Lower bound of the largest irreducible character degree of alternating group $A_n$

$$\newcommand\cd{\mathrm{cd}}$$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 $$\sqrt{\frac{n!/2}{k(n)}}$$?

• 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). Jul 23 at 21:47
• "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 Jul 24 at 0:10

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 $$2H_\mu, which is not always true. However, it is easy to see that we can in fact remove a corner square from $$\lambda$$ so that $$\mu\neq \mu'$$ ($$n\geq 3$$), so $$(n-1)!/(\dim N_\mu)= H_\mu. Thus it seems to me that the numbers $$(n!/2)/\mathrm{cd}_n$$ are strictly increasing for $$n\geq 3$$.