Corollary of Schur's Lemma - why abelian 
Corollary (of Schur's Lemma): Every irreducible complex representation of a finite abelian group G is one-dimensional.

My question is now, why has the group to be abelian? As far as I know, we want the representation $\rho(g)$ to be a $Hom_G(V,V)$, where $V$ is the representation space. Isn't this always the case (i.e. even if the $\rho(g)$ is not abelian) as it is by definition a function $G \rightarrow GL(V)$?

I had the same question, for maybe longer than a year, but because of a stupid mistake in understanding Schur's. Here it is, in case ( hoping ;) ) that someone else might do the same mistake:
Schur's lemma says something about >any< linear $\psi$ so that 
(*) $ \;\;\;\; \psi \rho (g) = \rho (g) \psi$ $\;$  $\forall g$
for some irreducible representation $\rho$. (Correct me if there are more assumptions.) $\psi$ is not assumed to be irreducible. If you would assume it is irreducible, it would follow from Schur's lemma that it's one dimensional, already at this stage. But you don't know that.
(In the corollary, you have $\psi = \rho $, so $\psi$ is irreducible now. But you still need (*), which you get from G's abelianness, as the others have pointed out.)
