Nl /vo.lt
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 significance Definition of the Eulerian Function
II. THE MOBIUS FUNCTION...... 4 Definition and calculation of the MObius function The Eulerian function of cyclic groups Frattini subgroup
III. THE EULERIAN FUNCTION OF DIHEDRAL GROUPS...... 14 Classification of subgroups Calculation of Mt*bius function
IV. THE EULERIAN FUNCTION OF P-GROUPS...... 35 Weisners Theorem
BIBLIOGRAPHY...... 50
iii LIST OF ILLUSTRATIONS
Figure Page 1. Mbuis function example...... 2. MObius function example...... 3. Subgroup structure of the integers modulo 12.
4. Subgroup structure of the integers modulo p . . . . .18
5. Subgroup structure of the integers modulo pq. . . . .20
6. Subgroup structure of the Quaternion group...... 4..21
7. Subgroup structure of D2p . . . .2518
8. Subgroup structure of D2p n . 0 20 . . . .39 9. Subgroup structure of D2pq...... 0 21
10. Subgroup structure of D2pqs...... 25 11. Subgroup structure of the elementary abelian group of order 4...... 39 12. Subgroup structure of the elementary abelian group of order 9...... 40 13. Subgroup structure of the elementary abelian group of order 8...... 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 1S2 1 > --- >> 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(S 3 ,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 FT < 0 > Figure 3 integers modulo 12 under addition. The subgroups of E12 are as follows (Figure 3): E1 2 < > ={,,1,i, ,~} < > = < > = {}. 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 pq = < X > [II -1< XP> < - {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: I On (E2) = p(HA12) Hn _ 12 [ 1_ ]+ 2 n IJ J [J + [ We now have a function for determining the number of n-bases that generate 112. For example: 8 01 (11 2 ) = 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> < < -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=1 1 to n. Using the definition of the Mbbius function we know: or K:HK< K p(KG) = 0 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 M But 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 of I 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 for 1 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, the I 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 Ir a, ar a, aj -1 a. ... r + #n(R a,...ar) Pi= * ..'Pr Pi .n.'pj 12 aj-1 ak- 1 par p a. .&P. .. p k r nk - j k al ar- [ a1-1 ... p. ... p + 1 r p ,- -1 ar-fl. n i=1 nr When each ai is equal to 1, we find that the Eulerian function forllr p ..Pris: [ Pi0 -- Pr] P i - -j 1 Pj+1 - Pr On (p .p 1 - ) = j n n + j = 1 k Pi *Pj -iPj +1.*.Pk-1Pk+1 -Pr n r nPi~C& Ln i= 1 CHAPTER BIBLIOGRAPHY 1. R. P. Burns, Groups, a Path to Geometry, Cambridge University 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, Cambridge University Press, Cambridge, 1978, 266-271. 13 CHAPTER III THE MOBIUS FUNCTION OF DIHEDRAL GROUPS The dihedral group D2 11 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 D 2 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 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 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 1 H * D2 n. Let H = some 1 from 1 to n-1. Let R = S = Given that the order of exists a minimal j such that r's,ri 1 s,....,rmi-i 1 s}. Let rYs represent the minimal reflection in H such that H = 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 D2 n 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 order n. For any integer, k, such that k divides n, there is an integer, 1, such that ki = n. Consider 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 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 = 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 and similarly, there exists an integer s 17 such that the order of 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. n Proof. 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)p5E 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 > 2S> -r .. r - -1i -1 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. are p subgroups of order 2pn - I Hence, including H, there 01 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,r 2 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}, 19 2 S5 = {e,r s}, ... , ... , . P- Sp+2 {e,r }, Sp+3 {e}- Hence, the Mbius functional values can be assigned as follows: p(D2 PD 2 p) = 1, u(< r >,D2 P) = p(c s >,D2 P) = rs >,D2P) = = p(< r 1 >,D2 P) = -1, u({e},D2 P)= P- The Eulerian function for D2 P is defined as follows: #m(D 2 p) =14(HD2 P) [ I I,3]= [ 2 - [ 3 - p[ 3 + p . Calculating #m(D 2 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 [P Hence, 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 (D2 P) we find: 1 (D2 =2= p3 - - p[2 = -(2p) (2p- 1) - -(p) (p- ) - p = .2 - Thus verifying the formula for m = 2. 20 D2pn < r > n - D2Pn -1 D2 Pn -1 ...... D < 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, 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 D2 Pn. Therefore, we are concerned only with the subgroups of the form D2pn, p(D 2 pnD2 p") = 1, 21 D2pq D p of these r rq D q of these < s > < rs > < r 2 s ><...... < rP'- 1 > {e} Figure 9 /z( ( u(H,D 2 P) = -1 where H Lv D2 p-ln The Eulerian Function for D2 Pn can now be written as follows: 2 Om(D2 p") = [ p"] 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 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 D2 P. 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. 0 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; al 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; al subgroups of order 2q, as well as D2pq, we find that: pi(< r(1 >, D2pq) = q. 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 D2 P 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(HD 2 pq) = 0 e < 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, 2 ID 2pI = 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] -[2qq [ 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 - ]PS pqs + 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 2 H D2 ...- I, )...p--- 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 these order q < rp > F7-1 D2 ps -- q of order q these < r > q 'order s D2P D q of these D2q .ps of these D2Su . ps of these 1 rrqs > -s order p < r ps > -s order q < r pq > I-p order 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 ,... P 1 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 1dihedral subgroups, 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 [ 3 Type 3 dihedral subgroups; [ ]3Type 2 dihedral subgroups; and [ 4 3Type 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 D2 Pi 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 order number G 2 pP2 - - -Pk I Hi for 1 < i < k 2 pi. Pi-iPi+1...P2k Pi Hij for 1 K D2 p' for 1 K i K k 2pi pP... Pi-iPi+i .. Pk {e ,rs} 2 1 for 0 i -PIP2 - -Pk 1 Hence, we have: . (G)= (G,G) [II + (HG Pi 1=n Pipj + + p(H ,G) Hij ] .. + i ,j=1 i * j +~~~ p(er}G) |e,r s}| ] k 2 2 -[ p121---.Pk [ P ... Pi-xPi+1...Pk Dp 11=1 k P + [ 2p ipn Jpk sn4 PI P2 'Pi-l Pi +i a aPk ns [2-3 29 Next, we calculate the Mbius function on the cyclic subgroups of D2 p - . 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 3I++I pa(HiG)[I'I3 pi j=1 k + p(Hyj ,G) [IHii I]pipj + ... + Ij =1 i * j + j({e,r s},G) I{erS}I k + pa(< r >,G) [ + k (< rP'P >,G) < nrP + ... + ... + + p({e,G)j. 31 k _ 2p, P2 --- Pk _ 2p, ....pi -1 pi +1 . . . Pk P i=1 2 + p . P... Pi -P+1.. Pk PD pi [2Pp2...k [ + 2pi P.2..Pi..Pi+1...Pjj.. P21...pk] P2 i ,j=1. i #j k -I 2 . ] .P P2--- Pi- .i+-1Pk - ] P P2---Pk =1 k P1 ....Pi -IPi+1 ... Pj.-1Pj+1 ... Pk pi +...- i ,j =1. i1 jJ + + 111n1] PI P2*.Pk. 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 aj - 1 ak 2pa' p aj- - *a 2p- 1 ...p ...p 1 j k 1 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 ak 1 k 32 The calculation of the Mtbius function for Dai ak 2p1... 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 1 j m ... p 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. M5bius function Subgroup Type order value < r > p a, pak_ a, ai-1 ak < r Pi > p .p . ... pkPi for 1< < '~rpi>rpi > p a1 ...... ai-1 ..~ a-p1... . Paj-I ak -Pi Pj ak i k for1 <:i,j < k k PI --- Pk a1 -1 ak-i + <>IPk1 k The Eulerian function for D a, ak can now be written as 2p1... Pk 33 f ollows: k at ai ak]- a, ak - 2p 1 ... k 2p ... p a ... p a [- 1 k 1i k pi I J n aj-I -1 akp 2p ... p. aj [ a1 i pk PiPj 1( i ,j < k n ka- ak- +2pp~ 2 a1 -1 ...- piap S k Pi ...-Pi-1Pi +1 - - -Pk n ... p a - a ai-I ak p 1 - a - 1 k + P['*2Pi op k p1 n J 1 n a1 ai- aj-1 ak p .. . p. ... P. .. .p I 1 3 pi pi n I ij I k 1 ai ak-l + p a, n kI Pi ..-PI-lpi+1 ---Pk- n CHAPTER BIBLIOGRAPHY 1. R. P. Burns, Groups, a Path to Geometry, Cambridge University 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, Cambridge University 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 nfo p. 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 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) =1.0 n 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. 1 =xIMI '= (xp) Let M E !. For all x e G, 1 = x G - 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 vectors p 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 1 p-I 2 (n - 1 )(Pn - P) 2 (p2 1)(P 2 P) p3 3 (pn 1 )(n _p)(Pn P2) 3 3 (p3 - 1)(p - p)(p - p) In general k (n 1(Pn P .(n -Pk-1) kk 1 )(kk-1 _ (pn 1)(Pn-1 1 )(Pn- k - 1) (Pk k-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 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 1 H:1i 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: HflA = 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} Thus 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) = 0 H: 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 pn 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 all H: 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 2 n = 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 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 =2p Thus the formula is verified for n = 2. Let us now proceed by induction. Assume that for some k > 0 and for some k elementary 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 k and 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 is p k 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 ~ k p - 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 - 1 k p( ) H: H f A = 1} {1} H k -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+1 k+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- m On (G/= p(Hl/$(G),G/4(G) ) fn Hk/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 p k-rn -r- m_ p (Hk/f(G),G/f(G)) k/ m G Calculating p(Hk /t (G) , G/t (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 P k =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-k k_ r- rn n(G) = [k-rnm CHAPTER BIBLIOGRAPHY 1. R. P. :Burns, Groups, a Path to Geometry, Cambridge University 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, Cambridge University Press, Cambridge, 1978, 266-271. 5., L. Weisner, Abstract Theory of Inversion of Finite Series, Trans. Amer. Math. Soc. 38 (1935), 474-84. 49 BIBLIOGRAPHY 1. R. P. Burns, Groups, a Path to Geometry, Cambridge University 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, Cambridge University Press, Cambridge, 1978, 266-271. 5. L. Weisner, Abstract Theory of Inversion of Finite Series, Trans. Amer. Math. Soc. 38 (1935), 474-84. 50