On the Number of Maximal Subgroups in a Finite Group

Home , ATLAS of Finite Groups, Maximal subgroup

J. Group Theory 18 (2015), 535–551 DOI 10.1515/jgth-2015-0001 © de Gruyter 2015

On the number of maximal subgroups in a ﬁnite group

L.-K. Lauderdale Communicated by Nigel Boston

Abstract. For a ﬁnite group G, let m.G/ denote the set of maximal subgroups of G and .G/ denote the set of primes which divide G . In [4], it is proven that m.G/ .G/ when G is cyclic and if G is noncyclic m.G/j j .G/ p, wherej p .G/j D j is thej smallest prime that divides G . In thisj paper,j we j producej C two new lower2 bounds for m.G/ , both of which considerj j all of the primes in .G/. j j

1 Introduction

For any ﬁnite group G, let .G/ denote the set of primes which divide G and let j j m.G/ denote the set of maximal subgroups of G. Lower bounds for the number of maximal subgroups have previously been investigated. In [4], three lower bounds are proven. In general, m.G/ .G/ and equality holds if and only if G is j j j j cyclic. Under further assumptions on the structure of G, this lower bound was improved. If G is noncyclic, then m.G/ .G/ p, where p .G/ is the j j j j C 2 smallest prime that divides G . Additionally, if G has a noncyclic Sylow subgroup j j and q .G/ is the smallest prime such that Q Syl .G/ is noncyclic, then it is 2 2 q proven that m.G/ .G/ q: (1.1) j j j j C Observe that given any set of distinct primes q1; q2; : : : ; qn where q1 is the ¹ º smallest, there exists a ﬁnite group G with

.G/ q1; q2; : : : ; qn and m.G/ .G/ q1: D ¹ º j j D j j C

For example, G Zq1 Zq1q2 qn achieves the lower bound in (1.1). However, D the previously stated lower bounds can be restrictive because they consider at most one prime in .G/. For instance, let G S10 Z11 Z143, where S10 D is the symmetric group on ten letters. Then .G/ 6 and G has a noncyclic j j D Sylow 2-subgroup. Therefore q 2 and equation (1.1) yields m.G/ 8, how- D j j ever m.G/ 4002. The lower bound in (1.1) can be drastically improved with j j D minimal assumptions on the structure of the group. 536 L.-K. Lauderdale

In the present paper, we improve the existing lower bound for m.G/ by par- j j titioning .G/ into three sets. We consider the set of primes p for which G is not p-solvable, denoted .G/, and the set of primes q for which G is q-solvable. We further subdivide the latter set of primes into the set of primes which correspond to cyclic Sylow subgroups of G, denoted .G/, and the set of primes which correspond to noncyclic Sylow subgroups of G, denoted .G/. Under these con- siderations we obtain two new lower bounds for m.G/ . In addition to the sets j j .G/, .G/ and .G/, the first lower bound includes a further assumption on G. We assume that the nonabelian composition factors of G satisfy a technical property, namely Property B, which is defined in Definition 2.5.

Theorem 1.1. Let G be a finite group. If the nonabelian composition factors of G satisfy Property B, then X X m.G/ .G/ p p: (1.2) j j j j C C p Ä.G/ p .G/ 2 2 Property B is not a very restrictive property and we conclude this paper by pro- viding some examples of simple groups that satisfy Property B (see Section 8). However, not all finite simple groups satisfy Property B and knowing the composition factors of a group may increase the possible lower bound. For a group G, we introduce below the invariant ˛.G/ which takes into account more finely the influence of the composition factors on our problem. The following theorem does not require any special condition on the composition factors.

Theorem 1.2. If G is a finite group, then X X m.G/ .G/ .p 1/ ˛.G/ .p 1/: (1.3) j j j j C C C C p Ä.G/ p .G/ 2 2 Notice that Theorem 1.2 implies Theorem 1.1 when ˛.G/ 1. The invariant ˛.G/ is closely connected to the values ˛.C/ as C runs over the nonabelian composition factors of G. In the penultimate section (see Section 7), we analyze ˛.G/ and see that ˛.C/ tends to infinity for many families of finite simple groups C. Although, we provide many examples of finite simple groups which satisfy both ˛.C/ > 1 and Property B, our results do not use the classification of finite simple groups. Both Theorem 1.1 and Theorem 1.2 improve the lower bound in equation (1.1). Reconsidering the group G S10 Z11 Z143, the lower bound in (1.3) yields D m.G/ 4001, a vast improvement over the lower bound in (1.1). In most cases j j Theorem 1.2 gives a stronger result than Theorem 1.1. However, it is possible On the number of maximal subgroups in a finite group 537 for the lower bound in (1.2) to be sharper than the lower bound in (1.3) because not all finite groups G satisfy ˛.G/ 1. For instance, ˛.A6/ < 1 where A6 is the alternating group on six letters, and ˛.PSL.2; 7// < 1. It can be shown that both A6 and PSL.2; 7/ satisfy Property B and thus Theorem 1.1 is applicable. Theorem 1.1 gives the sharper bound for G PSL.2; 7/ Z11 Z143. In this D case, the lower bound in (1.3) yields m.G/ 21, but the lower bound in (1.2) j j yields m.G/ 28. j j

2 Deﬁnitions

To improve the existing lower bounds for m.G/ , we ﬁrst partition .G/ into three j j sets: .G/, .G/ and .G/.

Definition 2.1. Let G be a finite group. Define .G/ as the set of p .G/ such that G is p-solvable and P Syl .G/ 2 2 p is cyclic. Define .G/ as the set of p .G/ such that G is p-solvable and P Syl .G/ 2 2 p is noncyclic. Define .G/ as the set of p .G/ such that p divides the order of a nonabel- 2 ian chief factor of G.

To aid in counting some of the maximal subgroups of G with index a .G/-number, we make the following deﬁnitions:

Deﬁnition 2.2. Let S be a nonabelian simple group and assume M is a maximal subgroup of S. We say that M satisﬁes Property A if N .M /S Aut.S/. Let Aut.S/ D mA.S/ denote the set of maximal subgroups of S that satisfy Property A.

Definition 2.3. Let G be a finite group and suppose C is a nonabelian composition P P factor of G. Define .C/ p .C/.p 1/ and .G/ p .G/.p 1/. D 2 C D 2 C Definition 2.4. Let G be a finite group and suppose C is a nonabelian composition mA.C/ factor of G. Set ˛.C/ j j . If G is not solvable, set D .C/ ˛.G/ min ˛.C/ C is a nonabelian composition factor of G D ¹ W º and if G is solvable, then set ˛.G/ 0. D Definition 2.5. Let S be a nonabelian finite simple group. We say that S satisfies Property B if m.G/ .S/, for all subgroups G such that S G Aut.S/. j j Ä Ä 538 L.-K. Lauderdale

3 Preliminary results

In this section, we collect a few results which will aid in the proof of the main theorems.

Lemma 3.1. Let p be a prime number and suppose that P is a noncyclic p-group. Let P=ˆ.P / be the elementary abelian group of rank k, where k 2 and ˆ.P / denotes the Frattini subgroup of P . Then pk 1 m.P / : j j D p 1 In particular, m.P / p 1. j j C Proof. Since P=ˆ.P / is an elementary abelian group of rank k, we have pk 1 m.P=ˆ.P // : j j D p 1 Because ˆ.P / is contained in every maximal subgroup of P , pk 1 m.P / m.P=ˆ.P // : j j D j j D p 1 pk 1 Observe that p 1 is smallest when k 2. In this case, D pk 1 p 1; p 1 D C so that m.P / p 1. j j C Theorem 3.2 (Maschke). Let G be a ﬁnite group and let F be a ﬁeld whose char- acteristic does not divide G . If V is any FG-module and U is any submodule j j of V , then V has a submodule W such that V U W . D ˚ Proof. See for example [2, Section 18.1, Theorem 1].

Theorem 3.3. For a ﬁnite cyclic group G, m.G/ .G/ . j j D j j Proof. See for example [4, Theorem 3.1].

Lemma 3.4. Suppose that G is a ﬁnite group, and let .G/ p1; p2; : : : ; pn . D ¹ º For each i 1; 2; : : : ; n , let Pi Syl .G/. If G P1 P2 Pn, then 2 ¹ º 2 pi D m.G/ m.P1/ m.P2/ m.Pn/ . j j D j j C j j C C j j Proof. See [4, Lemma 2.6]. On the number of maximal subgroups in a ﬁnite group 539

We conclude this section with some number theoretic results which will aid us in bounding both the number of maximal subgroups of a group that satisfy Property A as well as groups which satisfy Property B.

Lemma 3.5. Let n 3 be an integer. Suppose S is a set of n distinct primes. Assume that p and q are the largest and smallest elements of S, respectively. Then p n q. C Proof. Observe that all elements of S are at least two units apart from each other, except for possibly the two smallest elements of S which are one unit apart. Therefore p q 2.n 2/ 1, which is at least n as n 3. So p q n and C p n q, as desired. C Lemma 3.6. Let l 3 be an integer. If p1; p2; : : : ; p are prime numbers such l that p1 < p2 < < p , then l l X p l .pi 1/: l C i 1 D Proof. Since l 3, observe that p > pi 1 for all i 1; 2; : : : ; l 1 . Hence l C 2 ¹ º l 1 X p l > p .pi 1/ l l C C i 1 D and thus l X p l .pi 1/; l C i 1 D as desired.

Lemma 3.7. Suppose n 9 is an odd integer. If n is not a prime number, then X n .p 1/ 2: C C p n j Proof. If n is a prime power, say n pk, then n p2 p 3 and the statement D C holds. So we assume at least two distinct primes divide n and proceed by induction on the number of prime divisors of n. Note if the statement holds for the product of all distinct primes dividing n, then it will hold for n as well. Thus it sufﬁces to prove the statement when n is square free. To that end, suppose n p1p2 p , D l where p1 > p2 > > p . Then p1 5 and .p1 1/.p2 1/ 5, so that l 2 X p1p2 .pi 1/ 2: C C i 1 D 540 L.-K. Lauderdale

Now assume l 1 X p1p2 pl 1 .pi 1/ 2: C C i 1 D Because .p1p2 pl 1 1/.pl 1/ 2,

n p1p2 pl p1p2 pl 1 pl 1 D C C and we have

n p1p2 pl 1 pl 1 C C l 1 X .pi 1/ 2 p 1 C C C l C i 1 D l X .pi 1/ 2 D C C i 1 XD .p 1/ 2; D C C p n j as desired.

Lemma 3.8. If q 2f 8, then D X 2.q 1/ .p 1/: C C p .q3 q/ j Proof. To prove this result, we divide the proof into three cases. First suppose q 1 is prime. Then q 1 is not a prime number and C X q 1 .p 1/ 2 C C p .q 1/ j by Lemma 3.7. Adding .q 1/ 2 to both sides of the inequality above yields C C X 2.q 1/ .p 1/: C C p .q3 q/ j Thus assume that q 1 is prime. Then q 1 is not prime and C X q 1 .p 1/ 2 C C C p .q 1/ j C On the number of maximal subgroups in a ﬁnite group 541 by Lemma 3.7. Adding .q 1/ 2 to both sides of the inequality above yields C X 2.q 1/ .p 1/: C C p .q3 q/ j Finally, suppose that q 1 nor q 1 is prime. Then Lemma 3.7 implies C X q 1 q 1 .p 1/ 4 C C C C p .q2 1/ j and X 2.q 1/ .p 1/: C C p .q3 q/ j

4 m.G/ for p-solvable groups j j For a finite solvable group, we will make use of the well-known fact that every maximal subgroup has prime power index. For p .G/, we count the number 2 of maximal subgroups of G by counting the number of maximal subgroups whose index in G is a power of p. We denote mp.G/ as the set of maximal subgroups of G whose index in G is a power of p. The first two lemmas of this section are standard results which will aid us in bounding mp.G/ from below. j j Lemma 4.1. Let G be a finite p-solvable group, where p .G/. If H is a Hall 2 p0-subgroup of G and K is a subgroup which contains NG.H /, then K is self- normalizing.

g g Proof. Let g G satisfy K K. Then H is a Hall p0-subgroup of K. Since all 2 D g n Hall p0-subgroups of K are conjugate, there exists an n K such that H H. 1 2 D So gn NG.H / and g NG.H /n K. Therefore K is self-normalizing. 2 2 Â Lemma 4.2. Let N be a solvable normal subgroup of a ﬁnite group G. Suppose that H is a Hall subgroup of N . Then G NG.H /N . D Proof. Let g G. Since H g N g N and H g has the same order of a Hall 2 Â D subgroup of N , H g is a Hall subgroup of N . Therefore H and H g are N -conjugate, say H g n H for some n N . Then D 2 1 gn NG.H / and g NG.H /n NG.H /N : 2 2 Â Therefore G NG.H /N , as desired. D Let p .G/ for a ﬁnite group G. Recall that Op.G/ is the largest normal 2 p-subgroup of G. The following theorem gives a lower bound on mp.G/ . j j 542 L.-K. Lauderdale

Theorem 4.3. Suppose G is a finite group and assume p .G/ .G/. Then 2 [ mp.G/ 1 and mp.G/ p 1 when p .G/. j j j j C 2 Proof. Suppose this is false. Let G be a counterexample of minimum order and assume P Syl .G/. Since G has a maximal subgroup which contains a Hall 2 p p -subgroup, mp.G/ 1. Therefore mp.G/ < p 1 and P is noncyclic. Also, 0 j j j j C it follows from Lemma 3.1 that G is not a p-group. First, suppose that P E G. Then ˆ.P / E G and P=ˆ.P / is noncyclic. By minimality, ˆ.P / 1 and so P is elementary abelian. Let H be a Hall p -subgroup D 0 of G. Since ŒG H p2, H cannot be a maximal subgroup of G, otherwise it W would have at least p2 conjugates. Let M be a maximal subgroup containing H and set P1 M P . Since P1 E G, it is H-invariant. By Maschke’s Theorem D \ (Theorem 3.2), there exists a nontrivial proper H-invariant subgroup P2 of P such that P P1 P2. Now H must act non-trivially on at least one of P1 or P2. D Without loss of generality, assume H acts non-trivially on P1. Let M be a maximal subgroup which contains NG.H /P2. Then M is self-normalizing by Lemma 4.1, and so the p conjugates of M will not contain P1 and are elements of mp.G/. If L is a maximal subgroup which contains P1H , then L mp.G/ and is distinct from 2 the aforementioned p maximal subgroups of mp.G/. Hence mp.G/ p 1, j j C a contradiction. It follows that P E G and Op.G/ is a proper subgroup of P . Let J be a Hall 6 p -subgroup of O .G/, where O .G/ is the subgroup K of G such that 0 p; p0 p; p0 K= Op.G/ Op .G= Op.G//. Then J 1 and D 0 ¤ G NG.J / Op; p .G/ NG.J /.J Op.G// NG.J / Op.G/; D 0 D D where the first equality holds by Lemma 4.2. Notice that NG.J / is a proper subgroup of G because Op .G/ 1 by minimality. If M is a maximal subgroup 0 D of G which contains NG.J /, it is self-normalizing by Lemma 4.1. So M has at least p conjugates in mp.G/, all of which do not contain Op.G/. By minimality, mp.G= Op.G// 1 and so there is at least one maximal subgroup in mp.G/ j j which contains Op.G/. Therefore mp.G/ p 1, a final contradiction. j j C Observe that the theorem above proves both Theorem 1.1 and Theorem 1.2 for all finite solvable groups. Suppose G is a finite solvable group. Then ˛.G/ 0 D and the nonabelian composition factors G satisfy Property B. Thus, Theorem 4.3 yields at least X X .G/ .p 1/ .G/ p j j C C D j j C p Ä.G/ p Ä.G/ 2 2 distinct maximal subgroups of G. On the number of maximal subgroups in a finite group 543

5 m.G/ for nonsolvable groups j j Throughout this section we will make use of the fact that every minimal normal subgroup of a ﬁnite group is the direct product of isomorphic simple groups. For a nonsolvable group, we will count the maximal subgroups by separating them into those that contain a minimal normal subgroup N and those that do not contain N .

Lemma 5.1. Let N be a nonabelian minimal normal subgroup of a finite group G and suppose that N S1 S2 S , where Si is a nonabelian simple group D l and Si Sj for all i; j 1; 2; : : : ; l . Let S Si and fix some 'i S Si to Š 2 ¹ º Š W ! be an isomorphism. Let M be a maximal subgroup of S and define

QM '1.M / '2.M / ' .M /: D l

If M satisfies Property A, then G N NG.QM /. D Proof. As N E G, it follows that N NG.QM / is a subgroup of G. Let g G and 1 2 observe that gQM g is a proper subgroup of N . Define the map g N N 1 W ! by g .n/ gng for all n N . Then g Aut.N /. So g will send minimal D 2 2 normal subgroups of N to minimal normal subgroups of N , thus g permutes the minimal normal subgroups of N . Let Sl be the symmetric group on l objects. Then 1 there exists a Sl such that gS 1.i/g Si , for all i 1; 2; : : : ; l . 2 D 1 2 ¹ 1 º Define the map g;i S S by g;i .s/ ' .g' 1.i/.s/g / for all s S. W ! D i 2 Then g;i Aut.S/ for all i 1; 2; : : : ; l . Since Aut.S/ N .M /S, there 2 2 ¹ º D Aut.S/ exists some sg; 1.i/ S such that 2 1 1 1 sg; 1.i/M sg; 1.i/ 'i .g' 1.i/.M /g /: D

Applying 'i , we obtain the equality

1 1 'i .sg; 1.i/M sg; 1.i// g' 1.i/.M /g : D

Set ng .'1.sg; 1.1//; '2.sg; 1.2//; : : : ; 'l .sg; 1.l/// N . Then D 2 1 1 1 gQM g g' 1.1/.M /g g' 1.l/.M /g D 1 1 '1.sg; 1.1/M sg; 1.1// 'l .sg; 1.l/M sg; 1.l// D 1 ng .'1.M / ' .M //n D l g 1 ng QM n : D g 1 1 1 So n gQM g ng QM and thus n g NG.QM / and g ng NG.QM /. g D g 2 2 Since ng N , g N NG.QM / and G N NG.QM /. 2 2 D 544 L.-K. Lauderdale

Lemma 5.2. Let N be a nonabelian minimal normal subgroup of a finite group G and suppose that N S1 S2 S , where Si is a nonabelian simple group D l and Si Sj for all i; j 1; 2; : : : ; l . Let S Si and assume M is a maximal Š 2 ¹ º Š subgroup of S which satisfies Property A. Fix some 'i S Si to be an iso- W ! morphism and define QM '1.M / '2.M / 'm.M /. Then NG.QM / is D a maximal subgroup of G.

Proof. Since N is a minimal normal subgroup of G and QM N , it follows that NG.QM / is a proper subgroup of G. Suppose K is a proper subgroup of G which contains NG.QM /. Then N is not contained in K by Lemma 5.1. So there exists some i 1; 2; : : : ; l such that K does not contain Si , but K Si 'i .M /. 2 ¹ º \ D Observe that G acts transitively on S1;S2;:::;S because N is a minimal ¹ l º normal subgroup and N acts trivially on S1;S2;:::;S . Therefore K acts tran- ¹ l º sitively on S1;S2;:::;Sl because G KN . For each j 1; 2; : : : ; l , there is ¹ º 1 D 2 ¹ 1 º some kj K such that kj Si k Sj and K Sj kj .K Si /k . Therefore 2 j D \ D \ j K Sj Sj and so K Sj 'j .M / as 'j .M / K Sj . So \ ¤ \ D Â \ K N QM and NG.QM / N QM : \ D \ D Hence

G KN NG.QM /N and K NG.QM /.K N/ NG.QM /: D D D \ D Therefore NG.QM / is a maximal subgroup of G. The following two lemmas count the number of maximal subgroups which do not contain N , the ﬁrst lemma using Property A and the second lemma using Property B.

Lemma 5.3. Suppose N is a nonabelian minimal normal subgroup of a ﬁnite group G and assume N S1 S2 S , where Si is a simple group and D l Si Sj for all i; j 1; 2; : : : ; l . Then there are at least mA.S1/ maximal Š 2 ¹ º j j subgroups of G which do not contain N .

Proof. Let S Si , i 1; 2; : : : ; l . Fix some 'i S Si to be an isomorphism. Š 2 ¹ º W ! Set n mA.S/ , so that S has n maximal subgroups, say M1;M2;:::;Mn, such D j j that N .Mj /S Aut.S/ for all j 1; 2; : : : ; n . Deﬁne Aut.S/ D 2 ¹ º QM '1.Mj / '2.Mj / ' .Mj /: j D l Then G N NG.QM / by Lemma 5.1 and NG.QM / is a maximal subgroup D j j of G by Lemma 5.2. By the reasoning in Lemma 5.2, NG.QM / N QM j \ D j and so the maximal subgroups NG.QMj / are all distinct. Therefore G has at least n maximal subgroups which do not contain N , as desired. On the number of maximal subgroups in a ﬁnite group 545

Lemma 5.4. Suppose N is a nonabelian minimal normal subgroup of a ﬁnite group G and assume N is of the form N S1 S2 S , where Si Sj D l Š for all i; j 1; 2; : : : ; l . If S1 satisﬁes Property B, then there are at least .S1/ 2 ¹ º maximal subgroups of G which do not contain N .

Proof. Let p be the largest prime in .N / and suppose P is a Sylow p-subgroup of N . Then G N NG.P / by the Frattini Argument. Because P N and N D is a minimal normal subgroup of G, NG.P / G and there exists a maximal ¤ subgroup L of G containing NG.P /. Observe that N L because otherwise 6Â G NL L. Therefore L is maximal subgroup of G which does not contain N . D Â For i 1; 2; : : : ; l , deﬁne Ri Si L. Observe that L acts transitively on 2 ¹ º D \ S1;S2;:::;S . Thus Ri and Rj are L-conjugate for i; j 1; 2; : : : ; l and have ¹ l º 2 ¹ º the same order. Since N is the direct product of isomorphic simple groups, Si is not contained in L for all i 1; 2; : : : ; l . Also Si Ri 1 and Ri is not a normal 2 ¹ º ¤ ¤ subgroup of Si . It follows that L is not a normal subgroup of G, and thus L is a self-normalizing maximal subgroup of G. Because Si acts by left multiplication on the set of left cosets of Ri in Si , there exists a nontrivial homomorphism from Si to the symmetric group on ŒSi Ri W elements and Si must be isomorphic to a subgroup of the alternating group on ŒSi Ri elements. Thus p ŒSi Ri . Let Ti be the image of the projection map W Ä W from N L to Si . We claim that Ri Ti for all i, so that \ D l ŒG L ŒN N L ŒS1 S2 S R1 R2 R p : W D W \ D l W l

Clearly, one has Ri Ti . To prove the reverse containment, observe Ri E N L Â \ because Si E N . So Ri E Ti as Ri maps to itself under this map. Since Si is simple, Ti is a proper subgroup of Si . Then U is a proper subgroup of N , where U T1 T2 T . Therefore U is normalized by L and UL is a proper D l subgroup of G. Since L is maximal, U L and Ti Ri for all i. Hence Ri Ti Â Â D and ŒG L pl and we obtain at least pl distinct maximal subgroups of G which W do not contain N . If l 2, then G has at least p .N / maximal subgroups which do not con- j j tain N as p > .N / by Lemma 3.5. Since p .N / p .S1/ .S1/ by j j j j D j j Lemma 3.6, there are at least .S1/ maximal subgroups which do not contain N . So assume that l 1. Then N S1 and N CG.N / E G. So G=CG.N / is iso- D D morphic to H, where N H Aut.N /. Since N satisﬁes Property B, there are at Ä Ä least m.H / .N / .S1/ maximal subgroups of G which do not contain N , j j D as desired.

We conclude this section with two theorems that allow us to count some of the maximal subgroups of a nonsolvable group by their index. 546 L.-K. Lauderdale

Theorem 5.5. If G is a ﬁnite group, then it has at least .G/˛.G/ maximal subgroups of G with index a .G/-number.

Proof. Suppose this is false. Let G be a counterexample of minimum order. Assume N is a minimal normal subgroup of G such that N S1 S2 S , Š l where Si is a simple group and Si Sj for all i; j 1; 2; : : : ; l . First, suppose Š 2 ¹ º that N is solvable. Then .G/ .G=N / and ˛.G/ ˛.G=N /. By minimality, D D G=N has at least .G/˛.G/ maximal subgroups with index a .G/-number. Thus G has at least .G/˛.G/ maximal subgroups which contain N and have index a .G/-number, a contradiction. Therefore N is not solvable. By minimality, G=N has at least .G=N /˛.G=N / maximal subgroups with index a .G=N /-number. It follows that the group G has at least .G=N /˛.G=N / maximal subgroups which contain N and have index a .G/-number. In addition, there are at least mA.S1/ maximal subgroups which j j do not contain N by Lemma 5.3. Observe that

mA.S1/ mA.S1/ ˛.N / ˛.S1/ j j j j; D D .S1/ D .N / and thus mA.S1/ .N /˛.N /. In total, G has at least j j D .G=N /˛.G=N / .N /˛.N / .G/˛.G/ C maximal subgroups with index a .G/-number, a ﬁnal contradiction.

Theorem 5.6. Let G be a ﬁnite group. If the nonabelian composition factors of G satisfy Property B, then there exists at least .G/ maximal subgroups of G with index a .G/-number.

Proof. Suppose this is false. Let G be a counterexample of minimum order and assume N is a minimal normal subgroup of G. First, suppose that N is solvable. Then .G/ .G=N / and G=N has at least .G/ maximal subgroups with index D a .G/-number, by minimality. Thus G has at least .G/ maximal subgroups which contain N and have index a .G/-number, a contradiction. Therefore N is not solvable. Suppose N S1 S2 S , where Si is D l a nonabelian simple group and Si Sj for all i; j 1; 2; : : : ; l . By minimal- Š 2 ¹ º ity, G=N has at least .G=N / maximal subgroups with index a .G=N /-number. So G has at least .G=N / maximal subgroups which contain N and have index a .G/-number. By Lemma 5.4, G has at least .S1/ maximal subgroups which do not contain N . In total, G has at least

.G=N / .S1/ .G=N / .N / .G/ C D C maximal subgroups with index a .G/-number, a ﬁnal contradiction. On the number of maximal subgroups in a ﬁnite group 547

6 Proof of main theorems

Proof of Theorem 1.2. We obtain at least .G/ P .p 1/ distinct max- j j C p Ä.G/ C imal subgroups of G with index a .G/ .G/-number2 by Theorem 4.3. By [ Theorem 5.5, G has at least .G/˛.G/ maximal subgroups of G with index a .G/-number. Therefore, X m.G/ .G/ .p 1/ .G/˛.G/; j j j j C C C p Ä.G/ 2 as desired.

Proof of Theorem 1.1. We obtain at least .G/ P .p 1/ distinct max- j j C p Ä.G/ C imal subgroups of G with index a .G/ .G/-number2 by Theorem 4.3. From [ Theorem 5.6, we infer that G has at least .G/ maximal subgroups of G with index a .G/-number. Therefore, X X X m.G/ .G/ .p 1/ .G/ .G/ p p; j j j j C C C D j j C C p Ä.G/ p Ä.G/ p .G/ 2 2 2 as desired.

We conclude this section by proving the bound in Theorem 1.2 is best possible for solvable groups. Note the same example proves that the lower bound in Theo- rem 1.1 is best possible for solvable groups. There exist nonsolvable groups which demonstrate the lower bound in Theorem 1.2 is best possible. For example, suppose G is a ﬁnite simple group such that G Aut.G/ so that m.G/ mA.G/. D D Then .G/ .G/ and m.G/ mA.G/ .G/˛.G/, thus G achieves D;D j j D j j D the lower bound in Theorem 1.2.

Proposition 6.1. Given any ﬁnite set of primes and any subset of , there N exists a ﬁnite solvable group G with .G/ , .G/ and D D N X m.G/ .G/ .p 1/ .G/˛.G/: j j D j j C C C p Ä.G/ 2

Proof. Suppose that p1; p2; : : : ; pn and p1; p2; : : : ; p , 0 l n. D ¹ º N D ¹ l º Ä Ä Consider the noncyclic group

G Zp1 Zp1 Zp2 Zp2 Zpl Zpl Zpl 1 Zpl 2 Zpn ; D C C 2 2 2 which is of order p1p2 pl pl 1pl 2 pn. Then .G/, .G/ and C C D N D ˛.G/ 0. By Lemma 3.1, D m.Zp Zp / pi 1; i 1; 2; : : : ; l ; j i i j D C 2 ¹ º 548 L.-K. Lauderdale

and observe that m.Zp / 1 for i l 1; : : : ; n . Thus j i j D 2 ¹ C º l X m.G/ .G/ .pi 1/; j j D j j C C i 1 D by Lemma 3.4, and the lower bound in Theorem 1.2 is best possible.

7 Remarks on ˛.G/

In this section, we give some groups G which satisfy ˛.G/ > 1.

Lemma 7.1. Let An denote the simple alternating group on n letters. If n 9, then ˛.An/ > n: 21 10 127 Otherwise ˛.A5/ , ˛.A6/ , ˛.A7/ 3 and ˛.A8/ . D 13 D 13 D D 21 Proof. If 5 n 11, the result holds by the Atlas of Finite Groups [5]. So we Ä Ä assume that n 12. Let Sn denote the symmetric group on n letters. For each n integer k with 1 k < 2 , .Sk Sn k/ An .Ak An k/:2 is a maximal Ä \ D subgroup of An by [3]. Each of these maximal subgroups of An have ! nŠ n ŒAn .Sn Sn k/ An W \ D kŠ.n k/Š D k n conjugates in An. Set r.n/ to be the largest integer strictly less than 2 and ! ! ! n n n s.n/ : D 1 C 2 C C r.n/

Let M be one of the aforementioned s.n/ maximal subgroups of An. Since the automorphism group of An is Sn and M is normalized by an element of Sn An, n it follows that N .M /An Aut.An/ and mA.An/ s.n/. Aut.An/ D j j It remains to show that s.n/ > n .An/ for n 12. To that end, note n ! X n 2n: i D i 0 D Therefore ´ 2n 1 1 1 n ; for n even; s.n/ 2 n=2 D 2n 1 1; for n odd; On the number of maximal subgroups in a ﬁnite group 549

n 2 1 n n 2 and so s.n/ 2 for all n because 2 n=2 > 2 . Further, observe that there n are less than 2 primes in .An/. Since each of these primes is no larger than n, we have X ÂnÃ n .An/ n .p 1/ < n .n 1/: D C 2 C p .An/ 2 Consequently, Â Ã n n 2 n .An/ < n .n 1/ 2 s.n/; 2 C Ä Ä where the penultimate inequality holds because n 12. Lemma 7.2. Let q 2f . If q 8, then D q2 q 1 ˛.PSL.2; q// C C : 2.q 1/ C If q 4, then ˛.PSL.2; q// 21 . D D 13 Proof. Let S PSL.2; q/. If q 4, the result holds by the result holds by the D D Atlas of Finite Groups [5]. So assume that q 8. Then D2.q 1/, D2.q 1/ and f Ì C Z2 Zq 1 are maximal subgroups of S by Dickson’s Theorem (see [1]). Since S is simple, 3 f 1 D2.q 1/ has Œq q 2.q 1/ 2 .q 1/ conjugates, W D C 3 f 1 D2.q 1/ has Œq q 2.q 1/ 2 .q 1/ conjugates, C W C D 3 f Ì Œq q Z2 Zq 1 q 1. W D C Thus S has at least q2 q 1 distinct maximal subgroups. The outer automor- C C phism group of S is generated by a ﬁeld automorphism ' of order f , which ﬁxes all of the maximal subgroups in the three aforementioned conjugacy classes 2 3 of maximal subgroups. Therefore mA.S/ q q 1. Since S q q, j j C C j j D Lemma 3.8 implies q2 q 1 q2 q 1 ˛.S/ P C C C C ; p .S/.p 1/ 2.q 1/ 2 C C as desired.

Lemma 7.3. Let q pf , where p is an odd prime number. If q 13, then D q2 ˛.PSL.2; q// : 2.q 1/ C Otherwise ˛.PSL.2; 7// 8 , ˛.PSL.2; 9// 10 and ˛.PSL.2; 11// 67 . D 15 D 13 D 25 Proof. This proof is omitted due to its similarities to the proof of Lemma 7.2. 550 L.-K. Lauderdale

We conclude this section by giving values of ˛.G/, where G is a sporadic simple group. The following values given in the table below were calculated from the Atlas of Finite Groups [5].

G ˛.G/ ˛.G/ G ˛.G/ ˛.G/ b c b c 309 703;131;313 M11 25 12 Fi22 47 14; 960; 240 2;850 10 10 M12 25 114 Fi23 > 10 > 10 M 1;948 59 Fi > 1010 > 1010 22 33 240 44;414 c 495;529 M23 57 779 M L 33 15; 016 1;507;537 6;862;629 M24 57 26; 448 He 39 175; 964 13;016 367;882;130 J1 53 245 Ru 65 5; 659; 725 16;964 291;965;788 J2 21 807 Suz 47 6; 212; 038 68;410 8;935;530;772 J3 33 2; 073 HN 53 168; 594; 920 10 10 83;897;408 J4 > 10 > 10 O0N 85 987; 028 424;818;005 10 10 Co3 57 7; 452; 947 Th > 10 > 10 3;581;796;533 10 10 Co2 57 62; 838; 535 Ly > 10 > 10 10 10 10 10 Co1 > 10 > 10 B > 10 > 10 68;410 10 10 HS 33 2; 073 M > 10 > 10

8 Remarks on Property B

In this section, we give some examples of ﬁnite simple groups which satisfy Property B.

Lemma 8.1. If n 5, then An satisfies Property B. Proof. If n 6, the result holds by the Atlas of Finite Groups [5]. So assume D that n 6. Since Aut.An/ Sn, the result follows from Lemma 7.1. ¤ D Lemma 8.2. Suppose q pf , where p is a prime number. If q 4, then PSL.2; q/ satisfies Property B. D On the number of maximal subgroups in a finite group 551

Proof. If q 7; 9 , then the result holds by the Atlas of Finite Groups [5]. So 2 ¹ º assume that q 7; 9 . Since Out.PSL.2; q// Z.2;q 1/ Zf , the result follows 62 ¹ º Š from Lemma 7.2 and Lemma 7.3.

Acknowledgments. This paper is a part of the author’s Ph.D. thesis at the Univer- sity of Florida under the direction of Dr. Alexandre Turull. The author wishes to express the deepest appreciation for his continued support. The author would also like to thank the referee for his/her promptness and valuable comments.

Bibliography

[1] L. E. Dickson, Linear Groups: With an Exposition of the Galois Field Theory, Dover Publication, New York, 1958. [2] D. S. Dummit and R. M. Foote, Abstract Algebra, 3rd ed., John Wiley & Sons, Hoboken, 2004. [3] J. Fisher and J. McKay, The nonabelian simple groups g, g < 106 – Maximal subgroups, Math. Comput. 32 (1978), 1293–1302. j j [4] L.-K. Lauderdale, Lower bounds on the number of maximal subgroups in a ﬁnite group, Arch. Math. 101 (2013), 9–15. [5] S. P. Norton R. A. Parker J. H. Conway, R. T. Curtis and R. A. Wilson, Atlas of Finite Groups. Maximal Subgroups and Ordinary Characters for Simple Groups, Clarendon Press, Oxford, 1985.

Received January 25, 2014; revised December 20, 2014.

Author information L.-K. Lauderdale, Department of Mathematics, University of Texas at Tyler, Tyler, TX, 75799, USA. E-mail: [email protected]