TITLE: Does every finite nontrivial group have two distinct irreducible representations over the complex numbers of equal degree?
QUESTION [18 upvotes]: 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.

REPLY [6 votes]: 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.