Let be a finite group and NC( ) the set of the numbers of conjugates of noncyclic proper subgroups of . We prove that (1) if |NC( )| ≤ 2, then is solvable, an

Cui Zhang

Hoff Da Silva

J Park

Jiangtao Shi

CiteSeerX

Hindawi Publishing Corporation
The Scientific World Journal
Volume 2013, Article ID 430870, 3 pages
http://dx.doi.org/10.1155/2013/430870
Research Article
On the Set of the Numbers of Conjugates of Noncyclic Proper
Subgroups of Finite Groups
Jiangtao Shi1 and Cui Zhang2
1 School of Mathematics and Information Science, Yantai University, Yantai 264005, China
2Department of Applied Mathematics and Computer Science, Technical University of Denmark, 2800 Lyngby, Denmark
Correspondence should be addressed to Jiangtao Shi; shijt2005@163.com
Received 3 August 2013; Accepted 11 September 2013
Academic Editors: J. Hoff da Silva and J. Park
Copyright © 2013 J. Shi and C. Zhang. This is an open access article distributed under the Creative Commons Attribution License,
which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Let 𝐺 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).
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 group𝐺 having exactly four conj-
ugacy classes of nonnormal noncyclic proper subgroups is
nonsolvable if and only if 𝐺 ≅ 𝑃𝑆𝐿(2, 5) or 𝑆𝐿(2, 5) (see [2]).
Let 𝐺 be a group and NC(𝐺) the set of the numbers of
conjugates of noncyclic proper subgroups of𝐺. It is clear that
a group 𝐺 with NC(𝐺) = 0 is either a cyclic group or a
minimal noncyclic group, and a group𝐺withNC(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 ≅ 𝐷
2𝑝
𝑛 be a
dihedral group of order 2𝑝𝑛, where 𝑛 ≥ 1 and 𝑝 is an odd
prime. ThenNC(𝐷
2𝑝
𝑛) = {𝑝, 𝑝
2
, . . . , 𝑝
𝑛
}, so 1 ∉ NC(𝐷
2𝑝
𝑛).
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. Let 𝐺 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
Theorem 1.
Corollary 2. Let 𝐺 be a group with |NC(𝐺)| ≤ 3. Then 𝐺 is
nonsolvable if and only ifNC(𝐺) = {5, 6, 10} or {14, 78, 91}.
Corollary 3. Let 𝐺 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).
2 The Scientific World Journal
Let 𝐺 be a group andNC∗(𝐺) the set of the numbers of
conjugates of nonnormal noncyclic proper subgroups of 𝐺.
ObviouslyNC∗(𝐺) ⊆ NC(𝐺).
Arguing as in the proof of Theorem 1, we can obtain the
following result.
Theorem 4. Let 𝐺 be a group. If |NC∗(𝐺)| ≤ 2, then 𝐺 is
solvable.
Remark 5. If we assume that 𝐺 is a nonsolvable group with
|NC∗(𝐺)| = 3, we cannot get that Φ(𝐺) = 𝑍(𝐺). For
example, let 𝐺 ≅ 𝑃𝑆𝐿(2, 5) × Z
𝑝
, where 𝑝 ≥ 7 is a prime. It is
easy to see that |NC∗(𝐺)| = 3. ButΦ(𝐺) = 1 and𝑍(𝐺) = Z
𝑝
.
Let 𝐺 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 ofTheorem 1, we
can also obtain the following result.
Theorem 6. Let 𝐺 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]). Let 𝐺 be a group. If all nonnormal
maximal subgroups of𝐺have the same order, then𝐺 is solvable.
Lemma 8 (see [4]). Let 𝐺 be a nonsolvable group having
exactly two classes of nonnormal maximal subgroups of the
same order; then𝐺/𝑆(𝐺) ≅ 𝑃𝑆𝐿(2, 7), where 𝑆(𝐺) is the largest
solvable normal subgroup of 𝐺.
Lemma9 (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 that 𝐺 is a group with 𝑛 = 1, and then 𝐺 is a
𝑝-group for some prime 𝑝;
(2) suppose that 𝐺 is a nonsolvable group with 𝑛 = 2, and
then 𝐺/Φ(𝐺) ≅ (Z3𝑖
2
⋊ 𝑃𝑆𝐿(2, 7)) × Z
𝑗
7
, where 𝑖, 𝑗 =
0, 1, . . ., and Z3𝑖
2
⋊ 𝑃𝑆𝐿(2, 7) is a semidirect product of
the normal subgroup Z3𝑖
2
and the subgroup 𝑃𝑆𝐿(2, 7);
(3) suppose that 𝐺 is a nonsolvable group with 𝑛 = 3, and
then 𝐺/𝑆(𝐺) ≅ 𝐴
6
; 𝑃𝑆𝐿(2, 𝑞), 𝑞 = 11, 13, 23, 59, 61;
𝑃𝑆𝐿(3, 3); 𝑈
3
(3); 𝑃𝑆𝐿(5, 2); 𝑃𝑆𝐿(2, 2𝑓), and 𝑓 is a
prime; 𝑃𝑆𝐿(2, 7) × 𝑃𝑆𝐿(2, 7) × ⋅ ⋅ ⋅ × 𝑃𝑆𝐿(2, 7).
3. Proof of Theorem 1
The proof of Theorem 1 follows from the following two
lemmas.
Lemma 10. Let 𝐺 be a group. If |NC(𝐺)| ≤ 2, then 𝐺 is
solvable.
Proof. Assume that 𝐺 is nonsolvable. Then by [7, Exercise
10.5.7], all maximal subgroups of𝐺 are noncyclic. LetMS(𝐺)
be the set of the numbers of conjugates ofmaximal subgroups
of 𝐺. It follows thatMS(𝐺) ⊆ NC(𝐺). Then |MS(𝐺)| ≤ 2.
(1) Suppose that 1 ∈ MS(𝐺). Since 𝐺 is nonsolvable,
𝐺 must have nonnormal maximal subgroups. Let 𝑀
be any nonnormal maximal subgroup of 𝐺; one has
|𝐺 : 𝑁
𝐺
(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 that 1 ∉ 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) ≅ Z3𝑖
2
⋊
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(Z3𝑖
2
⋊ 𝑃𝑆𝐿(2, 7))| > 2. It follows that |NC(𝐺)| >
2, a contradiction.
Thus, our assumption is not true, so 𝐺 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 ∉ MS (𝐺) . (1)
Otherwise, assume that 1 ∈ 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 thatNC(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 Lemma 9 (1), 𝐺 cannot have exactly one class of
maximal subgroups of the same order.
If 𝐺 has exactly two classes of maximal subgroups of the
same order, according to Lemma 9 (2), one has 𝐺/Φ(𝐺) ≅
Z3𝑖
2
⋊ 𝑃𝑆𝐿(2, 7) since 𝐺 has no normal maximal subgroups,
where 𝑖 = 0, 1, . . .. Since |NC(Z3𝑖
2
⋊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
to𝐴
6
or𝑃𝑆𝐿(2, 𝑞), 𝑞 = 11, 13, 23, 59, 61 or𝑃𝑆𝐿(3, 3) or𝑈
3
(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
𝐴
6
or 𝑃𝑆𝐿(2, 𝑞), 𝑞 = 11, 23, 59, 61 or 𝑃𝑆𝐿(3, 3) or 𝑈
3
(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).
The Scientific World Journal 3
Note that 1 ∉ MS(𝐺) and |MS(𝐺)| = |NC(𝐺)| = 3. It
follows that 1 ∉ NC(𝐺), so 𝑆(𝐺) is cyclic. We claim that
Φ (𝐺) = 𝑆 (𝐺) . (2)
Otherwise, assume that Φ(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,
𝑁
𝐺
(𝑆(𝐺))/𝐶
𝐺
(𝑆(𝐺)) ≲ Aut(𝑆(𝐺)). That is, 𝐺/𝐶
𝐺
(𝑆(𝐺)) =
𝑆(𝐺)𝑀/𝐶
𝐺
(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/𝐶
𝐺
(𝑆(𝐺)) ≅
(𝑆(𝐺)𝑀/𝑆(𝐺))/(𝐶
𝐺
(𝑆(𝐺))/𝑆(𝐺)). Here 𝑆(𝐺)𝑀/𝑆(𝐺) ≅ 𝑀.
Therefore, one has 𝐶
𝐺
(𝑆(𝐺))/𝑆(𝐺) = 1 or 𝐶
𝐺
(𝑆(𝐺))/𝑆(𝐺) =
𝑆(𝐺)𝑀/𝑆(𝐺) = 𝐺/𝑆(𝐺). If 𝐶
𝐺
(𝑆(𝐺))/𝑆(𝐺) = 1, it follows
that 𝑆(𝐺)𝑀/𝑆(𝐺) ≲ Aut(𝑆(𝐺)) is abelian, a contradiction.
If 𝐶
𝐺
(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, soΦ(𝐺) = 𝑆(𝐺).
It follows that 𝐺/Φ(𝐺) ≅ 𝑃𝑆𝐿(2, 5) or 𝑃𝑆𝐿(2, 13).
If Φ(𝐺) = 1, then 𝐺 ≅ 𝑃𝑆𝐿(2, 5) or 𝑃𝑆𝐿(2, 13).
Next, suppose that Φ(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 ≅ Z
𝑝
.
That is, Φ(𝐺/𝑇) ≅ Z
𝑝
. Then (𝐺/𝑇)/Z
𝑝
≅ (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)
areZ
2
, we have that𝐺/𝑇 ≅ 𝑃𝑆𝐿(2, 5)×Z
𝑝
or𝑃𝑆𝐿(2, 13)×Z
𝑝
.
Note that |NC(𝑃𝑆𝐿(2, 5) × Z
𝑝
)| > 3 and |NC(𝑃𝑆𝐿(2, 13) ×
Z
𝑝
)| > 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 ≅ Z
2
. Then
(𝐺/𝐿)/Z
2
≅ (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). Let𝑀 be a subgroup of𝐿 such that𝐿/𝑀 ≅ Z
2
.Then
(𝐺/𝑀)/Z
2
≅ (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 that𝐺/𝑀 ≅ 𝑆𝐿(2, 5)×Z
2
or 𝑆𝐿(2, 13)×Z
2
; this
contradicts that all maximal subgroups of 𝐺 are nonnormal.
Thus, |Φ(𝐺)| = 2. It follows that𝐺 ≅ 𝑆𝐿(2, 5) or 𝑆𝐿(2, 13).
Lemmas 10 and 11 combined together give Theorem 1.
Acknowledgments
The authors are grateful to the referees who gave valuable
comments and suggestions. Jiangtao Shi was supported by
NSFC (Grant nos. 11201401 and 11361075). Cui Zhang was
supported by H.C. Ørsted Postdoctoral Fellowship at DTU
(Technical University of Denmark).
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On the Set of the Numbers of Conjugates of Noncyclic Proper Subgroups of Finite Groups

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

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