# dimension of a irreducible representation

Let G be a group and A an abelian subgroup of G. I want to prove that for each irreducible representation $$p$$ we have that $$dim(p)\leq [G:A]$$.
I proved that for each subspace $$p(A)-$$invariant $$W$$, $$W_0= \sum_{g \in G}p(g)(W)$$ is a p(G)-invariant of V (hint of the professor) and that for a abelian finite group, each irreducible representation has dimension 1.
But how can i go on? I wish i can tell more of what i did in this question but i honestly have no idea how to proceed.I was thinking in the canonical decomposition of a representation, considering the subspaces W,but after some writing it seems non-sense. Any help are welcome.
Ps:my representation theory course uses only knowlegde of theory of groups, not modules or other stuff.

• Are you working in the “““easy””” case (finite groups, algebraically closed fields with chacteristics coprime with $|G|$), or are there subtleties? Jul 24, 2020 at 14:59
• The field needs to be algebraically closed, otherwise it's false. As $Z_3$ over $\mathbb{Q}$ demonstrates. As for your question, do you really need all $g\in G$ in the definition of $W_0$? In other words, what is $\dim(W_0)$? Jul 24, 2020 at 15:07
• I assume that you are not familiar with induced representations and Frobenius reciprocity? Because the result follows very neatly from that. Jul 25, 2020 at 11:12
• forgot to mention, the field is C and the group is finite, Jul 25, 2020 at 13:53
• and no, i cant use induced representations for this, is literally the next content of the course Jul 25, 2020 at 13:54