ON p-SOLVABILITY AND AVERAGE CHARACTER DEGREE IN A FINITE GROUP

Abstract

Assume that G is a finite group, N is a nontrivial normal subgroup of G and p is an odd prime. Let $\mathrm{Irr}_p(G)=\{\chi \in \mathrm{Irr}(G) : \chi (1)=1~\mathrm{or}~ p \mid \chi (1)\}$ and $\mathrm{Irr}_p(G|N)=\{\chi \in \mathrm{Irr}_p(G) : N \not \leq \mathrm{ker}\,\chi \}$. The average character degree of irreducible characters of $\mathrm{Irr}_p(G)$ and the average character degree of irreducible characters of $\mathrm{Irr}_p(G|N)$ are denoted by $\mathrm{acd}_p(G)$ and $\mathrm{acd}_p(G|N)$, respectively. We show that if $\mathrm{Irr}_p(G|N) \neq \emptyset $ and $\mathrm{acd}_p(G|N) < \mathrm{acd}_p(\mathrm{PSL}_2(p))$, then G is p-solvable and $O^{p'}(G)$ is solvable. We find examples that make this bound best possible. Moreover, we see that if $\mathrm{Irr}_p(G|N) = \emptyset $, then N is p-solvable and $P \cap N$ and $PN/N$ are abelian for every $P \in \mathrm{Syl}_p(G)$.

1 Introduction

In this paper, G is a finite group and p is a prime divisor of $|G|$ . Let $\mathrm{Irr}(G)$ denote the set of (complex) irreducible characters of G. For a normal subgroup N of G and $\theta \in \mathrm{Irr}(N)$ , let $\mathrm{Irr}(G|N) = \{\chi \in \mathrm{Irr}(G) : N \not \leq \mathrm{ker}\,\chi \}$ and $\mathrm{Irr}(\theta ^G)$ denote the set of the irreducible constituents of the induced character $\theta ^G$ . The average character degree of G is denoted by $\mathrm{acd}(G)$ (see [5, 8]) and it is defined by

$$ \begin{align*}\mathrm{acd}(G) =\frac{ \Sigma_{\chi \in \mathrm{Irr}(G)}\chi(1)}{|\mathrm{Irr}(G)|}.\end{align*} $$

By $\mathrm{acd}(G|N)$ , we mean the average character degree of the irreducible characters in $\mathrm{Irr}(G|N)$ (see [3]). In [1], it has been shown that if $\mathrm{acd}(G|N) < \mathrm{\max }(\mathrm{acd}(\mathrm{PSL}_2(p)),16/5)$ , then G is p-solvable.

We write

$$ \begin{align*} \mathrm{Irr}_p(G)&=\{\chi \in \mathrm{Irr}(G): \chi(1)=1 ~\mathrm{or}~ p \mid \chi(1) \}\\ \mathrm{Irr}_p(G|N)&= \mathrm{Irr}_p(G) \cap \mathrm{Irr}(G|N) \\ \mathrm{Irr}_p(\theta^G)&=\mathrm{Irr}_p(G) \cap \mathrm{Irr}(\theta^G) \quad \mbox{for every } \theta \in \mathrm{Irr}(N). \end{align*} $$

Let $\mathrm{acd}_p(G)$ , $\mathrm{acd}_p(G|N)$ and $\mathrm{acd}_p(\theta ^G)$ be the average degree of irreducible characters belonging to $\mathrm{Irr}_p(G)$ , $\mathrm{Irr}_p(G|N)$ and $\mathrm{Irr}_p(\theta ^G)$ , respectively. For $\Delta \subseteq \mathrm{Irr}(G)$ ,

$$ \begin{align*} \mathrm{acd}_p(\Delta)=\frac{\Sigma_{ \chi \in \Delta \cap \mathrm{Irr}_p(G)}\chi(1)}{|\Delta \cap \mathrm{Irr}_p(G)|}. \end{align*} $$

Nguyen and Tiep [7] have shown that if either $p \geq 5$ and $\mathrm{acd}_p(G) <\mathrm{acd}_p(\mathrm{PSL}_2(p))$ or $p \in \{2,3\}$ and $\mathrm{acd}_p(G) <\mathrm{acd}_p(\mathrm{PSL}_2(5))$ , then G is p-solvable and $O^{p'}(G)$ is solvable, where $O^{p'}(G)$ is the minimal normal subgroup of G whose quotient is a $p'$ -group. Akhlaghi [2] proved that if N is a nontrivial normal subgroup of G with $\mathrm{Irr}_2(G|N) \neq \emptyset $ and $\mathrm{acd}_2(G|N) <5/2$ , then G is solvable.

We continue this investigation and show that considering the appropriate bound for $\mathrm{acd}_p(G|N)$ instead of $\mathrm{acd}_p(G)$ leads us to the p-solvability of G.

Let $f(p)=\mathrm{acd}_p(\mathrm{PSL}_2(p))$ if $p \geq 5$ and otherwise, let $f(p)=\mathrm{acd}_p(\mathrm{PSL}_2(5))$ . So,

$$ \begin{align*} f(p)=\left\{\begin{array}{ll} (p+1)/2 & \text{if } p \geq 5,\\ 7/3 & \text{if } p=3,\\ 5/2 & \text{if } p=2. \end{array} \right. \end{align*} $$

Theorem 1.1. Let $ 1 \neq N \unlhd G $ and $p $ be an odd prime divisor of $|G|$ . If $G/N$ is not p-solvable, then $\mathrm{acd}_p(\lambda ^G) \geq f(p)$ for every $\lambda \in \mathrm{Irr}(N)$ with $\mathrm{Irr}_p(\lambda ^G) \neq \emptyset $ .

Theorem 1.2. Let p be an odd prime and $1 \neq N \unlhd G$ with $\mathrm{acd}_p(G | N)<f(p)$ . Then:

1. (i) either G is p-solvable and $O^{p'}(G)$ is solvable;

2. (ii) or $\mathrm{Irr}_p(G|N)=\emptyset $ , N is p-solvable and for every $P \in \mathrm{Syl}_p(G)$ , $P \cap N$ and $PN/N$ are abelian.

Example 1.3. Let N be a cyclic group of order $2$ , $p $ be an odd prime and let $G=\mathrm{PSL}_2(p) \times N$ . If $p \geq 5$ , then $\mathrm{acd}_p(G|N)=\mathrm{acd}_p(\mathrm{PSL}_2(p))$ . Also, if $p=5$ , then $\mathrm{acd}_3(G|N)=\mathrm{acd}_3(\mathrm{PSL}_2(5))$ . This example shows that the bound given in Theorem 1.2 is the best possible.

Let $\mathrm{Irr}_p(G^{\sharp }) = \mathrm{Irr}_p(G)-\{1_G \}$ and $\mathrm{acd}(G^{\sharp }) ={\Sigma _{\chi \in \mathrm{Irr}_p(G^{\sharp })} \chi (1) }/{|\mathrm{Irr}_p(G^{\sharp })|}$ . By setting $G=N$ in Theorem 1.2, we arrive at the following corollary.

Corollary 1.4. If $\mathrm{acd}_p(G^{\sharp })<f(p) $ , then G is p-solvable and $O^{p'}(G)$ is solvable.

We can see that $\mathrm{acd}_3(\mathrm{Alt}_4^{\sharp })=5/3<7/3 $ and the Sylow $3$ -subgroup of $\mathrm{Alt}_4$ is not normal in $\mathrm{Alt}_4$ . This shows that the assumption $\mathrm{acd}_p(G^{\sharp })<f(p) $ does not guarantee normality of the Sylow p-subgroup of G.

2 The main results

We first state some lemmas that will be used in the proof of Theorems 1.1 and 1.2. For a nonempty finite subset of real numbers X, by $\mathrm{ave}(X)$ , we mean the average of X.

Lemma 2.1 [1, Lemma 3].

Let X be a nonempty finite subset of real numbers and $\{A_1,\ldots ,A_t\}$ be a partition of X. If d is a real number such that $\mathrm{ave}( A_i) \geq d$ (respectively $<d$ ) for $1 \leq i \leq t$ , then $\mathrm{ave}( X) \geq d$ (respectively $<d$ ).

Lemma 2.2 [7, Theorem B].

Let p be a prime divisor of $|G|$ . If $\mathrm{acd}_p(G) < f(p)$ , then G is p-solvable and $O^{p'}(G)$ is solvable.

Lemma 2.3 [6, Theorem A].

Let Z be a normal subgroup of a finite group G, $ \lambda \in \mathrm{Irr}(Z)$ and let $P/Z \in \mathrm{Syl}_p(G/Z)$ . If $ \chi (1)/\lambda (1)$ is coprime to p for every $ \chi \in \mathrm{Irr}(G)$ lying over $\lambda $ , then $P/Z$ is abelian.

We are ready to prove Theorems 1.1 and 1.2.

Proof of Theorem 1.1.

We complete the proof by induction on $|G|+|N|$ . Take $\lambda \in \mathrm{Irr}(N)$ with $\mathrm{Irr}_p(\lambda ^G) \neq \emptyset $ . Let E be a maximal normal subgroup of G such that $N \leq E$ and $G/E$ is not p-solvable. Then, $G/E$ admits the unique minimal normal subgroup $M/E$ and it is easy to check that $M/E$ is not p-solvable. Assume that $\{\mu _1, \ldots , \mu _t\} \subseteq \mathrm{Irr}(\lambda ^E)$ such that every element of $\mathrm{Irr}(\lambda ^E)$ is conjugate to exactly one of the elements in $\{\mu _1, \ldots , \mu _t\}$ . If $N \neq E$ , then from the hypothesis, $\mathrm{Irr}_p(\mu _i^G)=\emptyset $ or $\mathrm{acd}_p(\mu _i^G) \geq f(p)$ , for $1 \leq i \leq t$ . As $\mathrm{Irr}(\lambda ^G) =\dot {\cup }_{i=1}^t \mathrm{Irr}(\mu _i^G)$ and $\mathrm{Irr}_p(\lambda ^G) \neq \emptyset $ , we conclude that $\mathrm{Irr}_p(\mu _j^G) \neq \emptyset $ for some j with $1 \leq j \leq t$ . So, it follows from Lemma 2.1 that $\mathrm{acd}_p(\lambda ^G) \geq f(p) $ , as desired. Next, suppose that $N=E$ . If $\lambda $ is extendible to $\chi \in \mathrm{Irr}(G)$ , then Gallagher’s theorem [4, Corollary 6.17] implies that $\mathrm{Irr}(\lambda ^G)=\{\chi \mu : \mu \in \mathrm{Irr}(G/N)\}$ and for every $\mu _1,\mu _2 \in \mathrm{Irr}(G/N)$ with $\mu _1 \neq \mu _2$ , we have $\chi \mu _1 \neq \chi \mu _2$ . Thus, either $p \mid \chi (1)$ and $\mathrm{acd}_p(\lambda ^G)=\chi (1) \mathrm{acd}(G/N)$ or $p \nmid \chi (1)$ and $\mathrm{acd}_p(\lambda ^G)=\chi (1) \mathrm{acd}_p(G/N)$ . Obviously, $\mathrm{acd}(G/N) \geq 1$ . So, in the former case, $\mathrm{acd}_p(\lambda ^G) \geq p> f(p)$ , as needed. Since $G/N$ is not p-solvable, Lemma 2.2 yields $\mathrm{acd}_p(G/N) \geq f(p)$ . Hence, if $p \nmid \chi (1)$ , then $\mathrm{acd}_p(\lambda ^G)=\chi (1) \mathrm{acd}_p(G/N) \geq f(p)$ , as desired. Finally, suppose that $\lambda $ is not extendible to G. Then, for every $\chi \in \mathrm{Irr}(\lambda ^G)$ , $\chi (1)> \lambda (1) \geq 1$ . This means that $p \mid \chi (1) $ for every $\chi \in \mathrm{Irr}_p(\lambda ^G)$ . Therefore, $\mathrm{acd}_p(\lambda ^G) \geq p> f(p)$ . Now, the proof is complete.

Proof of Theorem 1.2.

First, assume that $\mathrm{Irr}_p(G|N) \neq \emptyset $ . As $\mathrm{acd}_p(G|N) < f(p) < p $ , we see that $\mathrm{Irr}_p(G|N)$ contains a linear character $\chi $ . Then, $\chi _N \neq 1_N $ and as $\chi (1)=1$ , we have $\chi _N \in \mathrm{Irr}(N)$ . This implies that N admits some linear characters which are extendible to G and they are nonprincipal. Assume that $\{\mu _1,\ldots , \mu _t\}$ is the set of all linear characters of N which are extendible to G and are nonprincipal. Since the $\mu_i$ s are extendible to G, none of them are G-conjugate. If $1 \leq i \neq j \leq t$ and there exists $\chi \in \mathrm{Irr}(\mu _i^G) \cap \mathrm{Irr}(\mu _j^G)$ , then $\mu _i$ and $\mu _j$ are irreducible constituents of $\chi _N$ . It follows from Clifford’s correspondence that $\mu _i$ and $\mu _j$ are G-conjugate, which is a contradiction with our former assumption on the $\mu _i$ s. This shows that

(2.1) $$ \begin{align} \mathrm{Irr}(\mu_i^G) \cap \mathrm{Irr}(\mu_j^G)=\emptyset \quad\mathrm{for~} 1 \leq i \neq j \leq t. \end{align} $$

Let $1 \leq i \leq t$ . Our assumption on the $\mu _i$ guarantees the existence of a linear character $\chi _i \in \mathrm{Irr}(G) $ such that $(\chi _i)_N=\mu _i$ . By Gallagher’s theorem [4, Corollary 6.17], $\mathrm{Irr}(\mu _i^G)=\{\chi _i \varphi : \varphi \in \mathrm{Irr}(G/N)\}$ and for distinct characters $\varphi _1,\varphi _2 \in \mathrm{Irr}(G/N)$ , $\chi _i\varphi _1 \neq \chi _i \varphi _2$ . Since  $\chi _i(1)=1$ ,

(2.2) $$ \begin{align} \mathrm{Irr}_p(\mu_i^G)=\{\chi_i \varphi: \varphi \in \mathrm{Irr}_p(G/N)\}. \end{align} $$

As $\mu _i \neq 1_N$ , $\chi _i \in \mathrm{Irr}(G|N)$ . Therefore,

$$ \begin{align*} \mathrm{Irr}_p(\mu_i^G) \subseteq \mathrm{Irr}_p(G|N). \end{align*} $$

In view of (2.1), $\bigcup _{i=1}^t\mathrm{Irr}(\mu _i^G)$ is disjoint. Take

$$ \begin{align*} \mathfrak{A}=\mathrm{Irr}_p(G|N) - \dot{\cup}_{i=1}^t\mathrm{Irr}(\mu_i^G). \end{align*} $$

If $\chi \in \mathrm{Irr}(G|N)$ is linear, then $\chi _N \neq 1_N$ and $\chi _N(1)=\chi (1)=1$ . Thus, $\chi _N \in \mathrm{Irr}(N)$ is nonprincipal. It follows from our assumption on the $\mu _i$ that $\chi _N \in \{\mu _1,\ldots , \mu _t\}$ . Therefore, $\chi \in \mathrm{Irr}(\mu _j^G)$ for some $1 \leq j \leq t$ . This implies that $\chi (1) \geq p$ for every $\chi \in \mathfrak {A}$ . Therefore,

(2.3) $$ \begin{align} \mathrm{acd}_p(\mathfrak{A}) \geq p> f(p). \end{align} $$

By (2.1) and (2.2), $|\dot {\cup }_{i=1}^t \mathrm{Irr}_p(\mu _i^G)|=t|\mathrm{Irr}_p(G/N)|$ and

$$ \begin{align*} \mathrm{acd}_p(\dot{\cup}_{i=1}^t \mathrm{Irr}(\mu_i^G))&= \frac{\Sigma_{i=1}^t \Sigma_{\chi \in \mathrm{Irr}_p(\mu_i^G)} \chi(1) }{|\dot{\cup}_{i=1}^t \mathrm{Irr}_p(\mu_i^G)|}\\ &= \frac{\Sigma_{i=1}^t \Sigma_{\varphi \in \mathrm{Irr}_p(G/N)} (\chi_i \varphi)(1)}{t|\mathrm{Irr}_p(G/N)|}\\\nonumber &= \frac{t \Sigma_{\varphi \in \mathrm{Irr}_p(G/N)} \varphi(1)}{t|\mathrm{Irr}_p(G/N)|}=\mathrm{acd}_p(G/N). \end{align*} $$

If $\mathrm{acd}_p(G/N) \geq f(p)$ , then

(2.4) $$ \begin{align} \mathrm{acd}_p(\dot{\cup}_{i=1}^t \mathrm{Irr}(\mu_i^G)) \geq f(p). \end{align} $$

Note that $\mathrm{Irr}_p(G|N) = (\dot {\cup }_{i=1}^t\mathrm{Irr}_p(\mu _i^G)) \dot {\cup } \mathfrak {A}$ . It follows from (2.3), (2.4) and Lemma 2.1 that $\mathrm{acd}_p(G|N) \geq f(p)$ , which is a contradiction. This implies that $\mathrm{acd}_p(G/N) <f(p)$ . As $\mathrm{acd}_p(G|N) < f(p)$ and $\mathrm{Irr}_p(G) =\mathrm{Irr}_p(G|N) \dot {\cup } \mathrm{Irr}_p(G/N)$ , we deduce from Lemma 2.1 that $\mathrm{acd}_p(G) < f(p)$ . Hence, Lemma 2.2 implies that G is p-solvable and $O^{p'}(G)$ is solvable, as desired.

Now, assume that $\mathrm{Irr}_p(G|N)=\emptyset $ . Working towards a contradiction, suppose that there exists $\theta \in \mathrm{Irr}(N) $ such that $p \mid \theta (1)$ . We have $\theta (1) \mid \chi (1)$ for every $\chi \in \mathrm{Irr}(\theta ^G)$ . Thus, $p \mid \chi (1)$ for every $\chi \in \mathrm{Irr}(\theta ^G)$ . Clearly, $\theta \neq 1_N$ . So, $\chi \in \mathrm{Irr}_p(\theta ^G) \subseteq \mathrm{Irr}_p(G|N) $ . This means that $\mathrm{Irr}_p(G|N) \neq \emptyset $ , which is a contradiction. This implies that $p \nmid \theta (1)$ for every $\theta \in \mathrm{Irr}(N)$ . It follows from the Ito–Michler theorem [4, Corollary 12.34] that N has a normal and abelian Sylow p-subgroup. Thus, N is p-solvable. Now, assume that $1_N \neq \theta \in \mathrm{Irr}(N)$ and $\chi \in \mathrm{Irr}(\theta ^G)$ . Hence, $\chi \in \mathrm{Irr}(G|N) $ . As $\mathrm{Irr}_p(G|N) = \emptyset $ , we deduce that $p \nmid \chi (1)$ . Thus, $p \nmid \chi (1)/\theta (1)$ . It follows from Lemma 2.3 that $G/N$ has an abelian Sylow p-subgroup. This completes the proof.