ON THE CHARACTERISATION OF ALTERNATING GROUPS BY CODEGREES

Abstract

Let G be a finite group and $\mathrm {Irr}(G)$ the set of all irreducible complex characters of G. Define the codegree of $\chi \in \mathrm {Irr}(G)$ as $\mathrm {cod}(\chi ):={|G:\mathrm {ker}(\chi ) |}/{\chi (1)}$ and let $\mathrm {cod}(G):=\{\mathrm {cod}(\chi ) \mid \chi \in \mathrm {Irr}(G)\}$ be the codegree set of G. Let $\mathrm {A}_n$ be an alternating group of degree $n \ge 5$. We show that $\mathrm {A}_n$ is determined up to isomorphism by $\operatorname {cod}(\mathrm {A}_n)$.

1 Introduction

Let G be a finite group and $\mathrm {Irr}(G)$ the set of all irreducible complex characters of G. For any $\chi \in \mathrm {Irr}(G),$ define the codegree of $\chi $ as $\mathrm {cod}(\chi ) := {|G:\mathrm {ker}(\chi )|}/{\chi (1)}$ and the codegree set of G as $\mathrm {cod}(G) := \{\mathrm {cod}(\chi ) \mid \chi \in \mathrm {Irr}(G)\}.$ We refer the reader to the authors’ previous paper [8] for the current literature on codegrees.

The following conjecture appears in the Kourovka Notebook of Unsolved Problems in Group Theory [12, Question 20.79].

Codegree Version of Huppert’s Conjecture. Let H be a finite nonabelian simple group and G a finite group such that $\mathrm {cod}(H) = \mathrm {cod}(G).$ Then $G \cong H$ .

In [8], the authors verified the conjecture for all sporadic simple groups. In this paper, we provide a general proof verifying this conjecture for all alternating groups of degree greater than or equal to $5$ .

Theorem 1.1. Let $\mathrm {A}_n$ be an alternating group of degree $n \ge 5$ and G a finite group. If $\mathrm {cod}(G)= \mathrm {cod}(\mathrm {A}_n)$ , then $G \cong \mathrm {A}_n$ .

Throughout the paper, we follow the notation used in Isaacs’ book [10] and the ATLAS of Finite Groups [6].

2 Proof of Theorem 1.1

First, we note that the cases $n=5,6$ and $7$ have already been proven in [1, 2], so in the following, we always assume that $n>7.$ Now, let G be a minimal counterexample and N be a maximal normal subgroup of G. So $\mathrm {cod}(G)=\mathrm {cod}(\mathrm {A}_n)$ and $G/N$ is simple. By [8, Lemma 2.5], $\mathrm {cod}(G/N)\subseteq \mathrm {cod}(\mathrm {A}_n)$ . Then, by [9, Theorem B], $G/N\cong \mathrm {A}_n$ so $N\neq 1$ since $G\not \cong \mathrm {A}_n$ .

Step 1: N is a minimal normal subgroup of G.

Suppose L is a nontrivial normal subgroup of G with $L < N$ . Then by [8, Lemma 2.6], $\mathrm {cod}(G/N) \kern1.3pt{\subseteq}\kern1.3pt \mathrm {cod}(G/L) \kern1.3pt{\subseteq}\kern1.3pt \mathrm {cod}(G)$ . However, ${\mathrm {cod}(G/N)\kern1.3pt{=}\kern1.3pt\mathrm {cod}(\mathrm {A}_n)\kern1.3pt{=}\kern1.3pt\mathrm {cod}(G)}$ , so equality must be attained in each inclusion. Thus, $\mathrm {cod}(G/L)=\mathrm {cod}(\mathrm {A}_n)$ which implies that $G/L \cong \mathrm {A}_n$ since G is a minimal counterexample. This is a contradiction since we also have $G/N\cong \mathrm {A}_n,$ but $L < N$ .

Step 2: N is the only nontrivial, proper normal subgroup of G.

Otherwise, we assume M is another proper nontrivial normal subgroup of G. If N is included in M, then $M=N$ or $M=G$ since $G/N$ is simple, which is a contradiction. Then $N\cap M=1$ and $G=N\times M$ . Since M is also a maximal normal subgroup of G, we have $N\cong M\cong \mathrm {A}_n$ . Choose $\psi _1\in \operatorname {Irr}(N)$ and $\psi _2\in \operatorname {Irr}(M)$ such that $\operatorname {cod}(\psi _1)=\operatorname {cod}(\psi _2)=\max (\operatorname {cod}(\mathrm {A}_n))$ . Set $\chi =\psi _1\cdot \psi _2\in \operatorname {Irr}(G)$ . Then $\operatorname {cod}(\chi )=(\max (\operatorname {cod}(\mathrm {A}_n)))^2\notin \operatorname {cod}(G)$ , which is a contradiction.

Step 3: $\chi $ is faithful, for each nontrivial $\chi \in \mathrm {Irr}(G|N):=\mathrm {Irr}(G)-\mathrm {Irr}(G/N$ ).

From the proof of [8, Lemma 2.5],

$$ \begin{align*} \operatorname{Irr}(G/N) = \{ \hat{\chi}(gN) = \chi(g) \mid \chi \in \operatorname{Irr}(G) \text{ and } N \le \operatorname{ker}(\chi)\}. \end{align*} $$

By the definition of $\mathrm {Irr}(G|N)$ , it follows that if $\chi \in \mathrm {Irr}(G|N),$ then $N \not \leq \mathrm {ker}(\chi ).$ Thus, since N is the unique nontrivial, proper, normal subgroup of G, $\mathrm {ker}(\chi ) = G$ or $\mathrm {ker}(\chi )~=~1$ . Therefore, $\mathrm {ker}(\chi ) = 1$ for all nontrivial $\chi \in \mathrm {Irr}(G|N).$

Step 4: N is an elementary abelian group.

Suppose that N is not abelian. Since N is a minimal normal subgroup, by [7, Theorem 4.3A(iii)], $N=S^n$ , where S is a nonabelian simple group and $n\in \mathbb {Z}^+$ . By [14, Lemma 4.2] and [11, Theorem 4.3.34], there is a nontrivial character $\chi \in \mathrm {Irr}(N)$ which extends to some $\psi \in \mathrm {Irr}(G).$ Now, ker $(\psi )=1$ by Step 3, so cod $(\psi )=|G|/\psi (1)=|G/N|\cdot |N|/\chi (1).$ However, by assumption, $\mathrm {cod}(G)= \mathrm {cod}(\mathrm {A}_n)= \mathrm {cod}(G/N)$ . Thus, cod $(\psi ) \in \mathrm {cod}(G)= \mathrm {cod}(G/N),$ so $\mathrm {cod}(\psi )=|G/N|/\phi (1)$ for some $\phi \in \mathrm {Irr}(G/N).$ Hence, $|G/N|$ is divisible by $\mathrm {cod}(\psi )$ which contradicts the fact that $\mathrm {cod}(\psi )=|G/N|\cdot |N|/\chi (1),$ as $\chi (1)\neq |N|.$ Thus, N must be abelian.

Now to show that N is elementary abelian, let a prime p divide $|N|.$ Then N has a p-Sylow subgroup K, and K is the unique p-Sylow subgroup of N since N is abelian, so K is characteristic in N. Thus, K is a normal subgroup of $G,$ so $K=N$ as N is minimal. Thus, $|N|=p^n.$ Now, take the subgroup $N^p=\{n^p \mid n \in N\}$ of $N,$ which is proper by Cauchy’s theorem. Since $N^p$ is characteristic in $N,$ it must be normal in $G,$ so $N^p$ is trivial by the uniqueness of $N.$ Thus, every element of N has order p and N is elementary abelian.

Step 5: $\mathbf {C}_G(N) = N.$

First note that since N is normal, $\mathbf {C}_G(N) \trianglelefteq G.$ Additionally, since N is abelian by Step 4, $N \leq \mathbf {C}_G(N)$ . By the maximality of $N,$ we must have $\mathbf {C}_G(N) = N$ or $\mathbf {C}_G(N) = G.$ If $\mathbf {C}_G(N) = N,$ we are done.

If not, then $\mathbf {C}_G(N) = G,$ so N must be in the centre of $G.$ Then since N is the unique minimal normal subgroup of G by Step 2, $|N|$ must be prime. If not, there always exists a proper nontrivial subgroup K of $N,$ and K is normal since it is contained in $\mathbf {Z}(G),$ contradicting the minimality of $N.$ Hence, we have $N\le \mathbf {Z}(G)$ which implies that $\mathbf {Z}(G)\cong N$ . This is because N is a maximal normal subgroup of G so if not, we would have $\mathbf {Z}(G)=G$ , implying G is abelian which is a contradiction. Thus, N is isomorphic to a subgroup of the Schur multiplier of $G/N$ by [10, Corollary 11.20].

Now, we note that it is well known that for $n>7$ , the Schur multiplier of $\mathrm {A}_n$ is $\mathbb {Z}_2,$ so $G\cong 2.\mathrm {A}_n$ [17]. From [3, Theorem 4.3], $2.\mathrm {A}_n$ always has a faithful irreducible character $\chi $ of degree $2^{\lfloor (n-2)/2\rfloor }$ . Recall that by Step 2, there is only one nontrivial proper normal subgroup of $G \cong 2.\mathrm {A}_n$ . In particular, $N \cong \mathbb {Z}_2$ is the only nontrivial proper normal subgroup of G. Thus, $|\mathrm {ker}(\chi )|~=~1$ or $2$ . Then $\operatorname {cod}(\chi )={|2.\mathrm {A}_n:\mathrm {ker}(\chi )|}/{\chi (1)}$ . If $|\mathrm {ker}(\chi )|=1,$ then $\operatorname {cod}(\chi )={n!}/{2^{\lfloor (n-2)/2\rfloor }},$ and if $|\mathrm {ker}(\chi )|=2,$ then $\operatorname {cod}(\chi )={(n!/2)}/{2^{\lfloor (n-2)/2\rfloor }}= {n!}/{2^{\lfloor n/2\rfloor }}.$ In either case, for any prime $p\neq 2, |\operatorname {cod}(\chi )|_p=|n!|_p=|\mathrm {A}_n|_p$ . However, $\operatorname {cod}(\chi )\in \operatorname {cod}(\mathrm {A}_n)$ since $\operatorname {cod}(G)=\operatorname {cod}(\mathrm {A}_n)$ . Therefore, there is a character degree of $\mathrm {A}_n$ which is a power of $2$ .

However, from [13], for $n>7, \mathrm {A}_n$ only has a character degree equal to a power of $2$ when $n=2^d+1$ for some positive integer d. In this case, $2^d=n-1\in \mathrm {cd}(\mathrm {A}_n)$ so we need ${|\mathrm {A}_n|}/{n-1}={|2.\mathrm {A}_n|}/{2^{\lfloor (n-2)/2\rfloor }}$ or ${|2.\mathrm {A}_n|}/{2^{\lfloor n/2\rfloor }}$ . Hence,

$$ \begin{align*} \frac{1}{n-1}=\frac{2}{2^{\lfloor (n-2)/2\rfloor}}=\frac{1}{2^{\lfloor (n-2)/2\rfloor-1}} \quad\mbox{or}\quad \frac{1}{2^{\lfloor n/2\rfloor-1}} \end{align*} $$

so $n-1=2^{\lfloor (n-2)/2\rfloor -1}$ or $2^{\lfloor n/2\rfloor -1}$ . However, the only integer solution to either of these equations occurs when $n=9$ and $9-1=8=2^3=2^{\lfloor 9/2\rfloor -1}$ . In this case, we check the ATLAS [6] to find that the codegree sets of $A_9$ and $2.A_9$ do not have the same order. This is a contradiction, so $\mathbf {C}_G(N)=N$ .

Step 6. Let $\lambda $ be a nontrivial character in $\mathrm {Irr}(N)$ and $\vartheta \in \mathrm {Irr}(I_G(\lambda )|\lambda ),$ the set of irreducible constituents of $\lambda ^{I_G(\lambda )},$ where $I_G(\lambda )$ is the inertia group of $\lambda $ in G. Then ${|I_G(\lambda )|}/{\vartheta (1)} \in \mathrm {cod}(G).$ Also, $\vartheta (1)$ divides $|I_G(\lambda )/N|$ and $|N|$ divides $|G/N|.$ Lastly, $I_G(\lambda ) < G,$ that is, $\lambda $ is not G-invariant.

Let $\lambda $ be a nontrivial character in $\operatorname {Irr}(N)$ and $\vartheta \in \operatorname {Irr}(I_G(\lambda )|\lambda )$ . Let $\chi $ be an irreducible constituent of $\vartheta ^G.$ By [10, Corollary 5.4], $\chi \in \operatorname {Irr}(G)$ , and by [10, Definition 5.1], we have $\chi (1) = ({|G|}/{|I_G(\lambda )|}) \cdot \vartheta (1)$ . Moreover, $\operatorname {ker}(\chi ) = 1$ by Step 2, and thus $\operatorname {cod}(\chi ) = {|G|}/{\chi (1)} = {|I_G(\lambda )|}/{\vartheta (1)}$ , so ${|I_G(\lambda )|}/{\vartheta (1)} \in \operatorname {cod}(G)$ . Now, since N is abelian, $\lambda (1) = 1$ , so we have $\vartheta (1) = \vartheta (1)/\lambda (1)$ which divides ${|I_G(\lambda )|}/{|N|}$ , so $|N|$ divides ${|I_G(\lambda )|}/{\vartheta (1)}$ . Moreover, $\operatorname {cod}(G) = \operatorname {cod}(G/N),$ and all elements in $\operatorname {cod}(G/N)$ divide $|G/N|$ , so $|N|$ divides $|G/N|$ .

Next, we want to show $I_G(\lambda )$ is a proper subgroup of G. To reach a contradiction, assume $I_G(\lambda ) = G$ . Then $\operatorname {ker}(\lambda ) \unlhd G$ . From Step 2, $\operatorname {ker}(\lambda ) = 1,$ and from Step 4, N is a cyclic group of prime order. Thus, by the Normaliser–Centraliser theorem, $G / N= \mathbf {N}_{G}(N) / \mathbf {C}_{G}(N) \leq \operatorname {Aut}(N)$ so $G / N$ is abelian, which is a contradiction.

Step 7: Final contradiction.

From Step 4, N is an elementary abelian group of order $p^m$ for some prime p and integer $m\geq 1$ . By the Normaliser–Centraliser theorem, $\mathrm {A}_n \cong G/N = \mathbf {N}_G(N)/\mathbf {C}_G(N) \leq \mathrm {Aut}(N)$ and $m>1$ . Note that in general, $\mathrm {Aut}(N)\cong \mathrm {GL}(m,p)$ . By Step 6, $|N|$ divides $|G/N|,$ so $|N|=p^m$ divides $|\mathrm {A}_n|$ and $G/N\cong \mathrm {A}_n\lesssim \mathrm {GL}(m,p).$ We prove by contradiction that this cannot occur.

First, we claim that if $p^m$ divides $|\mathrm {A}_n|$ and $\mathrm {A}_n\lesssim (\mathrm {GL}(m,p),$ then p must equal $2$ . To show this, we note that for $p>2,$ by [4], if $p^m$ divides $|\mathrm {A}_n|,$ then $m < {n}/{2}$ . However, by [16, Theorem 1.1], if $n>6$ , the minimal faithful degree of a modular representation of $\mathrm {A}_n$ over a field of characteristic p is at least $n-2$ . Since embedding $\mathrm {A}_n$ as a subgroup of $\mathrm {GL}(m,p)$ is equivalent to giving a faithful representation of degree m over a field of characteristic p, we have $m\geq n-2$ . This is a contradiction since ${n}/{2}> n-2$ implies $n<4$ . Therefore, $p=2$ .

Now, let $p=2$ . As above, from [4], we obtain $|n!|_2\leq 2^{n-1}$ . Thus, if $2^m$ divides $|\mathrm {A}_n|,$ then $2^m \leq |\mathrm {A}_n|_2 \leq 2^{n-2}$ so $m\le n-2.$ We will deal first with $n>8$ and then treat the case $n=8$ later. For $n>8$ , [15, Theorem 1.1] shows that the minimal faithful degree of a modular representation of $\mathrm {A}_n$ over a field of characteristic $2$ is at least $n-2$ . Therefore, we must have $m\geq n-2$ , so we have equality, $m=n-2.$

Let $\lambda \in \mathrm { Irr}(N), \vartheta \in \mathrm {Irr}(I_G(\lambda )|\lambda )$ and $T := I_G(\lambda )$ . Then $1<|G:T|<|N|=2^{n-2}$ for $|G:T|$ is the number of all conjugates of $\lambda $ . By Step 5, ${|T|}/{\vartheta (1)}\in \mathrm {cod}(G)$ and moreover $|N|$ divides ${|T|}/{\vartheta (1)}$ . Since $|N|=|N|_2=|\mathrm {A}_n|_2$ and $\mathrm {cod}(G)=\mathrm {cod}(\mathrm {A}_n)$ , it follows that $|{|T|}/{\vartheta (1)}|_2=|N|.$ Thus, $|{|T/N|}/{\vartheta (1)}|_2=1$ so the $2$ -parts of $|T/N|$ and $\vartheta (1)$ are equal. Thus, for every $\vartheta \in \mathrm {Irr}(T\mid \lambda )$ , we have $|\vartheta (1)|_2=|T/N|_2.$ However, $|T/N|= \sum _{\vartheta \in \mathrm {Irr}(T\mid \lambda )} \vartheta (1)^2.$ Hence, if $|\vartheta (1)|_2=2^k\geq 2$ for every $\vartheta \in \mathrm {Irr}(T\mid \lambda ),$ we would have $|T/N|_2=2^{2k}$ , which contradicts the fact that $|\vartheta (1)|_2=|T/N|_2.$ Therefore, $|T/N|_2=1.$ Thus, since $|G/N|_2\geq |N|=2^{n-2}$ , we have $|G:T|_2=|G/N:T/N|_2\geq 2^{n-2},$ so $|G:T|\geq 2^{n-2}=|N|,$ which is a contradiction.

Now we turn to the case $n=8$ . We have $p=2$ and $m=4,5$ or $6$ . In this case, $\mathrm {A}_8 \cong \mathrm {GL}(4,2)$ and $2^6$ divides $|\mathrm {A}_8|$ . We look at each possibility for m in turn. If $m=6,$ then $|N|_2=|\mathrm {A}_8|_2.$ For this case, the same argument as above holds since $6=8-2$ , and we reach a contradiction.

Second, let $m=5$ . As above, $|G:T|<|N|=2^5$ and ${|T|}/{\vartheta (1)}\in \mathrm {cod}(G)$ such that $2^5$ divides ${|T|}/{\vartheta (1)}$ . Further, $|{|T/N|}/{\vartheta (1)}|_2\leq 2$ so $|T/N|_2\leq 4$ and $|G/N:T/N|_2\geq 16$ . Thus, $16$ divides $|G/N:T/N|$ and $|G/N:T/N|< 32.$ However, we may check the index of all subgroups of $G/N\cong \text {A}_8$ using [6] and find that none of them satisfy these two properties.

Third, let $m=4$ . Then $G/N \cong \mathrm {A}_8 \cong \mathrm {GL}(4,2)$ and $N=(\mathbb {Z}_2)^4$ so G is an extension of $\mathrm {GL}(4,2)$ by $N.$ We may computationally calculate the codegree set for any such group using MAGMA [5]. There are only four such nonisomorphic extensions and we find that none of them have the same codegree set as $\mathrm {A}_8$ . (The MAGMA code is available at https://github.com/zachslonim/Characterizing-Alternating-Groups-by-Their-Codegrees.) In every case, $|N|=p^m$ produces a contradiction, so $N=1$ and $G\cong ~\mathrm {A}_n$ .