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Question: Let $G$ be a group. For any two representations $V,V'$ of $G$ over $\mathbb C$, let $Hom_G (V,V')$ denote the space of all linear maps $h: V\rightarrow V'$ such that $h\rho'_g = \rho_g h\forall g\in G$. I want to prove that if $V$ and V' are irreducible and $V\cong V'$ then $Hom_G(V,V')$ is 1-dimensional.

Please tell me if I have got it right: By Schur's lemma, Any element $f\in Hom_G(V,V)$ is of the form $\lambda.Id_V$, where $\lambda\in \mathbb C$, so dim $Hom_G(V,V) = 1.$ Now, let $T:V\rightarrow V'$ be an isomorphism, so that $T\rho'_g = \rho_g T$, hence $T\neq0$, and $T\in Hom_G(V,V')$. So $Hom_G(V,V')\neq0.$ Now, for any $h\in Hom_G(V,V')$, we have$$V\xrightarrow{h}V'\xrightarrow{T^{-1}}V$$, so that $T^{-1}h\in Hom_G(V,V)$, and hence $T^{-1}h=\alpha.Id_V$ for some $\alpha\in\mathbb C\Rightarrow h=\alpha.T\circ Id_V=\alpha T$. Hence any element in $Hom_G(V,V')$ is a scalar multiple of $T$, and so dim $Hom_G(V,V')=1$.

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This is a part of Schur lemma, as a remark. – awllower Jun 16 '13 at 13:21
up vote 1 down vote accepted

Looks fine. $\phantom{space filler to have more space}$

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