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Suppose $V$ and $W$ are both representations of a group $G$, where $V$ and $W$ are $k$-vector spaces. Define $\mathrm{Hom}(V,W)$ to be the space of $k$-linear maps $V \to W$. My notes say that:

$\mathrm{Hom}_G(V,W) = \{ \phi \in \mathrm{Hom}(V,W): g \phi = \phi \}$, and we have a linear projection $\mathrm{Hom}(V,W) \to \mathrm{Hom}_G(V,W)$ given by $ \displaystyle \phi \mapsto \frac{1}{|G|} \sum_{g \in G} g \phi$

I'm confused by this. $g \phi = \phi$ isn't enough for $\phi$ to be a $G$-linear map, is it? But yet the projection map fixes those maps that satisfy $g \phi = \phi$, which suggests that this isn't a mistake.

Thanks

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    @QiaochuYuan Can you also convert this question to an answer? Again posting to [the chat](http://chat.stackexchange.com/rooms/9141/the-crusade-of-answers)?2013-06-09

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This CW answer intends to remove the question from the unanswered queue.


As Qiaochu Yuan remarked in the comments the action of $g$ on homomorphisms is as follows $(g\phi)(x):=g\phi(g^{-1}x)$ Now if you have $g\phi=\phi$ then you have $(g\phi)(x)=g\phi(g^{-1}x)=\phi(x)$ for all $x$. Multiplying this by $g^{-1}$ from the left you get $\phi(g^{-1}x)=g^{-1}\phi(x)$. This is obviously the same as $\phi$ being in $\operatorname{Hom}_G(V,W)$.