Group representation scalar product Let $\rho: G \rightarrow GL(V)$ be a finite dimensional complex representation of the group $G$. Show that there is an inner product on $V$ such that $G$ acts by unitary matrices.
My approach so far was to define the sought after inner product as follows:
$\sum\nolimits_{g \in G} < \rho(g) u, \rho(g) v>$ where $u,v \in V$, and $<,>$ is the standard scalar product on $V$. I need to show that this equals $\sum\nolimits_{g \in G} <u,v> = |G|<u,v>$ where $|G|$ is the cardinality of $G$. 
This is where I'm stuck and I don't know how to show that the left sight equals the right side. Anyone got a hint for me?
Cheers!

We need to define a $G$-invariant inner product by using an initial inner product. So pick any basis for the space and let $\langle\cdot,\cdot\rangle$ be the inner product it induces. This is pretty arbitrary. Now define
$$\langle u,v\rangle_G:=\frac{1}{|G|}\sum_{g\in G}\langle\rho(g)u,\rho(g)v\rangle.$$
It is standard to put the normalization factor $\frac{1}{|G|}$ in front, though ultimately it isn't necessary. The reason is that if $\langle\cdot,\cdot\rangle$ was $G$-invariant to begin with then $\langle\cdot,\cdot\rangle=\langle\cdot,\cdot\rangle_G$, which is aesthetically pleasing. The task is then to show that $G$ acts by unitary transformations with respect to $\langle\cdot,\cdot\rangle_G$, i.e.
$$\langle\rho(h)u,\rho(h)v\rangle_G=\langle u,v\rangle_G.$$
You need to do this by expanding the inner products (using its definition above), then employing the inherent symmetry in the situation (the summands are the exact same, only permuted).
