FINITE GROUPS WHOSE CHARACTER CODEGREES ARE CONSECUTIVE INTEGERS

Abstract

Let G be a finite group. We investigate the structure of finite groups whose irreducible character codegrees are consecutive integers.

1. Introduction

Throughout this paper, G always denotes a finite group. As usual, $\mathrm {Irr}(G)$ denotes the set of complex irreducible characters of G and $\mathrm {cd}(G)=\{\chi (1)\mid \chi \in \mathrm {Irr}(G)\}$ the set of character degrees. A number of papers, such as [5–7], have studied the influence of the set $\mathrm {cd}(G)$ on the structure of $G.$ In particular, Huppert in [3, Theorem 32.1] considered finite groups whose irreducible character degrees are consecutive integers and showed that if $\mathrm {cd}(G)=\{1,2,\ldots , k-1, k\},$ then G is solvable if and only if $k\leqslant 4,$ and that if $k>4,$ then $k=6$ and $G=HZ(G),$ where $H\cong \mathrm {SL}(2,5).$

Inspired by these results, we consider the analogous problem related to the character codegrees. The concept of character codegrees was first introduced by Qian et al. in [9] as follows. For $\chi \in \mathrm {Irr}(G),$ the codegree of $\chi $ is defined to be

$$ \begin{align*} \mathrm{cod}\,\chi=\frac{|G:\ker\chi|}{\chi(1)}. \end{align*} $$

Recently many papers have studied character codegrees (see, for instance, [4, 8, 10]). Let $\mathrm {Cod}(G)=\{\mathrm {cod}\,\chi \mid \chi \in \mathrm {Irr}(G)\}$ be the set of irreducible character codegrees of $G.$ The aim of this paper is to investigate finite groups whose irreducible character codegrees are consecutive integers. We have the following result.

Theorem 1.1. Let G be a group with $\mathrm {Cod}(G)=\{1,2,\ldots , n-1, n\},$ where n is a positive integer. Then $n\leqslant 3$ and one of the following holds:

1. (1) if $n=1,$ then $G=1$ ;

2. (2) if $n=2,$ then G is an elementary abelian $2$ -group;

3. (3) if $n=3,$ then $G=N\rtimes H$ is a Frobenius group with an elementary abelian $3$ -group as its kernel, $N=G'$ and H is cyclic of order $2.$

2. Preliminaries

We begin with the following basic lemma concerning character codegrees, which will be used frequently in our proofs.

Lemma 2.1 [9, Lemma 2.1].

Let G be a group and $\chi \in \mathrm {Irr}(G).$

1. (1) If N is a normal subgroup of $G,$ then $\mathrm {Cod}(G/N) \subseteq \mathrm {Cod}(G).$

2. (2) If N is subnormal in G and $\phi \in \mathrm {Irr}(N)$ is a constituent of $\chi _N,$ then $\mathrm {cod}\,\phi \mid \mathrm {cod}\,\chi .$

Next we recall the concept of the codegree graph, which was first introduced in [9]. The codegree graph $\Gamma (G)$ is a graph whose vertex set $V(G)$ is the set of all primes dividing $\mathrm {cod}\,\chi $ for some $\chi \in \mathrm {Irr}(G)$ and there is an edge between two distinct primes p and q if $pq$ divides $\mathrm {cod}\,\chi $ for some $\chi \in \mathrm {Irr}(G).$ We present the following facts on the codegree graph $\Gamma (G).$

Lemma 2.2 [9, Theorems A and E].

Let G be a group and $\pi (G)$ be the set of prime divisors of $|G|.$

1. (1) $\pi (G)$ coincides with $V(G),$ the vertex set of $\Gamma (G).$

2. (2) For any subset $\Delta \subseteq \pi (G)$ with $|\Delta |\geqslant 3,$ there are two distinct primes $p,q\in \Delta $ so that there is an edge between p and $q.$

3. (3) $\Gamma (G)$ is not connected if and only if G is a Frobenius group or a $2$ -Frobenius group.

3. Proof of Theorem 1.1

We start by proving the following result concerning number theory, which plays a very important role in determining the integer n when $\mathrm { Cod}(G)=\{1,2,\ldots , n-1, n\}.$

Proposition 3.1. Let n be an integer and $r,q,p$ be three consecutive primes so that $2<r<q<p\leqslant n$ and p is the largest prime less than or equal to $n.$ Then $n<2p$ and $n<rq.$

Proof. Assume that $n\geqslant 2p.$ Then $p<2p\leqslant n.$ By Bertrand’s postulate, there exists a prime, say $s,$ so that $p<s<2p.$ This contradicts the hypothesis that p is the largest prime less than or equal to $n.$ Hence, $n<2p.$

Now assume that $n\geqslant rq.$ Applying Bertrand’s postulate again, we see that ${q<p<2q}$ and so $q<p<2q<3q\leqslant rq\leqslant n.$ By [1, Theorem 1.3], there is a prime between $2q$ and $3q.$ This is a contradiction. Thus, $n<rq.$

Proposition 3.1 enables us show that the integer n will not be too large.

Proposition 3.2. Let G be a group with $\mathrm {Cod}(G)=\{1,2,\ldots , n-1, n\},$ where n is a positive integer. Then $n\leqslant 6$ and $n\not =5.$

Proof. Assume that $n\geqslant 7.$ Then there are three consecutive primes $r,q,p$ as defined in Proposition 3.1. Thus, $n<2p$ and $n<rq.$ Consider the subset $\Delta =\{r,p,q\}\subseteq V(G)=\pi (G).$ By Lemma 2.2(2), there exists $\chi \in \mathrm {Irr}(G)$ so that $pq\mid \mathrm {cod}\,\chi $ , $rq\mid \mathrm {cod}\,\chi $ or $rp\mid \mathrm {cod}\,\chi .$ It follows from Proposition 3.1 that $n\geqslant \mathrm { cod}\,\chi \geqslant \min \{pq,rq,rp\}>n.$ This is a contradiction. Thus, $n\leqslant 6.$ Similarly, by Lemma 2.2(2), $n\not = 5.$

With the above proposition, to prove Theorem 1.1, we only need to classify the groups when $1\leqslant n\leqslant 3$ and show that $n\not =4,6.$ Notice that if $n\leqslant 6$ , the codegree graph $\Gamma (G)$ is not connected. Then by Lemma 2.2(3), G is a Frobenius group or a 2-Frobenius group. So we need to understand the structure of Frobenius groups. In particular, we give the following proposition.

Proposition 3.3. Let $G=N\rtimes H$ be a Frobenius group with kernel $N.$ Suppose that $\pi (N)=\{p_1, p_2,\ldots ,p_s\}.$ Then the following statements hold.

1. (1) If $\phi \in \mathrm {Irr}(N),$ then $\mathrm {cod}\,\phi \mid \mathrm {cod}\,\chi $ for some $\chi \in \mathrm {Irr}(G).$ In particular, $\prod _{i=1}^{s}p_i\mid \mathrm {cod}\,\chi $ for some $\chi \in \mathrm {Irr}(G).$

2. (2) $\mathrm {Cod}(G){\kern-0.5pt}={\kern-0.5pt}\mathrm {Cod}(G/N)\bigcup \{\mathrm {cod}(\phi ^G)\mid 1_N{\kern-0.5pt}\not ={\kern-0.5pt}\phi {\kern-0.5pt}\in{\kern-0.5pt} \mathrm {Irr}(N)\}.$ Furthermore, $\mathrm {cod}(\phi ^G)$ divides $|N|$ if $1_N\not =\phi \in \mathrm {Irr}(N).$

Proof. (1) The first part follows from Lemma 2.1(2) immediately. Notice that N is nilpotent. Then $N=P_1\times P_2\times \cdots \times P_s,$ where $P_i$ is a Sylow $p_i$ -subgroup of $N.$ Let $1_{P_i}\not =\lambda _i\in \mathrm {Irr}(P_i)$ and set $\phi =\lambda _1\times \lambda _2\times \cdots \times \lambda _s.$ Then $\phi \in \mathrm {Irr}(N)$ and $ \mathrm {cod}\,\phi =\prod _{i=1}^{s}\mathrm {cod}\,\lambda _i$ with $p_i\mid \mathrm { cod}\,\lambda _i.$ Hence, by Lemma 2.1(2), $ \prod _{i=1}^{s}p_i\mid \mathrm {cod}\,\chi $ for some $\chi \in \mathrm {Irr}(G),$ as required.

(2) It is well known that $\mathrm {Irr}(G)=\mathrm {Irr}(G/N)\bigcup \{\phi ^G\mid 1_N\not =\phi \in \mathrm {Irr}(N)\}.$ Thus, the first part is true. Notice that $\phi ^G(1)=|G:N|\phi (1).$ Then

$$ \begin{align*} \displaystyle \mathrm{cod}(\phi^G)=\frac{|G:N||N|}{|G:N|\phi(1)|\mathrm{ker}\,\phi^G|}=\frac{|N|}{\phi(1)|\mathrm{ker}\,\phi^G|} \end{align*} $$

divides $|N|,$ as required.

Proposition 3.4. Let G be a group with $|\pi (G)|=3.$ Suppose that $\mathrm {Cod}(G)\subseteq \{1,2,3,4,5,6\}.$ Then G is not a Frobenius group.

Proof. We work by contradiction. Assume that $G=N\rtimes H$ is a Frobenius group with kernel $N.$ By Lemma 2.2(1) and (2), $\pi (G)=\{2,3,5\}$ and $6\in \mathrm {Cod}(G).$ First we consider the case when $|\pi (N)|=2.$ Since N is nilpotent, it follows from Proposition 3.3(1) that $\pi (N)=\{2,3\}$ and N is a direct product of an elementary abelian 2-group and an elementary abelian 3-group. Notice that the complement H is a cyclic 5-group and $\mathrm {Cod}(H)\subseteq \mathrm {Cod}(G).$ Then H must be cyclic of order 5. Since there is $\phi \in \mathrm {Irr}(N)$ so that $\mathrm {cod}(\phi ^G)=|N|/|\mathrm {ker}\,\phi ^G|=6,$ we have $\mathrm {ker}\,\phi ^G<N$ and so $G/\mathrm {ker}\,\phi ^G=N/\mathrm { ker}\phi ^G\rtimes H\mathrm {ker}\phi ^G/\mathrm {ker}\phi ^G\cong C_6\rtimes C_5\cong C_{30}.$ Hence, $30\in \mathrm {Cod}(G),$ a contradiction.

Assume now that $|\pi (N)|=1.$ By Proposition 3.3(2), $\pi (N)=\{5\}$ and so N is elementary abelian. Since there is $\phi \in \mathrm {Irr}(N)$ so that $\mathrm {cod}(\phi ^G)=|N|/|\mathrm {ker}\,\phi ^G|=5,$ we have $\mathrm {ker}\,\phi ^G<N.$ Let $\overline {G}=G/\mathrm {ker}\,\phi ^G.$ Then $\overline {G}=\overline {N}\rtimes \overline {H}$ is a Frobenius group with kernel $\overline {N}\cong C_5$ and $\overline {H}\cong H$ . Let $\overline {Q}$ be a Sylow 3-subgroup of $\overline {H}.$ Then $\overline {Q}$ is cyclic of order 3. It follows that $\overline {N}\overline {Q}$ is a Frobenius group of order 15. This is a contradiction since such a group does not exist.

Both cases are impossible. The proof is completed.

For convenience, here we introduce the notation of 2-Frobenius groups. If G is a 2-Frobenius group, then there are normal subgroups $N, M$ of G so that $G/N$ is a Frobenius group with kernel $M/N$ , and M is a Frobenius group with kernel $N.$ We write $G=\mathrm {Frob}_2(G,M,N)$ to denote such a 2-Frobenius group.

Proof of Theorem 1.1.

We first introduce two basic facts.

1. (A) $\mathrm {cod}\,\chi> \chi (1)$ if $1_{G}\not = \chi \in \mathrm {Irr}(G)$ .

2. (B) If G is abelian, then $\mathrm {cod}\,\chi $ is equal to the order of $\chi $ in the group $\mathrm {Irr}(G)\cong G.$

There is nothing to prove when $n=1.$ Assume that $n\geqslant 2.$ Applying fact (A), together with $2\in \mathrm {Cod}(G),$ we see that there exists a linear character $\chi \in \mathrm {Irr}(G)$ such that $\mathrm {cod}\,\chi =2$ and hence $\chi \in \mathrm {Irr}(G/G').$ Then it follows from fact (B) that $2\mid |G:G'|.$

If $n=2,$ then by facts (A) and (B), G is an elementary abelian $2$ -group and (2) follows.

Assume that $n=3.$ Then by Lemma 2.2(1) and (3), $\pi (G)=\{2,3\}$ and G is a Frobenius group or a 2-Frobenius group. First suppose that $G=N\rtimes H$ is a Frobenius group with kernel $N.$ Since $2\mid |G:G'|,$ it follows that H is a 2-group and N is a 3-group. By Proposition 3.3, $\mathrm {Cod}(N)=\{1,3\}$ and $\mathrm {Cod}(G/N)=\mathrm {Cod}(H)=\{1,2\}.$ Therefore, N is an elementary abelian 3-group and H is an elementary abelian 2-group. Notice that the complement H must be cyclic or a generalised quaternion group (see [2, Theorem 9.2.10]). Hence, H is cyclic of order 2. Since $6\not \in \mathrm { Cod}(G),$ we have $G'=N.$ To complete the proof of (3), we only need to show that G is not a 2-Frobenius group. Assume that $G=\mathrm {Frob}_2(G,M,N)$ is a 2-Frobenius group. It follows from Proposition 3.3 that $\mathrm {Cod}(G/N)=\mathrm {Cod}(M)=\{1,2,3\}.$ Similarly, $G/N\cong C_3^s\rtimes C_2$ and $M\cong C_3^t\rtimes C_2$ for some positive integers s and $t.$ This is a contradiction. Hence, (3) follows.

Now we show that $n\not =4.$ If $n=4,$ then $\pi (G)=\{2,3\}$ and G is a Frobenius group or a 2-Frobenius group. First assume that $G=N\rtimes H$ is a Frobenius group with kernel $N.$ Then by a proof similar to that above, N is an elementary abelian 3-group and H is a 2-group with $\mathrm {Cod}(H)=\{1,2,4\}.$ Together with the fact that the complement H must be cyclic or a generalised quaternion group, we have $H\cong C_4$ or $Q_8.$ Notice that there exists $\phi \in \mathrm {Irr}(N)$ so that $\mathrm {cod}(\phi ^G)=|N|/|\mathrm {ker}\,\phi ^G|=3.$ It is obvious that $\mathrm {ker}\,\phi ^G<N.$ Then $G/\mathrm {ker}\,\phi ^G=N/\mathrm {ker}\,\phi ^G\rtimes H\mathrm {ker}\,\phi ^G/\mathrm {ker}\,\phi ^G\cong C_3\rtimes C_4$ or $C_3\rtimes Q_8$ is a Frobenius group, which is a contradiction. Thus, G cannot be a Frobenius group. Assume that $G=\mathrm {Frob}_2(G,M,N)$ is a 2-Frobenius group. Since $G/N$ is Frobenius and $\mathrm {Cod}(G/N) \subseteq \mathrm {Cod}(G),$ we have $\mathrm {Cod}(G/N)=\{1,2,3\}.$ Similarly, $\mathrm {Cod}(M)=\{1,2,3\}.$ This cannot happen by statement (2) of this theorem. Hence, $n\not =4.$

By Proposition 3.2, it remains to show that $n\not =6.$ If $n=6,$ then $\pi (G)=\{2,3,5\}$ and G is a Frobenius group or a 2-Frobenius group. It follows by Proposition 3.4 that G is not a Frobenius group. We may assume that $G=\mathrm {Frob}_2(G,M,N)$ is a 2-Frobenius group. By Proposition 3.3(1) and (2), $\mathrm {Cod}(G/N) \subseteq \mathrm {Cod}(G)$ and $\mathrm {Cod}(M/N) \subseteq \mathrm {Cod}(M) \subseteq \mathrm { Cod}(G).$ Since both $G/N$ and M are Frobenius groups, it follows by Proposition 3.4 that $|\pi (G/N)|=|\pi (M)|=2.$ Write $\overline {G}=G/N$ and then $\overline {G}=\overline {M}\rtimes \overline {K},$ where $\overline {K}$ is the Frobenius complement. First consider the case when $\pi (\overline {K})\not =\{2\}.$ As $\overline {K}$ is cyclic and $\mathrm {Cod}(\overline {K}) \subseteq \mathrm {Cod}(G),$ we have $G'\leqslant M$ and $\overline {K}$ is cyclic of order 3 or 5. Notice that $2\mid |G:G'|.$ Then $\overline {K}\cong C_3$ and $\overline {M}$ is a 2-group. Since M is Frobenius and $\overline {M}$ is isomorphic to its complement, it follows that $\overline {M}$ is cyclic or a generalised quaternion group and hence $\overline {M}\cong C_2, C_4$ or $Q_8.$ But $\overline {G}\cong \overline {M}\rtimes C_3$ is Frobenius, which is impossible. Assume now that $\pi (\overline {K})=\{2\}.$ It is obvious that $\pi (M)=\{3,5\}.$ Let $M=N\rtimes H$ be a Frobenius group with kernel $N.$ By a similar proof, there exists $\phi \in \mathrm {Irr}(N)$ so that $G/\mathrm { ker}\phi ^G\cong N/\mathrm {ker}\phi ^G\rtimes H\cong C_{15}.$ It follows that $15\in \mathrm {Cod}(G),$ a contradiction. Hence, $n\not =6.$ The proof is completed.