n A = {1}

2Thus the theorem is verified when 1G = p2 . For the purpose

of induction let us assume that the theorem is true for all

G such that IGI < k. It is necessary to show that if G is

a group such that the order of G is pk+1 then:

H(HG) = 0H: H n A = 1

for any subgroup A of G such that A t G.

Let A = U1 , U2 ... U = {1} be as previously defined. For

all Ui such that i * n the quotient group G/Uj is of order

less than p k+1 and therefore:

0 = u(H,G/Ui) = Ip(HG) = 0.H < G/Uj H n A = Uj

Since the above is true for all Ui when 1 < i < (n - 1) it

can be concluded that 4(HG) = 0.H n A = {1}

44

{1,2,4,6} 1,)5,6,7} {\1,3,6,8}

{1,2,3,5} {1,3,4,7} {1,4,5,8}

{1,2} {1,3} {1,4} {I,5} {1,6} {1,7} 1,8}

Figure 13

Example 4.3. Figure 13, above, illustrates the

previous theorem on the elementary abelian group:

12 x 112 x 112 = {(OOO), (0,0,1), (0,1,0), (1,0,0),

(0,1,1), (1,0,1), (1,1,0), (1,1,1)}

= { 1, 2, 3, 4, 5, 6, 7, 8}, respectively.

The subgroup structure is depicted. Let A = {1,2,7,8}.

The subgroups {1,3}, {1,4}, {1,5}, {1,6}, and {1} all

intersect A at the identity and the sum of their Mtbius

functional values is equal to zero.

45

It is now possible, through the use of the previous

theorem to calculate an explicit formula for the M~bius

functional values of any given subgroup of an elementary

abelian group. The next theorem provides the formula.

THEOREM 4.3. Let G be an elementary abelian group of

order p n and let Hm be a subgroup of G such that the order of

n-in)Hm is pm, then: pt(Hm, G) = ( 1 )n-m P( 2

Proof. From Theorem 4.2 it is known that for any group G

and any subgroup A of G: H(HG) = 0.H: H n A = {1

H < G

Therefore, p({1}, G) = ~p(H,G) for allH: H fl A {1}{1} H G

subgroups A of G. It suffices to show that for any

elementary abelian group G such that the order of G is pn

( n

(*) pa({1},G) = (-1)n p 2~

When n = 1 the order of G is p. Hence:

p({1},G) = (-1) p = -1.

Thus the theorem is obviously true for n = 1. Next, when

2n = 2 the order of G isp . Let A < G such that the order

of A is p. From Theorem 4.2:

p({t}, G) = ~ p(HG).H: H l A = 1}{1} * H <G

p2There are p= p + 1 one dimensional subspaces in G.

The subgroup A intersects every one dimensional subspace,

46

aside from A itself, at the identity {1}. Each of these

subspaces is a maximal subgroup of G and therefore has a

Mbbius functional value of -1. Therefore,

p({1},G) = - (-1) (p + I -I) = p.

Using (*),2

p(1}G)= (-)2 p =2 p

Thus the formula is verified for n = 2. Let us now proceed

by induction. Assume that for some k > 0 and for some

kelementary abelian group G such that the order of G is p

the theorem is true for all k' < k. By induction, it is

necessary to show that:

k+1

p({1},G) = (-1)pk+1 2 2 for IGI =pk+1

Let G be an elementary abelian group whose order is p

kand let A be a subgroup of G such that the order of A is p .

In order to use Theorem 4.2, it is necessary to find all the

subgroups of G that intersect A at the identity. The only

subgroups that can intersect A at the identity are of order p.

Let H1 be a subgroup of G such that the order of H1 is p.

Since every subgroup of G is normal it is possible to form

the quotient group G/H1 The order of this quotient group

kis p Therefore, by the induction assumption:

p,({I},G/H) = (-1 )k 2

Since u(H 1,G) = p({I},G/H1), we have found that:

k(H = (- 4) 2p ).

47

It is now necessary to determine the number of subgroups of

order p that intersect A at the identity. Since there are

k+1 _ k_1

p - 1 one dimensional subspaces of which p are

subspaces of A. Therefore, there are:

pk+1 _1 pk ~ kp - p- =p

subspaces of dimension one that intersect A at the identity.

Therefore:

p({1}, G) = ~ y(H,G)H: H f A = 1}{1} H <G

- 1 k p( )

H: H f A = 1}{1} H <G

kk -k (2)

(p ) (-1)k p

pk(_1)k+1 p 1+2+3+...+(k-1)

(-1)k+1 p1+2+3...+(k+1)+k

k+1k+1 ( 2 )

=(-1) p

Thus the induction is complete. 0

Given any p-group G we are now able to find the

significant subgroups of G, the order of each significant

subgroup, and the the M~bius functional value of each of

these subgroups. It is now possible to write the Eulerian

function for any given p-group. Let G be a p-group such

that |GI = pr and let 4(G) be the Frattini subgroup of G

and I'P(G)I = pm. We begin by calculating the Eulerian

function on G/4(G).

48

k- mOn (G/= p(Hl/$(G),G/4(G) ) fnHk/4 (G): Hk < G

Where jiki = pand therefore IHk/(G)I = pk-r

For each k = m to r there are M7 ] subgroups isomorphic

to HIk/4(G). Therefore we can write:

n(G/(G))= /(G): I (Hk/.f(G),G/f(G)) 1 k

r

-r- m_

k/ m G

Calculating p(Hk /t (G) , G/t (G))

p (Hk/f(G),G/f(G))

pa(Hk /4(G) , G/4,(G)) = ()(r- m)- (k- m) p 2

r-k

(-1)- p2

Again we rewrite the Eulerian function as:

r n-k) k-..

#n (G/4,(G)) =r- m n- k p 2 Pk =M k

The Eulerian function for the p-group G differs from

the above function only by the order of the subgroups.

The generalized Eulerian function for the p-group G of

order p n is:

n(G) =k_ r- r n

[k-rnm

n-k

p k-rn

CHAPTER BIBLIOGRAPHY

1. R. P. :Burns, Groups, a Path to Geometry, CambridgeUniversity Press, Cambridge, 1985.

2. P. Hall, The Eulerian Functions of a Group, Quart. J.Math., 7 (1936), 134-151.

3. I. N. Herstein, Topics in Algebra, John Wiley & Sons,Toronto, 1975.

4. J. S. Rose, A Course on Croup Theory, CambridgeUniversity Press, Cambridge, 1978, 266-271.

5., L. Weisner, Abstract Theory of Inversion of FiniteSeries, Trans. Amer. Math. Soc. 38 (1935), 474-84.

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BIBLIOGRAPHY

1. R. P. Burns, Groups, a Path to Geometry, CambridgeUniversity Press, Cambridge, 1985.

2. P. Hall, The Eulerian Functions of a Group, Quart. J.Math., 7 (1936), 134-151.

3. I. N. Herstein, Topics in Algebra, John Wiley & Sons,Toronto, 1975.

4. J. S. Rose, A Course on Group Theory, CambridgeUniversity Press, Cambridge, 1978, 266-271.

5. L. Weisner, Abstract Theory of Inversion of FiniteSeries, Trans. Amer. Math. Soc. 38 (1935), 474-84.

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