Lecture 4

Solution: If $(\pi,V)$ is irreducible, we are done. If not, then by definition $(\pi,V)$ has a subrepresentation $(\pi,W)$ with $0<\operatorname{dim}W<\operatorname{dim}V$. If $(\pi,W)$ is irreducible, then we are done. If not, repeat the process. This will produce a chain of subrepresentations with strictly decreasing dimension. Since the positive integers are bounded from below by $1$, this process must stop after finitely many steps, giving the desired irreducible subrepresentation.

Solution: Let $h\in Z(G)$. Then by definition, $\pi(h)\in\operatorname{GL}(V)\subset\operatorname{Hom}(V)$. We claim that in fact $\pi(h)\in\operatorname{Hom}_{G}(V)$: for any $g\in G$, we have

hence $\pi(h)\in\operatorname{Hom}_{G}(V)$. By Schur’s lemma, we then have $\pi(h)=\lambda_{h}{\,\mathrm{Id}}$ for some $\lambda_{h}\in{\mathbb{C}}$. Since $\pi(h)\in\operatorname{GL}(V)$, $\pi(h)$ is invertible, hence $\lambda_{h}\neq 0$, i.e. $\lambda_{h}\in{\mathbb{C}}^{\times}$. Define $\psi:H\rightarrow{\mathbb{C}}^{\times}$ by $\psi(h)=\lambda_{h}$. It remains to verify that $\psi$ is a homomorphism: let $h_{1},h_{2}\in Z(G)$. Then

giving $\psi(h_{1}h_{2})=\psi(h_{1})\psi(h_{2})$.