# On the Set of the Numbers of Conjugates of Noncyclic Proper Subgroups of Finite Groups

Authors: Jiangtao Shi; Cui Zhang
Journal: The Scientific World Journal
 (2013)

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## Abstract

LetG be a finite group and 𝒩𝒞(G) the set of the numbers of conjugates of noncyclic proper subgroups of G. We prove that (1) if |𝒩𝒞(G)|≤2, then G is solvable, and (2) G is a nonsolvable group with |𝒩𝒞(G)|=3 if and only if G≅PSL(2,5) or PSL(2,13) or SL(2,5) or SL(2,13).


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## Body

## 1. Introduction

In this paper, all groups are assumed to be finite. It seems interesting to investigate the influence of some arithmetic properties of noncyclic proper subgroups on the solvability of groups. In [1], Li and Zhao proved that any group having at most three conjugacy classes of noncyclic proper subgroups is solvable, and a group G having exactly four conjugacy classes of noncyclic proper subgroups is nonsolvable if and only if G≅PSL(2,5) or SL(2,5). As a generalization of the above result, we showed that any group having at most three conjugacy classes of nonnormal noncyclic proper subgroups is solvable, and a group G having exactly four conjugacy classes of nonnormal noncyclic proper subgroups is nonsolvable if and only if G≅PSL(2,5) or SL(2,5) (see [2]).LetG be a group and 𝒩𝒞(G) the set of the numbers of conjugates of noncyclic proper subgroups of G. It is clear that a group G with 𝒩𝒞(G)=∅ is either a cyclic group or a minimal noncyclic group, and a group G with 𝒩𝒞(G)={1} is a group in which every noncyclic proper subgroup is normal. In [2], we also obtained a complete classification of groups G in which every noncyclic proper subgroup is nonnormal; all such groups G satisfy 1∉𝒩𝒞(G).By|𝒩𝒞(G)| we denote the order of 𝒩𝒞(G). Note that we cannot ensure that 1∈𝒩𝒞(G) for any solvable group G with |𝒩𝒞(G)|=n≥1. For example, let G≅D2pn be a dihedral group of order 2pn, where n≥1 and p is an odd prime. Then 𝒩𝒞(D2pn)={p,p2,…,pn}, so 1∉𝒩𝒞(D2pn). For the nonsolvable group of the smallest order PSL(2,5), it is easy to see that 𝒩𝒞(PSL(2,5))={5,6,10}, and so |𝒩𝒞(PSL(2,5))|=3.For the influence of|𝒩𝒞(G)| on the solvability of groups, we have the following result, the proof of which is given in Section 3.Theorem 1.
LetG be a group.(1)
If|𝒩𝒞(G)|≤2, then G is solvable.(2)
G is a nonsolvable group with |𝒩𝒞(G)|=3 if and only if G≅PSL(2,5) or PSL(2,13) or SL(2,5) or SL(2,13).The following two corollaries are direct consequences of Theorem1.Corollary 2.
LetG be a group with |𝒩𝒞(G)|≤3. Then G is nonsolvable if and only if 𝒩𝒞(G)={5,6,10} or {14,78,91}.Corollary 3.
LetG be a group and 𝒩𝒯(G) the set of the numbers of conjugates of nontrivial subgroups of G.(1)
If|𝒩𝒯(G)|≤2, then G is solvable.(2)
G is a nonsolvable group with |𝒩𝒯(G)|=3 if and only if G≅PSL(2,13).LetG be a group and 𝒩𝒞*(G) the set of the numbers of conjugates of nonnormal noncyclic proper subgroups of G. Obviously 𝒩𝒞*(G)⊆𝒩𝒞(G).Arguing as in the proof of Theorem1, we can obtain the following result.Theorem 4.
LetG be a group. If |𝒩𝒞*(G)|≤2, then G is solvable.Remark 5.
If we assume thatG is a nonsolvable group with |𝒩𝒞*(G)|=3, we cannot get that Φ(G)=Z(G). For example, let G≅PSL(2,5)×ℤp, where p≥7 is a prime. It is easy to see that |𝒩𝒞*(G)|=3. But Φ(G)=1 and Z(G)=ℤp.
LetG be a group and 𝒩𝒜(G) the set of the numbers of conjugates of nonabelian proper subgroups of G. Obviously 𝒩𝒜(G)⊆𝒩𝒞(G). Arguing as in the proof of Theorem 1, we can also obtain the following result.Theorem 6.
LetG be a group. If |𝒩𝒜(G)|≤2, then G is solvable.


## 2. Preliminaries

In this section, we collect some essential lemmas needed in the sequel.Lemma 7 (see [3]).
LetG be a group. If all nonnormal maximal subgroups of G have the same order, then G is solvable.Lemma 8 (see [4]).
LetG be a nonsolvable group having exactly two classes of nonnormal maximal subgroups of the same order; then G/S(G)≅PSL(2,7), where S(G) is the largest solvable normal subgroup of G.Lemma 9 (see [5, 6]).
LetG be a group having exactly n classes of maximal subgroups of the same order, where 1≤n≤3; then one of the following statements holds:(1)
suppose thatG is a group with n=1, and then G is a p-group for some prime p;(2)
suppose thatG is a nonsolvable group with n=2, and then G/Φ(G)≅(ℤ23i⋊PSL(2,7))×ℤ7j, where i,j=0,1,…, and ℤ23i⋊PSL(2,7) is a semidirect product of the normal subgroup  ℤ23i and the subgroup PSL(2,7);(3)
suppose thatG is a nonsolvable group with n=3, and then G/S(G)≅A6; PSL(2,q), q=11,13,23,59,61; PSL(3,3); U3(3);PSL(5,2); PSL(2,2f), and f is a prime; PSL(2,7)×PSL(2,7)×⋯×PSL(2,7).


## 3. Proof of Theorem1

The proof of Theorem1 follows from the following two lemmas.Lemma 10.
LetG be a group. If |𝒩𝒞(G)|≤2, then G is solvable.Proof.
Assume thatG is nonsolvable. Then by [7, Exercise 10.5.7], all maximal subgroups of G are noncyclic. Let ℳ𝒮(G) be the set of the numbers of conjugates of maximal subgroups of G. It follows that ℳ𝒮(G)⊆𝒩𝒞(G). Then |ℳ𝒮(G)|≤2.(1)
Suppose that1∈ℳ𝒮(G). Since G is nonsolvable, G must have nonnormal maximal subgroups. Let M be any nonnormal maximal subgroup of G; one has |G:NG(M)|=|G:M|. Since |ℳ𝒮(G)|≤2, we know that G has at most one class of nonnormal maximal subgroups of the same order. It follows that G is solvable by Lemma 7, a contradiction.(2)
Suppose that1∉ℳ𝒮(G). It follows that all maximal subgroups of G are nonnormal. By the hypothesis, G has at most two classes of maximal subgroups of the same order. Since G is nonsolvable and G has no normal maximal subgroups, one has G/Φ(G)≅ℤ23i⋊PSL(2,7) by Lemma 9 (1) and (2), where i=0,1,…. It is easy to see that 𝒩𝒞(G/Φ(G))⊆𝒩𝒞(G) and |𝒩𝒞(ℤ23i⋊PSL(2,7))|>2. It follows that |𝒩𝒞(G)|>2, a contradiction.
Thus, our assumption is not true, soG is solvable.Lemma 11.
A groupG is a nonsolvable group with |𝒩𝒞(G)|=3 if and only if G≅PSL(2,5) or PSL(2,13) or SL(2,5) or SL(2,13).Proof.
The sufficiency part is evident, and we only need to prove the necessity part.
By the hypothesis,|ℳ𝒮(G)|≤3. We claim that
(1)1∉ℳ𝒮(G).
Otherwise, assume that1∈ℳ𝒮(G). Then G has at most two classes of nonnormal maximal subgroups of the same order. Since G is nonsolvable, one has G/S(G)≅PSL(2,7) by Lemmas 7 and 8. It is easy to see that 𝒩𝒞(G/S(G))⊆𝒩𝒞(G) and |𝒩𝒞(PSL(2,7))|>3. It follows that |𝒩𝒞(G)|>3, a contradiction. Thus, 1∉ℳ𝒮(G).
Since|ℳ𝒮(G)|≤3, we have that G has at most three classes of maximal subgroups of the same order.
By Lemma9 (1), G cannot have exactly one class of maximal subgroups of the same order.
IfG has exactly two classes of maximal subgroups of the same order, according to Lemma 9 (2), one has G/Φ(G)≅ℤ23i⋊PSL(2,7) since G has no normal maximal subgroups, where i=0,1,…. Since |𝒩𝒞(ℤ23i⋊PSL(2,7))|>3, it follows that |𝒩𝒞(G)|>3, a contradiction.
Thus,G has exactly three classes of maximal subgroups of the same order. By Lemma 9 (3), G/S(G) might be isomorphic to A6 or PSL(2,q),q=11,13,23,59,61 or PSL(3,3) or U3(3) or PSL(5,2) or PSL(2,2f), and f is a prime or PSL(2,7)×PSL(2,7)×⋯×PSL(2,7). If G/S(G) is an isomorphism to A6 or PSL(2,q), q=11,23,59,61 or PSL(3,3) or U3(3) or PSL(5,2) or PSL(2,2f), and f is an odd prime or PSL(2,7)×PSL(2,7)×⋯×PSL(2,7). It is easy to see that |𝒩𝒞(G/S(G))|>3 by [8, 9], which implies that |𝒩𝒞(G)|>3, a contradiction. Thus, G/S(G)≅PSL(2,4)≅PSL(2,5) or PSL(2,13).
Note that1∉ℳ𝒮(G) and |ℳ𝒮(G)|=|𝒩𝒞(G)|=3. It follows that 1∉𝒩𝒞(G), so S(G) is cyclic. We claim that
(2)Φ(G)=S(G).
Otherwise, assume thatΦ(G)<S(G). Let M be a maximal subgroup of G such that S(G)≰M. Then G=S(G)M. It is obvious that S(G)∩M⊴M. Moreover, S(G)∩M⊴S(G), since S(G) is cyclic. It follows that S(G)∩M⊴G. Therefore, G/(S(G)∩M)=S(G)/(S(G)∩M)⋊M/(S(G)∩M). Let G-=G/(S(G)∩M),S-(G)=S(G)/(S(G)∩M), and M-=M/(S(G)∩M). By N/C-theorem, NG-(S-(G))/CG-(S-(G))≲Aut(S-(G)). That is, G-/CG-(S-(G))=S-(G)M-/CG-(S-(G)))≲Aut(S-(G)). Note that Aut(S-(G)) is abelian since S-(G) is cyclic. Moreover, M-≅S(G)M/S(G)=G/S(G) is a nonabelian simple group and S-(G)M-/CG-(S-(G))≅(S-(G)M-/S-(G))/(CG-(S-(G))/S-(G)). Here S-(G)M-/S-(G)≅M-. Therefore, one has CG-(S-(G))/S-(G)=1 or CG-(S-(G))/S-(G)=S-(G)M-/S-(G)=G-/S-(G). If CG-(S-(G))/S-(G)=1, it follows that S-(G)M-/S-(G)≲Aut(S-(G)) is abelian, a contradiction. If CG-(S-(G))/S-(G)=G-/S-(G), then S-(G)≤Z(G-). It follows that G-=S-(G)×M- and then M⊴G; this contradicts that all maximal subgroups of G are nonnormal. Thus, our assumption is not true, so Φ(G)=S(G).
It follows thatG/Φ(G)≅PSL(2,5) or PSL(2,13).
IfΦ(G)=1, then G≅PSL(2,5) or PSL(2,13).
Next, suppose thatΦ(G)≠1. Let p be any prime divisor of |Φ(G)|. We claim that p≯2. Otherwise, assume that p>2. Let T be a subgroup of Φ(G) such that Φ(G)/T≅ℤp. That is, Φ(G/T)≅ℤp. Then (G/T)/ℤp≅(G/T)/Φ(G/T)=(G/T)/(Φ(G)/T)≅G/Φ(G)≅PSL(2,5) or PSL(2,13). Since p>2 and Schur multipliers of both PSL(2,5) and PSL(2,13) are ℤ2, we have that G/T≅PSL(2,5)×ℤp or PSL(2,13)×ℤp. Note that |𝒩𝒞(PSL(2,5)×ℤp)|>3 and |𝒩𝒞(PSL(2,13)×ℤp)|>3. It follows that |𝒩𝒞(G)|>3, a contradiction. Thus, p≯2, so Φ(G) is a cyclic 2-group. If |Φ(G)|=2n>2, let L be a subgroup of Φ(G) such that Φ(G)/L≅ℤ2. Then (G/L)/ℤ2≅(G/L)/Φ(G/L)=(G/L)/(Φ(G)/L)≅G/Φ(G)≅PSL(2,5) or PSL(2,13). We have that G/L≅SL(2,5) or SL(2,13). Let M be a subgroup of L such that L/M≅ℤ2. Then (G/M)/ℤ2≅(G/M)/(L/M)≅G/L≅SL(2,5) or SL(2,13). Since Schur multipliers of both SL(2,5) and SL(2,13) are trivial, we have that G/M≅SL(2,5)×ℤ2 or SL(2,13)×ℤ2; this contradicts that all maximal subgroups of G are nonnormal. Thus, |Φ(G)|=2. It follows that G≅SL(2,5) or SL(2,13).Lemmas10 and 11 combined together give Theorem 1.



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