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Take all fields to be algebraically closed. Show that if $G$ is a finite group with trivial centre and $H$ is a subgroup of $G$ with non-trivial centre, then any faithful representation of $G$ is reducible after restriction of $H$

I think I have it, but I'd like someone to check my answer.

Let $ \rho : G \to \mathrm{GL}(V)$ be a faithful representation, where $G$ is finite with trivial centre and where $V$ is a $k$-vector space where $k$ is algebraically closed. Then the restriction $\theta = \rho \big|_H : H \to \mathrm{GL}(V)$ is a faithful representation of $H$. Suppose this representation were irreducible. We have some $ e \neq z \in Z(H)$, and so $\theta(z) \in \mathrm{Hom}_H(V,V)$. By Schur's Lemma, this means that $\mathrm{dim} \ \mathrm{Hom}_H(V,V) = 1$, and so $\theta(z) =\lambda \mathrm{id}_V$ for some for some $\lambda \in k \backslash \{1\}$ ($\lambda \neq 1$, since the representation is faithful). Since $\rho$ is faithful $G \cong \mathrm{Im}(\rho)$, but $Z(\mathrm{Im}(\rho)) \ni \theta(z) = \rho(z) \neq \mathrm{id}_V$, and so $z \in Z(G)$. The contradiction establishes the result.

Thanks!

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up vote 1 down vote accepted

Your argument is correct.

Just a nitpicky remark: the sentence "By Schur's Lemma, this means that $\mathrm{dim} \mathrm{Hom}_H(V,V) = 1$" is slightly confusing, since "this" does not refer to the previous statement. Rather, the fact about dim hom follows from your assumption of irreducibility, and not from the fact that $\theta(z)\in \mathrm{Hom}_H(V,V)$.

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