The sum of an Irreducible Representation I was hoping someone could help me with the following question.
Let $\rho$ be an irreducible presentation of a finite group $G.$  Prove
\begin{equation*}
\sum_{g \in G} \rho(g) = 0 
\end{equation*}
unless $\rho$ is the trivial representation of degree $1$.
I think I have to use Schur's Lemma which states the following. Let $\rho: G \longrightarrow GL(n,\mathbb{C})$ be a representation of G. Then $\rho$ is irreducible if and only if every $n \times n$ matrix $A$ which satisfies 
\begin{equation*}
\rho(g)A = A\rho(g) \ \ \ \forall \ g \in G
\end{equation*}
has the form $A = \lambda I_n \, $ with $\lambda \in \mathbb{C}$. 
But I am really not sure how the lemma can be applied to this question?
 A: Let $t=\sum_{g\in G}\rho(g)$, which is an linear endomorphism of $V$. The subset $t(V)$ of $V$ is a $G$-submodule of $V$, as you can easily check. Moreover, $G$ acts trivially on all elements of $t(V)$.
If $V$ is irreducible, then either $t(V)=0$ or $t(V)=V$. In the first case, we have that in fact $t=0$. In the second one, we see that $G$ acts trivially on all of $V$, so $V$ must be of dimension $1$.
A: I think you can use this variant of Schur's lemma, yes! If $A$ is your sum, then $\rho(g)A = A\rho(g) = A$ for all $g \in G$: use the fact that $G$ is a group and that $\rho$ is a homomorphism. Thus $A = \lambda I$. If $\rho$ is not the trivial representation, then there exists a $g$ such that $\rho(g) \neq I$. Now you have $\lambda\rho(g) = \lambda I$. If $\lambda \neq 0$, then does this make any sense?
Added. That $\rho(g)\sum_{x \in G} \rho(x) = \sum_{x \in G} \rho(x)$ is a special case of the following fact: if I have a commutative monoid $A$, a finite set $I$, an indexing function $f\colon I \to A$, and a bijection $\mu\colon J \to I$ then $\sum_{i \in I} f(i) = \sum_{j \in J} f(\mu(j))$. This is just a pedantic way of changing variables. Here both $I$ and $J$ are $G$, and $\mu(x) = gx$.
