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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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    $\begingroup$ See this paper of Noah Snyders: arxiv.org/pdf/math/0603239v3.pdf $\endgroup$
    – B R
    May 22, 2012 at 6:12
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    $\begingroup$ Did you happen to see my comment on that question? $\endgroup$
    – S. Carnahan
    May 22, 2012 at 7:57
  • $\begingroup$ @S. Carnahan: I did not. $\endgroup$
    – Will Sawin
    May 22, 2012 at 16:25

3 Answers 3

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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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There are infinitely many groups like that, since any Frobenius group of order d(d+1) where d+1 is a prime power has an irreducible representation of degree d.

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    $\begingroup$ In fact, these are almost the only such groups! From Snyder's paper that I mentioned above, the others are the trivial group, the two-element group, the cyclic group of order 3, or a nonabelian group of order 8. (See Theorem 6.2 for the last two.) $\endgroup$
    – B R
    May 22, 2012 at 6:24
  • $\begingroup$ By Gerry's and my version of the inequality, the abelian groups of order $4$ also count. $\endgroup$
    – Will Sawin
    May 22, 2012 at 16:12
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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.

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