Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
3  
See this paper of Noah Snyders: arxiv.org/pdf/math/0603239v3.pdf –  B R May 22 '12 at 6:12
2  
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

3 Answers 3

up vote 15 down vote accepted

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).

share|improve this answer

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.

share|improve this answer
1  
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.) –  B R May 22 '12 at 6:24
    
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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.