$\newcommand\cd{\mathrm{cd}}$Let $A_m$ and $A_n$ be two alternating groups and $15\le m+2 \le n$. Denote $\cd_m$ and $\cd_n$ as the largest irreducible character degree of $A_m$ and $A_n$, respectively. I want to show that

$$ \frac{m!/2}{\cd_m} < \frac{n!/2}{\cd_n}. $$

My thought is to use upper and lower bounds for the largest irreducible character degrees of alternating groups. I've found a paper giving the upper bound and now I need the lower bound.

Given an alternating group $A_n$, denote its number of conjugacy classes as $k(n)$; is there a tighter lower bound for its largest irreducible character degree than $\sqrt{\frac{n!/2}{k(n)}}$?

  • 1
    $\begingroup$ The alternating group and symmetric group are not so different that I would imagine a maximizer would be similar, and so the vast literature on asymptotics of Young tableaux counting could be useful to you (e.g. I think it is known that a maximizer will be of Vershik-Kerov/Logan-Shepp shape). $\endgroup$ Jul 23 at 21:47
  • 2
    $\begingroup$ "Highest degree of an irreducible representation of the alternating group $A_n$" is tabulated at oeis.org/A060955 – for the symmetric group, see oeis.org/A003040 $\endgroup$ Jul 24 at 0:10

1 Answer 1


My original answer was flawed. I am replacing it with the following.

The irreducible representations $M_\lambda$ of the symmetric group $S_n$ are indexed by partitions $\lambda$ of $n$. If $\lambda\neq \lambda'$ (the conjugate partition to $\lambda$), then the restriction $N_\lambda$ of $M_\lambda$ to $A_n$ remains irreducible. If $\lambda=\lambda'$ then $M_\lambda$ splits into two irreps of the same dimension.

Now $n!/\dim(M_\lambda)$ is the product $H_\lambda$ of the hook lengths of $\lambda$. Thus $n!/(\dim N_\lambda)$ is either the product of the hook lengths (when $\lambda\neq \lambda'$) or twice this product. We are interested in those $\lambda$ that minimize $n!/(\dim N_\lambda)$. Suppose that $\lambda$ minimizes $n!/(\dim N_\lambda)$ and $\lambda=\lambda'$. Thus this minimum value is $2H_\lambda$. Remove any corner square from $\lambda$, giving a partition $\mu$ of $n-1$. Then clearly $H_\mu<2H_\mu<2H_\lambda$. Now suppose that $\lambda\neq\lambda'$. If $\mu=\mu'$ then we need to show $2H_\mu<H_\lambda$, which is not always true. However, it is easy to see that we can in fact remove a corner square from $\lambda$ so that $\mu\neq \mu'$ ($n\geq 3$), so $(n-1)!/(\dim N_\mu)= H_\mu<H_\lambda=n!/(\dim N_\lambda)$. Thus it seems to me that the numbers $(n!/2)/\mathrm{cd}_n$ are strictly increasing for $n\geq 3$.

  • $\begingroup$ This is very helpful. I was wondering if you could point out a reference for your fact about moving a box of a self-conjugate partition and getting a non-self-conjugate partition with less than double the hook length product? $\endgroup$
    – Suoria
    Jul 25 at 17:50
  • $\begingroup$ See corrected answer. $\endgroup$ Jul 31 at 19:31

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