# Groups with irreducible representations of the largest possible dimension

I thought that the interesting question Gerry Myerson asked in the comments of this question deserved to be asked in a non-closed mathoverflow question.

What can we say about groups of order $n$ with an irreducible representation of dimension $d$ such that $(d+1)^2\geq n$?

To ask a concrete question, are there infinitely many such groups?

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See this paper of Noah Snyders: arxiv.org/pdf/math/0603239v3.pdf –  B R May 22 '12 at 6:12
Did you happen to see my comment on that question? –  S. Carnahan May 22 '12 at 7:57
@S. Carnahan: I did not. –  Will Sawin May 22 '12 at 16:25

Just to give correct references. Let $d$ be the degree of an irreducible character of a finite group $G≠1$. Then $|G|=d(d+e)$ for some $e > 0$ (that is because $d$ divides $|G|$ and $d^2 < |G|$). Therefore the condition $(d+1)^2 > |G|$ means $d(d+e)=|G|$ with $e=1$ or $2$. If $e=1$, then $G$ is a doubly transitive Frobenius group or of order 2 by Berkovich, Yakov Groups with few characters of small degrees. Israel J. Math. 110 (1999), 325–332. If $e=2$, then $G$ is a cyclic group of order 3 or non-Abelian group of order $8$ by Snyder, Noah Groups with a character of large degree. Proc. Amer. Math. Soc. 136 (2008), no. 6, 1893–1903. In general the order of $G$ is bounded by $((2e)!)^2$ by Snyder's paper. That estimate was greatly improved to $O(e^6)$ in Isaacs, I. M. Bounding the order of a group with a large character degree. J. Algebra 348 (2011), 264–275 (using the Classification of finite simple groups) and then to $e^6-e^4$ (if $e\ge 2$) by Durfee, Christina, Jensen, Sara A bound on the order of a group having a large character degree. J. Algebra 338 (2011), 197–206 (without the Classification). On the other hand, by Snyder's remark, a finite version of the Heisenberg group gives a low bound $e^4-e^3$ and $O(e^4)$ is conjecturally also the upper bound (see Isaacs' paper where this upper bound is proved in many cases).

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By Gerry's and my version of the inequality, the abelian groups of order $4$ also count. –  Will Sawin May 22 '12 at 16:12
There are at least a handful. The groups of order 1, 2, and 3, and $S_3$, and the group of the square, and as I noted at that earlier question, $A_4$, and one of the groups of order 20, and one of the groups of order 42.