I know that $\textrm{Hom}_G (V, V')$ is the set of $G$-equivariant maps from $V$ to $V'$, so what does $\textrm{Hom}_{\mathbb{C}}(V, V')^G$ mean? I'm supposed to show $\textrm{Hom}_{\mathbb{C}}(V, V')^G \simeq \textrm{Hom}_G (V, V')$, if that helps. I'm not looking for someone to do this problem, I just have no idea what this superscript $G$ is supposed to mean, and if it's something standard I'd rather not have to ask the professor. Thanks!

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    $\begingroup$ In contexts like these, $X^G$ means the subset of $X$ consisting of those elements fixed by a given action of $G$ on $X$. In your situation, where $X$ is $\text{Hom}_{\mathbb C}(V,V')$ the action of $G$ is by conjugation: $g\in G$ sends $f\in X$ to $g\circ f\circ g^{-1}$, where the first $g$ refers to the action of $g$ on $V'$ and the second to the action on $V$. (I'm working with left actions; if you prefer right actions, hold my comment up to a mirror.) $\endgroup$ – Andreas Blass Jan 30 at 2:22
  • $\begingroup$ If you like modules then you are given $V,V'$ two $\mathbb{C}[G]$-modules and if $f \in Hom_{\mathbb{C}-\text{modules}}(V,V')$, what does $f$ need to be an homomorphism of $\mathbb{C}[G]$ modules ? That $g . f(v) = f(g.v)$ equivalently that $f(v) = g.f(g^{-1}.v)$ $\endgroup$ – reuns Jan 30 at 2:33

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