Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $G$ be a finite $p$-group. Is it true that every irreducible representation of $G$ over an algebraically closed field of characteristic zero ($\mathbb{C}$, for example) must have dimension a power of $p$?

Proof or reference are appreciated.

share|improve this question
add comment

1 Answer 1

up vote 4 down vote accepted

The degrees of the simple complex representations of a group divide the order of the group, so yes. (In fact, they divide the index of the center of the group, which in your case is smaller.)

See, for example, Serre's book for a proof.

share|improve this answer
    
Serre has quite a few books. Probably you mean Linear representations of finite groups? –  Pete L. Clark Apr 11 '11 at 6:00
1  
Well, Serre's book on the subject :) –  Mariano Suárez-Alvarez Apr 11 '11 at 6:40
2  
In fact, the degrees divide the index of any abelian subgroup (since a p-group is nilpotent so all subgroups are subnormal). Also, as a side note, if all degrees of irreducible complex characters are powers of some prime p, then the group has an abelian normal p-complement, so it behaves almost like a p-group. –  Tobias Kildetoft Apr 11 '11 at 7:59
    
@Mariano & Jiangwe: In fact, it's even better: the square of the degree of each irreducible character of any finite nilpotent group $G$ divides $[G:Z(G)]$! –  Geoff Robinson Jul 13 '11 at 16:28
    
@Geoff: nice. Do you have a reference on that? –  Mariano Suárez-Alvarez Jul 13 '11 at 16:35
add comment

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.