ON A QUESTION OF MORETÓ

Abstract

We present a family of counterexamples to a question proposed recently by Moretó concerning the character codegrees and the element orders of a finite solvable group.

1 Introduction

Let G be a finite group, and write $\operatorname {\mathrm {Irr}}(G)$ to denote the set of irreducible complex characters of G. The concept of character codegree, originally defined as $|G|/\chi (1)$ for any nonlinear irreducible character $\chi $ of G, was introduced in [1] to characterise the structure of finite groups. However, a nonlinear character $\chi \in \mathrm {Irr}(G/N)$ , where N is a nontrivial normal subgroup of G, will have two different codegrees when it is considered as a character of G and of $G/N$ . To eliminate this inconvenience, Qian et al. in [9] redefined the codegree of an arbitrary character $\chi $ of G as

$$ \begin{align*}\chi^{c}(1)=|G:\operatorname{\mathrm{Ker}} \chi|/\chi(1).\end{align*} $$

Many properties of codegrees have been studied, including variations on Huppert’s $\rho $ - $\sigma $ conjecture, the relationship between codegrees and element orders, groups with few codegrees, and recognising simple groups using the codegree set.

The authors believe that among the above-mentioned results, the most interesting is the relation between codegrees and element orders (see [4, 7, 8]). Here we mention a result of Qian [7, Theorem 1.1] which says that if a finite solvable group G has an element g of square-free order, then G must have an irreducible character of codegree divisible by the order $o(g)$ of g. Isaacs [4] established the same result for an arbitrary finite group. Recently, Qian [8] strengthened his earlier result, showing that for every element g of a finite solvable group G, there necessarily exists some $\chi \in \text {Irr}(G)$ such that $o(g)$ divides $\chi ^{c}(1)$ .

Motivated by the results in [4, 7, 8], Moretó considered the converse relation of codegrees and element orders and proposed an interesting question [6, Question B]. He also mentioned that counterexamples, if they exist, seem to be rare.

Question 1.1. Let G be a finite solvable group and let $\chi \in \operatorname {\mathrm {Irr}}(G)$ . Does there exist $g\in G$ such that $\pi (\,\chi ^{c}(1))\subseteq \pi (o(g))$ ? Here, $\pi (n)$ denotes the set of prime divisors of a positive integer n.

In this note, we will construct a family of examples to show that this question has a negative answer in general. For notation and terminology of character theory, we refer to [3].

2 Counterexamples

We begin with some facts about automorphisms of extra-special p-groups, which are more or less well-known but we give a complete proof for the reader’s convenience.

Theorem 2.1. For any distinct primes $p,r$ with $r>3$ , choose an extra-special p-group P of order $p^{2r+1}$ , such that P has exponent p if $p>2$ , and P is the central product of $r-1$ dihedral groups of order $8$ and a quaternion group if $p=2$ .

1. (1) There exists a prime q dividing $p^{r}+1$ but not $p^{2}-1$ . In particular, r divides $q-1$ , so that the semi-direct product $C_{q}\rtimes C_{r}$ makes sense.

2. (2) $\operatorname {\mathrm {Aut}}(P)$ contains a subgroup A, which acts trivially on $Z(P)$ and is isomorphic to $C_{q}\rtimes C_{r}$ .

Proof (1)By Zsigmondy’s prime theorem (see [2, Theorem IX.8.3]) and the condition that $r>3$ , there exists a prime q dividing $p^{2r}-1$ but not $p^{i}-1$ for all ${i=1,\ldots , 2r-1}$ . It follows that $2r$ is the order of p modulo q, establishing (1).

(2) Write $S=\operatorname {\mathrm {Sp}}(2r,p)$ for $p>2$ and $S=\text {O}_{-}(2r,2)$ for $p=2$ . Let $\Lambda $ denote the subgroup of $\operatorname {\mathrm {Aut}}(P)$ consisting of those automorphisms of P acting trivially on $Z(P)$ . By the construction of P, it is well known that $\Lambda /\kern-1.5pt\operatorname{\mathrm {Inn}}(P)$ is isomorphic to S (see, for example, [10, Theorem 1]). Since the orders of $C_{q}\rtimes C_{r}$ and $\operatorname {\mathrm {Inn}}(P)$ are relatively prime (as p does not divide $qr$ ), it suffices to show that S contains a subgroup isomorphic to $C_{q}\rtimes C_{r}$ by the Schur–Zassenhaus theorem.

To do this, we consider the finite field ${\mathbb {F}}_{p^{2r}}$ of $p^{2r}$ elements and let $V={\mathbb {F}}_{p^{2r}}$ be the vector space over ${\mathbb {F}}_{p}$ of dimension $2r$ . We need to construct a nonsingular symplectic form $\langle \;, \rangle $ on V for $p>2$ and a nonsingular quadratic form Q on V for $p=2$ .

Fix an element $a\in {\mathbb {F}}_{p^{2}}-{\mathbb {F}}_{p}$ . Then $a\not \in {\mathbb {F}}_{p^{r}}$ since r is odd. Let $\operatorname {\mathrm {tr}}:{\mathbb {F}}_{p^{r}}\to {\mathbb {F}}_{p}$ be the trace map. It is easy to see that

$$ \begin{align*}\langle v,w\rangle=\operatorname{\mathrm{tr}}((a-a^{p^{r}})(vw^{p^{r}}-v^{p^{r}}w)),\quad v,w\in V,\end{align*} $$

defines a nonsingular symplectic form on V and the corresponding symplectic group is S for $p>2$ (see [5, Example 8.5]). When $p=2$ ,

$$ \begin{align*}Q(v)=\operatorname{\mathrm{tr}}(v^{1+p^{r}}),\quad v\in V,\end{align*} $$

defines a nonsingular quadratic form on V and the corresponding orthogonal group is $S=\text {O}_{-}(2r,2)$ .

Now, let $\Gamma _{0}(V)=\{v\mapsto av\ |\ 0\neq a\in V\}$ , consisting of multiplications, and let $\sigma :v\mapsto v^{p}$ be the Frobenius field automorphism of ${\mathbb {F}}_{p^{2r}}$ . Then $\sigma $ acts naturally on $\Gamma _{0}(V)$ , and we consider the corresponding semi-direct product $\Gamma _{0}(V)\rtimes \langle \sigma \rangle $ . Observe that $\Gamma _{0}(V)$ is cyclic of order $p^{2r}-1$ and the order of $\sigma $ is $2r$ . Recall that the prime q divides $p^{r}+1$ and let C be the unique subgroup of order q in $\Gamma _{0}(V)$ . It is clear that C is invariant under $\sigma $ , and furthermore, by elementary Galois theory, the fixed point of $\sigma ^{2}$ in C is trivial since q does not divide $p^{2}-1$ . So we can form the semi-direct product $C\rtimes \langle \sigma ^{2}\rangle $ , which is clearly isomorphic to $C_{q}\rtimes C_{r}$ . What remains is to show that $C\rtimes \langle \sigma ^{2}\rangle \le S$ .

A simple calculation shows that both the symplectic form $\langle \; ,\;\rangle $ and the quadratic form Q defined above are preserved by the map on V induced by multiplication by an element of order $p^{r}+1$ , and thus the unique cyclic subgroup of $\Gamma _{0}(V)$ of order $p^{r}+1$ must be contained in S. In particular, we have $C\le S$ . To prove $\sigma ^{2}\in S$ , we distinguish two cases. For $p=2$ , since the Galois group of ${\mathbb {F}}_{p^{r}}/{\mathbb {F}}_{p}$ can be identified with $\langle \sigma ^{2}\rangle $ (as r is odd), we conclude that $\sigma ^{2}$ must preserve the trace map from ${\mathbb {F}}_{p^{r}}$ to ${\mathbb {F}}_{p}$ and hence lies in S. For $p>2$ , we need to establish that $\langle v^{\sigma ^{2}},w^{\sigma ^{2}}\rangle =\langle v,w\rangle $ for all $v,w\in V$ . Let $b=a-a^{p^{r}}$ and $x=vw^{p^{r}}-v^{p^{r}}w$ . It suffices to prove $\operatorname {\mathrm {tr}}(bx^{\sigma ^{2}})=\operatorname {\mathrm {tr}}(bx)$ . Since $(bx)^{p^{r}}=(-b)(-x)=bx$ , we have $bx\in {\mathbb {F}}_{p^{r}}$ . It follows that $\operatorname {\mathrm {tr}}((bx)^{\sigma ^{2}})=\operatorname {\mathrm {tr}}(bx)$ . By the choice of a, we know that $b\in {\mathbb {F}}_{p^{2}}$ and hence b is fixed by $\sigma ^{2}$ . Thus, $\operatorname {\mathrm {tr}}(bx^{\sigma ^{2}})=\operatorname {\mathrm {tr}}(bx)$ and $\sigma ^{2}\in S$ , as required.

As an application of Theorem 2.1, we can now construct a family of counterexamples to Moretó’s question.

Example 2.2. In the notation of Theorem 2.1, let $G=P\rtimes A$ be the corresponding semi-direct product, so that G is solvable. Then there exists an irreducible character $\chi $ of G such that $\chi ^{c}(1)=p^{r+1}qr$ but G contains no element of order divisible by $pqr$ .

Proof. Note that $Z(P)$ is cyclic of order p, and thus we can choose a faithful linear character $\lambda $ of $Z(P)$ . Then it is well known that $\lambda ^{P}=p^{r}\theta $ for some $\theta \in \operatorname {\mathrm {Irr}}(P)$ . Since A acts trivially on $Z(P)$ , it fixes $\lambda $ and hence $\theta $ is A-invariant. It follows that $\theta $ extends to some $\chi \in \operatorname {\mathrm {Irr}}(G)$ by Gallagher’s theorem (see [3, Corollary 6.28]), so that $\chi (1)=\theta (1)=p^{r}$ . Also, let B be the unique subgroup of A of order q. Then $P/Z(P)$ is irreducible as an ${\mathbb {F}}_{p}[B]$ -module because $|P/Z(P)|=p^{2r}$ and the order of p modulo q is exactly $2r$ . From this, we conclude that $Z(P)$ is the unique minimal normal subgroup of G, and since $\theta $ is faithful, we have $\operatorname {\mathrm {Ker}}\chi \cap P=\operatorname {\mathrm {Ker}}\theta =1$ . Obviously, P is the Fitting subgroup of the solvable group G, and thus $\operatorname {\mathrm {Ker}}\chi $ contains no minimal normal subgroup of G, which forces $\operatorname {\mathrm {Ker}}\chi =1$ . Therefore, we have $\chi ^{c}(1)=|G|/\chi (1)=p^{r+1}qr$ .

Finally, if G has an element of order divisible by all the primes $p, q, r$ , then A contains an element of order $qr$ and thus A is cyclic, which is not the case by the choice of primes $q, r$ . The proof is now complete.