Let $Z(G)$ be the centre of G. Let $V$ be an irreducible $G-$space with matrix representation $\rho_v$.
Let $z \in Z(G)$, then I'm trying to show that $\rho_v(z)$ is multiplication by a root of unity.
Now, I thought about using Schur's Lemma ; Let $V$, $W$ be two irreducible $G-$spaces and let $f : V \rightarrow W$ be a homomorphism of $G-$spaces, then if $V$ and $W$ aren't isomorphic, then $f = 0$, and if $V = W$, then $f$ is mulitplication by a scalar.
So, to prove the first statement I thought I'd find a $G-$space homomorphism from $V$ to $V$, call it $\rho(z)$. We may check this is a homomorphism of $G$-spaces as follows; Let $g \in G, v \in V$, then $$ g\cdot(\rho(z)(v)) = \rho(gz)(v)$$ and $$\rho(z)(g \cdot v) = \rho(zg)(v)$$
But $zg = gz$ since $z \in Z(G)$, so they are equal.
Now, I can apply Schur's lemma to say that $\rho(z)$ is multiplication by a scalar. But where does the root of unity come from? Is it implicit that this constant is a root of unity?
Thank you for any help.
Best Answer
Assuming that $G$ is a finite group, say $|G|=n$, we have $g^n=1$, for all $g\in G$. In particular $z^n=1$, and thus $\rho(z)^n=\rho(z^n)=\rho(1)=I$ since $\rho$ is a group homomorphism. Since we also have $\rho(z)^n=\lambda^nI$, $\lambda^n=1$.