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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?

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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 $\mu(x) = gx$.

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  • $\begingroup$ Firstly how did you get $\rho(g)A = A\rho(g) = A$? As $\rho(g)A = \rho(g)(\rho(g_1) + \dots \rho(g_n)) = \rho(g)\rho(g_1) + \dots \rho(g)\rho(g_n) = \rho(gg_1) + \dots \rho(gg_n) \neq \rho(g_1g) + \dots \rho(g_ng) = (\rho(g_1) + \dots \rho(g_n))\rho(g) = A\rho(g).$ $\endgroup$ – Anon Apr 1 '12 at 20:46
  • $\begingroup$ @AlexKite Is there any difference between the sets $\{h\}_{h \in G}$ and $\{gh\}_{h \in G}$? $\endgroup$ – Dylan Moreland Apr 1 '12 at 20:51
  • $\begingroup$ No, I don't think there is? Also how do you get the @AlexKite sign by your comment? $\endgroup$ – Anon Apr 1 '12 at 20:54
  • $\begingroup$ So this implies $\rho(g)A = A\rho(g) = A$? $\endgroup$ – Anon Apr 1 '12 at 21:01
  • $\begingroup$ @AlexKite I just type @A in and your name pops up. People always get notified of comments on their answers though, so there's no need for you to do it. $\endgroup$ – Dylan Moreland Apr 1 '12 at 21:34
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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 $1$.

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