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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.

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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.

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Serre has quite a few books. Probably you mean Linear representations of finite groups? – Pete L. Clark Apr 11 '11 at 6:00
Well, Serre's book on the subject :) – Mariano Suárez-Alvarez Apr 11 '11 at 6:40
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

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