# Does every finite nontrivial group have two distinct irreducible representations over the complex numbers of equal degree?

Is it true that for any finite nontrivial group G, there exist two inequivalent irreducible representations of G over the complex numbers that have the same degree.

If so, is there an easy proof? If not, what is the smallest counterexample?

Note: Any counterexample group must be perfect, because if the abelianization is nontrivial, we get multiple irreducible representations of degree one. [EDIT: Further, as Colin Reid notes in the comment, a minimal counterexample must be a simple (non-abelian) group]. This whittles down our search considerably. The general expressions for the degrees of irreducible representations for the families of simple groups that I've checked suggests that there is plenty of repetition of degrees in these cases.

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A minimal counterexample would have to be simple, so if you can prove it for the simple groups, you are done. –  Colin Reid Oct 30 '11 at 1:45
Thanks for pointing it out! I forgot to say it explicitly, but that's the direction I was going with "checking the families of simple groups." I've edited the original question to make that clear. –  Vipul Naik Oct 30 '11 at 2:33
This is a nice question. I wonder if it's possible to give a proof that doesn't rely on the classification of simple groups. –  Faisal Oct 30 '11 at 2:44
@Faisal: My own guess is that the classification is impossible to avoid in this kind of question, but of course I can't prove that. See also Geoff Robinson's answer. –  Jim Humphreys Dec 22 '13 at 14:53

It seems that the answer is yes. A MathSciNet search brought up the paper

Y. Berkovich, D. Chillag, and M. Herzog, Finite groups in which the degrees of the nonlinear irreducible characters are distinct, Proc. Amer. Math. Soc. 115 (1992), 955–959.

In it you can find a characterization of groups whose nonlinear irreducible characters have distinct degrees. In particular, such a group can't be perfect (see Lemma 1), and so will always have multiple linear characters as was noted in the OP. The proof, however, relies on the classification of finite simple groups, so is not "easy".

Addendum: I took a closer look at the related literature and happened across the following interesting result, which I figured was worth sharing. (It can also be used to give an affirmative answer to the original question.)

Theorem. Let $G$ be a nontrivial finite group. If the character table of $G$ has a column or row containing distinct rational entries, then $G$ must be isomorphic to either $S_2$ or $S_3$.

The reference is

M. Bianchi, D. Chillag, A. Gillio, Finite groups with many values in a column or a row of the character table, Publ. Math. Debrecen 69 (2006), no. 3, 281–290.

The result from the classification of finite simple groups used in the Berkovich–Chillag–Herzog paper is also used here (in very much the same spirit).

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If you count the classification of finite simple groups as "already done", I presume it's pretty easy, no? Just check all the finite simple groups. –  Will Sawin Oct 30 '11 at 3:51
@Will Sawin: it might not be easy, because the finite simple groups include some infinite families, and it's not obvious that the question can be answered uniformly for all groups in a family. –  Noam D. Elkies Oct 30 '11 at 3:55
@Noam: To be a counterexample all irreducible characters have to be rationalvalued which means that every element $g$ must be conjugate to all elements generating the same cyclic group. I think that that quickly excludes most (all?) Chevalley groups and I guess the alternating groups. –  Torsten Ekedahl Oct 30 '11 at 10:25
@Noam: For groups of Lie type, the principal series representations are generically irreducible. Since they are all of the form $\mathrm{Ind}^G_B(\chi)$ with $\chi$ one-dimensional they all have the same dimension (for a given group). –  Lior Silberman Nov 2 '11 at 3:51

It has already been noted in earlier answers and comments that a finite non-trivial group of minimal order subject to having all complex irreducible characters of different degrees would have to be non-Abelian simple with all its characters rational-valued.

From the point of view of historical perspective, it is worth pointing out G. Seitz and W. Feit proved (using the classification of finite simple groups) that there only finitely many (in fact 5) possible non-Abelian composition factors, other than alternating groups, of a finite group with all characters rational valued. By a result of James and Kerber, for example, no alternating group has all its characters rational-valued. In fact, JG Thompson and I have given a precise description of the field generated by the character values of $A_{n}$ for every $n.$

Hence, using these facts, the question asked here reduces to checking the character tables of 5 non-Abelian simple groups.

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