SOME PRODUCTS OF SUBGROUPS AND VANISHING CONJUGACY CLASS SIZES

Abstract

In this note, we investigate some products of subgroups and vanishing conjugacy class sizes of finite groups. We prove some supersolubility criteria for groups with restrictions on the vanishing conjugacy class sizes of their subgroups.

1 Introduction

All groups considered in this note will be finite.

The structural influence of the conjugacy classes of a group is a question of interest in the Theory of Groups. In particular, the set of vanishing conjugacy class sizes has acquired special significance (see [3, 4, 8]). An element x of a group G is called a vanishing element if there exists some irreducible character $\chi $ of G such that $\chi (x) = 0$ ; the conjugacy class $x^G$ is called a vanishing conjugacy class. We denote by $\mathrm {Van}(G)$ the set of all elements of G which are vanishing in G.

During the past three decades, factorised groups whose factors are connected by means of certain permutability properties, namely mutually permutable products, have been extensively studied (see [1]). Recall that two subgroups A and B of a group G are called mutually permutable if A permutes with every subgroup of B and B permutes with every subgroup of A. We say that a group G is a mutually permutable product of the subgroups A and B if $G = AB$ and A and B are mutually permutable. Note that normal products are typical examples of mutually permutable products. However, the symmetric group of degree $3$ is an easy example of a group which is a nonnormal mutually permutable product.

Kong and Chen [9] studied normal products of groups in which every element of prime power order of each factor is vanishing. They proved the following result.

Theorem 1.1 [9, Theorem A].

Let $ G=AB $ be the product of two normal subgroups $ A $ and $ B $ of G. If $ |x^{G}| $ is square-free for every $ x\in \mathrm {Van}(A) \cup \mathrm {Van}(B) $ of prime power order, then $ G $ is supersoluble.

However, Felipe et al. in [7] proved the following theorem as a consequence of a more general result.

Theorem 1.2. Let $G = AB$ be the mutually permutable product of the subgroups $ A $ and $ B $ . If $ |x^{G}| $ is square-free for every $ x\in A\cup B $ of prime power order that is vanishing in G, then $ G $ is supersoluble.

We prove an analogous result but instead of assuming that all elements of $A\cup B$ of prime power order that are vanishing in G have square-free conjugacy class size, we assume that all elements of prime power order in $\mathrm {Van}(A) \cup \mathrm {Van}(B) $ have square-free conjugacy class size, extending Theorem 1.1.

Theorem 1.3. Let the group $G = AB$ be the mutually permutable product of the subgroups A and B. If $ |x^{G}| $ is square-free for every $ x\in \mathrm {Van}(A) \cup \mathrm {Van}(B) $ of prime power order, then $ G $ is supersoluble.

Theorem 1.1 is not true if only one of the factors is normal: the alternating group G of degree $4$ is a product of a normal subgroup A of order $4$ and a subgroup B of order $3$ which is not supersoluble and $\mathrm {Van}(A) = \mathrm {Van}(B) = \emptyset $ . However, an additional condition allows us to obtain the following extension.

Theorem 1.4. Let $ G=AB $ be a product of two subgroups $ A $ and $ B $ . Assume that A is normal in G and B permutes with every maximal subgroup of A. If $ |x^{G}| $ is square-free for every $ x\in \mathrm {Van}(A) \cup \mathrm {Van}(B) $ of prime power order, then $ G $ is supersoluble.

2 Proofs

Given a normal subgroup N of a group G and $x \in G$ , both conjugacy class sizes $|x^N|$ and $|x^{G/N}|$ divide $|x^G|$ . Moreover, there is a natural bijection between the set of irreducible characters of $G/N$ and the set of all irreducible characters of G with N in their kernel. These facts, together with [1, Lemmas 1.2.7 and 4.1.10], show that the hypotheses of our main theorems are inherited by factor groups.

The proofs of our main results also depend on the following lemmas.

Lemma 2.1 [4, Theorem 1.5].

Let $ G $ be a group. If $ |x^{G}| $ is square-free for every vanishing element of prime power order $x \in G$ , then G is supersoluble.

Lemma 2.2 [8, Theorem 1.1].

If G is nilpotent-by-supersoluble, then every nonvanishing element of G belongs to $F(G)$ , the Fitting subgroup of G.

Lemma 2.3. Let $ N $ be a nonabelian minimal normal subgroup of $ G $ . Then there exists a nontrivial element $ x $ of $ N $ such that:

1. (i) $ x $ is of prime power order;

2. (ii) $ x $ is a vanishing element of $ H $ for any subgroup $ H $ such that $ N \leq H \leq G $ ;

3. (iii) $ 4 $ divides $ |x^{G}| $ .

Proof. We may write $ N = S_{1}\times \cdots \times S_{n} $ , where $ S_{i}\cong S $ for some nonabelian simple group. Suppose that $ S $ has an irreducible character of $ p $ -defect zero for every prime $ p $ dividing $ |S| $ . Applying [3, Lemma 2.2], we conclude that every element of $ N $ is a vanishing element of any subgroup $ H $ such that $ N \leq H \leq G $ . Since $ S $ is not soluble, there exists a nontrivial $ p $ -element of $ S $ such that $ 4 $ divides $ |x^{S}| $ using [5, Corollary]. Hence, $ 4 $ divides $ |x^{G}| $ as required.

We may assume that there exists a prime $ p $ such that $ S $ does not have an irreducible character of $ p $ -defect zero. By [4, Corollary 2.1], $S\cong \mathrm {A}_{n} $ , $ n\geq 7 $ , or S is isomorphic to a sporadic simple group. If $ S\cong \mathrm {A}_{n} $ , $ n\geq 7 $ , then there exists a $ q $ -element $ x $ of $ S $ for some prime $ q\geq 5 $ such that $ 4 $ divides $ |x^{S}| $ by [4, Lemma 2.4]. Note that every nonabelian simple group $ S $ has an irreducible character of $ q $ -defect zero for every prime $ q\geq 5 $ dividing $ |S| $ (see [4, Corollary 2.1]). Hence, the result follows by the above argument.

Suppose that $ S $ is isomorphic to one of the sporadic simple groups given in [4, Corollary 2.1]. Using [4, Lemma 2.4], there exists a nontrivial element $ x $ of $ S $ of prime power order such that $ 4 $ divides $ |x^{S}| $ and there exists $ \chi \in {\mathrm {Irr}}(S) $ with the following properties: $ \chi $ extends to $ {\mathrm {Aut}}(S) $ and $ \chi $ vanishes on $ x^{S} $ . By [3, Proposition 2.3], $ \chi \times \cdots \times \chi \in {\mathrm {Irr}}(N) $ extends to any subgroup $ H $ such that $ N \leq H \leq G $ . Hence, $ x $ is a vanishing element of $ H $ and also $ 4 $ divides $ |x^{G}| $ . This concludes the proof.

Proof of Theorem 1.3.

Assume the result is false and let $ G $ be a minimal counterexample. By Lemma 2.1, A and B are proper subgroups of G. Since the hypotheses are inherited by factor groups and the class of all supersoluble groups is a saturated formation, it follows that G is a primitive group. Let $ N $ be the unique minimal normal subgroup of $ G $ . Then, $G/N$ is supersoluble. By [1, Theorem 4.3.11], $N \leq A$ or $N \leq B$ ; suppose without loss of generality that $N \leq A$ . Assume N is nonabelian. By Lemma 2.3, there exists a nontrivial element $x \in N$ of prime power order such that $4$ divides $|x^{G}|$ and x is vanishing in A. This contradiction implies that N is a q-elementary abelian group of order $ q^{n} $ for some prime q and $ n> 1 $ . In particular, G is nilpotent by supersoluble. By [1, Lemma 4.3.3, 4 and 5], $N \leq B$ or $B \cap N=1$ . Assume that $B \cap N=1$ . Since N is not cyclic, it follows that $N \leq C_{G}(B)$ . Hence, $B \leq C_{G}(N)=N \leq A$ , and $G=A$ which is not possible. Consequently, $ N\leq A\cap B $ .

Since $G/N=(A/N)(B/N)$ is the mutually permutable product of $A/N$ and $B/N$ , there exists a minimal normal subgroup $L/N$ of $G/N$ contained in $A/N$ or in $B/N$ . Suppose $L/N \leq A/N$ . Since $G/N$ is supersoluble, it follows that $L/N$ is a cyclic group of prime order p. Moreover, $p \neq q$ . Therefore, $L=N\langle x \rangle $ , where $|\langle x \rangle |=p$ . By Lemma 2.2, x is a vanishing element of A. Since $ G $ is a primitive group, there exists a complement of $ N $ , $ M $ say, such that $ MN=G $ and $ M\cap N=1 $ and $ M $ is maximal in $ G $ . Hence, $ L/N\cong \langle x \rangle $ is a minimal normal subgroup of $ M$ . Since $ \langle x \rangle $ is normal in $ M $ , $ C_{G}(\langle x \rangle )\leqslant N_{G}(\langle x \rangle ) = M $ or $ G $ . Since $ N\cap M=1 $ , $ N_{G}(\langle x \rangle ) = M $ and so $ |N| $ divides $ |x^{G}| $ , our final contradiction.

Proof of Theorem 1.4.

Suppose that the theorem is not true and let $ G $ be a minimal counterexample. Then G has a unique minimal normal subgroup N, N is not cyclic, G is a primitive group, and A and B are proper subgroups of G. Moreover, N is contained in A. Since A is a normal subgroup of G, it follows that $|x^A|$ divides $|x^G|$ . By Lemma 2.1, A is supersoluble and N is abelian. By [6, Theorem A.15.2], $C_{G}(N)=N$ and so $F(G) = N$ . Note that $F(A)$ is a normal nilpotent subgroup of G contained in $F(G) = N$ . Therefore, $F(A) = N$ .

Assume that N is a proper subgroup of A. Then there exists a minimal normal subgroup $L/N$ of $G/N$ such that $L/N \subseteq A/N$ . We can argue as in the proof of Theorem 1.3 to reach a contradiction. Consequently, $A = N$ , B is a maximal subgroup of G and $B \cap N = 1$ . Since N is not cyclic, N has a maximal subgroup $D \neq 1$ . By hypothesis, $DB$ is a subgroup of G. Hence, $D = N \cap DB$ is normalised by N and B. Thus, D is a normal subgroup of G, contradicting the fact that N is minimal normal in G.

The proof of the theorem is now complete.