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ON THE CHARACTERISATION OF ALTERNATING GROUPS BY CODEGREES

Published online by Cambridge University Press:  26 January 2024

MALLORY DOLORFINO
Affiliation:
Department of Mathematics, Kalamazoo College, Kalamazoo, Michigan, USA e-mail: [email protected]
LUKE MARTIN
Affiliation:
Department of Mathematics, Gonzaga University, Spokane, Washington, USA e-mail: [email protected]
ZACHARY SLONIM
Affiliation:
Department of Mathematics, University of California, Berkeley, Berkeley, California, USA e-mail: [email protected]
YUXUAN SUN
Affiliation:
Department of Mathematics and Statistics, Haverford College, Haverford, Pennsylvania, USA e-mail: [email protected]
YONG YANG*
Affiliation:
Department of Mathematics, Texas State University, San Marcos, Texas, USA
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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 $\operatorname {cod}(\mathrm {A}_n)$.

Type
Research Article
Copyright
© The Author(s), 2024. Published by Cambridge University Press on behalf of Australian Mathematical Publishing Association Inc.

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 [Reference Dolorfino, Martin, Slonim, Sun and Yang8] for the current literature on codegrees.

The following conjecture appears in the Kourovka Notebook of Unsolved Problems in Group Theory [Reference Khukrho and Mazurov12, 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 [Reference Dolorfino, Martin, Slonim, Sun and Yang8], 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 [Reference Isaacs10] and the ATLAS of Finite Groups [Reference Conway, Curtis, Norton, Parker and Wilson6].

2 Proof of Theorem 1.1

First, we note that the cases $n=5,6$ and $7$ have already been proven in [Reference Ahanjideh1, Reference Bahri, Akhlaghi and Khosravi2], 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 [Reference Dolorfino, Martin, Slonim, Sun and Yang8, Lemma 2.5], $\mathrm {cod}(G/N)\subseteq \mathrm {cod}(\mathrm {A}_n)$ . Then, by [Reference Hung and Moreto9, 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 [Reference Dolorfino, Martin, Slonim, Sun and Yang8, 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 [Reference Dolorfino, Martin, Slonim, Sun and Yang8, 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 [Reference Dixon and Mortimer7, Theorem 4.3A(iii)], $N=S^n$ , where S is a nonabelian simple group and $n\in \mathbb {Z}^+$ . By [Reference Moretó14, Lemma 4.2] and [Reference James and Kerber11, 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 [Reference Isaacs10, 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 $G\cong 2.\mathrm {A}_n$ [Reference Wilson17]. From [Reference Bessenrodt and Olsson3, 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 [Reference Malle and Zalesskii13], 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 [Reference Conway, Curtis, Norton, Parker and Wilson6] 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 [Reference Isaacs10, Corollary 5.4], $\chi \in \operatorname {Irr}(G)$ , and by [Reference Isaacs10, 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 [Reference Bessenrodt, Tong-Viet and Zhang4], if $p^m$ divides $|\mathrm {A}_n|,$ then $m < {n}/{2}$ . However, by [Reference Wagner16, 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 [Reference Bessenrodt, Tong-Viet and Zhang4], 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 $n>8$ , [Reference Wagner15, 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 [Reference Conway, Curtis, Norton, Parker and Wilson6] 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 [Reference Bosma and Playoust5]. 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 $G\cong ~\mathrm {A}_n$ .

Acknowledgements

The authors gratefully acknowledge the financial support of NSF and NSA, and also thank Texas State University for providing a great working environment and support. The authors would also like to thank Professor Richard Stanley for his help.

Footnotes

This research was conducted under NSF-REU grant DMS-1757233, DMS-2150205 and NSA grant H98230-21-1-0333, H98230-22-1-0022 by Dolorfino, Martin, Slonim and Sun during the Summer of 2022 under the supervision of Yang. Yang was also partially supported by a grant from the Simons Foundation (#918096).

References

Ahanjideh, N., ‘Nondivisibility among irreducible character co-degrees’, Bull. Aust. Math. Soc. 105 (2022), 6874.Google Scholar
Bahri, A., Akhlaghi, Z. and Khosravi, B., ‘An analogue of Huppert’s conjecture for character codegrees’, Bull. Aust. Math. Soc. 104(2) (2021), 278286.Google Scholar
Bessenrodt, C. and Olsson, J. B., ‘Prime power degree representations of the double covers of the symmetric and alternating groups’, J. London Math. Soc. 66(2) (2002), 313324.10.1112/S0024610702003575CrossRefGoogle Scholar
Bessenrodt, C., Tong-Viet, H. P. and Zhang, J., ‘Huppert’s conjecture for alternating groups’, J. Algebra 470 (2017), 353378.Google Scholar
Bosma, J. C. W. and Playoust, C., ‘The Magma algebra system I: the user language’, J. Symbolic Comput. 24 (1997), 235265.Google Scholar
Conway, J. H., Curtis, R. T., Norton, S. P., Parker, R. A. and Wilson, R. A., Atlas of Finite Groups (Clarendon Press, Oxford, 1985).Google Scholar
Dixon, J. D. and Mortimer, B., Permutation Groups, Graduate Texts in Mathematics, 163 (Springer, New York, 1996).Google Scholar
Dolorfino, M., Martin, L., Slonim, Z., Sun, Y. and Yang, Y., ‘On the characterisation of sporadic simple groups by codegrees’, Bull. Aust. Math. Soc., to appear. Published online (27 March 2023); doi:10.1017/S0004972723000187.Google Scholar
Hung, N. N. and Moreto, A., ‘The codegree isomorphism problem for finite simple groups’, Preprint, 2023, arXiv:2301.00446.Google Scholar
Isaacs, I. M., Character Theory of Finite Groups (Academic Press, New York, 1976).Google Scholar
James, G. and Kerber, A., The Representation Theory of the Symmetric Group (Addison-Wesley, Reading, MA, 1981).Google Scholar
Khukrho, E. I. and Mazurov, V. D., Unsolved Problems in Group Theory, The Kourovka Notebook, 20 (Russian Academy of Sciences, Novosibirsk, 2022).Google Scholar
Malle, G. and Zalesskii, A. E., ‘Prime power degree representations of quasi-simple groups’, Arch. Math. (Basel) 77 (2001), 461468.10.1007/PL00000518CrossRefGoogle Scholar
Moretó, A., ‘Complex group algebra of finite groups: Brauer’s problem 1’, Adv. Math. 208 (2007), 236248.10.1016/j.aim.2006.02.006CrossRefGoogle Scholar
Wagner, A., ‘The faithful linear representations of least degree of ${S}_n$ and ${A}_n$ over a field of characteristic 2’, Math. Z. 151 (1976), 127138.Google Scholar
Wagner, A., ‘The faithful linear representations of least degree of ${S}_n$ and ${A}_n$ over a field of odd characteristic’, Math. Z. 154 (1977), 104113.Google Scholar
Wilson, R. A., The Finite Simple Groups (Springer, London, 2009).Google Scholar