The sum of an Irreducible Representation

Question

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?

Answer 1

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 $1$.

Answer 2

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 $\mu(x) = gx$.

Answer 3

Given a finite group G ,let $r$ labels a non trivial representation of G let $$A \equiv \sum_{g \epsilon G}\rho(g) $$ Given any $h \epsilon G$, $$\rho(h)^\dagger A \rho(h)=\sum\rho(h^{-1}gh)=\sum\rho(g)=A$$ $A$ commutes with $\rho(g)$ for all $g$ hence by Schur's lemma $A=\lambda I$.$$trA=tr(\sum_g\rho(g))=\sum_g(tr(\rho(g))=\sum_c n_c\chi(c)=\lambda d$$ here $c$ denotes the conjugate class, $\chi(c)$ denotes the character of that class under the representation $r$ and d is the dimension of the representation.

By orthogonality of the character table we have ,$$\sum_c n_c\chi^s(c)^\ast \chi^r(c)=N(G)\delta^{rs}$$ Taking s to be the trivial representation gives $$\sum_c n_c \chi^r(c)=0 \implies trA=\lambda d =0 \implies \lambda=0\implies A=\sum_g\rho(g)=0$$