$\mathbb{C}[G]$-module homomorphism on finite dimensional modules and finite groups Nice to meet you folks! I'm currently a grad student reviewing some representation theory of finite groups for prelims next year, and I'm stuck proving a simple statement. Translating the question into one of ring theory, it can be restated as follows: Let $G$ be a finite group, $M$ a finite-dimensional $\mathbb{C}[G]$-module, and a homomorphism $\phi\in\text{Hom}_{\mathbb{C}}(M,\mathbb{C})$. Consider the map $\Gamma:M\to\mathbb{C}[G]$ given by
$$\Gamma(m)=\sum_{g\in G}\phi(g^{-1}m)g.$$
This example came up in Etingof's Introduction to Representation Theory, but I'm not sure why this is actually a $\mathbb{C}[G]$-module homomorphism. Could one of my betters offer a (hopefully elementary?) proof? I may have been overthinking it, as I've been trying to work with its irreducible linear representations, and seeing what comes from there.
 A: First check $\bf C$-linearity (this should be obvious). The only other step needed to check that $\Gamma$ is a homomorphism of ${\bf C}[G]$-modules is to check that it is $G$-equivariant. We write out
$$\quad \Gamma(h m)=\sum_{g\in G}\phi(g^{-1}h m)g $$
Consider replacing $g$ with $hg$ (this is merely a substitution) so that the sum looks different. What happens to the resulting expression? Remember $\phi$ takes $\bf C$-scalar values so commutes with all elements of ${\bf C}[G]$. Also keep in mind what $G$-equivariance means, what it is you're aiming for: it means that $\Gamma(hm)=h\Gamma(m)$ for all $h\in G$ and $m\in M$.
A: Just for the fun of it, let me put this in a broader context:
Let $A$ be a ring (with $1$), and let $B$ be a ring (with $1$) equipped with a morphism $A \to B$ (respecting $1$).  If $N$ is a left $A$-module,  then Hom_A(B,N) (Hom of left $A$-modules) is naturally a left $B$_module.  (We define $b \cdot \phi$, for $b \in B$ and $\phi$ in the Hom set,
by $(b \cdot \phi)(b') = \phi(bb')$; check this gives a left $B$-module structure.)
Now if $M$ is any $B$-module, 
then there is a canonical isomorphism
$$Hom_B(M, Hom_A(B,N) )  \cong Hom_A(M, N) $$
where in the second Hom, $M$ is regarded as a left $A$-module via the map $A \to B$.
The map is defined by $\psi \mapsto \bigl(m \mapsto \psi(m)(1)\bigr)$.  (To grok this, note that $\psi(m)$ is a hom from $B$ to $N$, so we can evaluate it on $1 \in B$ to get an element of $N$.)
It is an exercise to check it is an isomorphism.  (This is an example of an adjunction: $Hom_A(B,\text{--})$ is right adjoint to the forgetful functor from $B$-modules to $A$-modules.)

Now apply this in the case when $A = \mathbb C$, $N = \mathbb C$, and $B = \mathbb C[G]$.  It says that 
$$Hom_{\mathbb C[G]}(M, Hom_{\mathbb C}(\mathbb C[G],\mathbb C) ) 
\cong Hom_{\mathbb C}(M,\mathbb C).$$
Now we can identify $Hom_{\mathbb C}(\mathbb C[G],\mathbb C)$ with
$\mathbb C[G]$ as left $\mathbb C[G]$-modules, via the perfect pairing
$\mathbb C[G] \times \mathbb C[G] \to \mathbb C[G] \to \mathbb C$,
where the first map is just multiplication in $\mathbb C[G]$, and the second map is projection onto the coefficient of $1$.
(In general, $Hom(B,A)$ and $B$ are different $B$-modules, but 
we are in a special case where there is a canonical projection $B \to A$
so that the map $B \times B \to B \to A$ becomes a perfect pairing.  See Frobenius algebras for more on the axiomatics of the situation.)
Putting everything together, we obtain an isomorphism
$$Hom_{\mathbb C[G]}(M,\mathbb C[G]) \cong Hom_{\mathbb C}(M,\mathbb C),$$
which, if you go from right to left, is the construction given in the OP.

The point of this is that this construction is less ad hoc than it looks; it is a consequence of the fact that group rings are Frobenius algebras.  Let me add that the I've found the general adjunction described above to be one of the most useful (in my own work) of all the  algebraic tools I know.    
