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$G$ is a finite group with a subgroup $H$. Let $\rho_1:G \to GL(V)$ and $\rho_2:H \to GL(U)$ be representations. $Z=\mathbb{C}[G]^H$, i.e., $Z$ is the centralizer of $H$ in $\mathbb{C}[G]$.

How do I show the representation $$\phi:Z \to End(Hom_H(U,V))$$ given by $\phi_g:f \to \rho_1(g) \circ f$ is irreducible?

If I am not mistaken, $Z$ might be infinite and $Hom_H(U,V)$ might be infinite dimensional.

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  • $\begingroup$ I am sorry. I fixed the notation. $\endgroup$ – saubhik Apr 10 '15 at 17:36
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The claim isn't true. Maybe you can clarify the claim after this post.

There is a natural isomorphism $\hom(\Bbb C,V)\cong V$. Every vector $v\in V$ can be identified canonically with the linear map $\lambda\mapsto \lambda v$ (this is $\Bbb C\to V$). Moreover, it is a natural isomorphism of representations of $G$ if $G$ acts trivially on $\Bbb C$.

So suppose that $H=1$ is the trivial subgroup of $G$. Then $Z=\Bbb C[G]^H=\Bbb C[G]$. Furthermore we may note that $\hom_H(U,V)\cong V$ naturally if $U=\Bbb C$ is the trivial representation. Therefore the map

$$Z\to{\rm End}(\hom_H(U,V))$$

is just the already-in-place action of $\Bbb C[G]\to{\rm End}(V)$. If we choose $V$ to be reducible, then the above representation must also be reducible, since it's equivalent.

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