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 $G \rightarrow GL(V)$?
 A: The trick here is, that for a abelian group every element is a intertwining operator. This means let $h \in G$, then $\rho(h)\rho(g)\rho(h^{-1})=\rho(g)$ for all $g$ and therefore by Schur's lemma $\rho(h)=\lambda id$. Since your representation was assumed to be irreducible it follows that it is one dimensional. Note that we used the commutativity of the group here in an essential way. This is no longer true for non abelian groups. 
A: Any finite group is isomorphic to a direct product of its irreducible representations, acting on a direct sum of vector spaces. If all irreducible representations are one-dimensional then this faithful representation consists of diagonal matrices which commute. Whence the group is abelian.
A: 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.)
