Q:

All $\mathbb C$-representations of a finite group $G$ have a $G$-invariant inner product

Let $G$ be a finite group, and suppose we have a complex representation $V$ of $\mathbb C$.  Let $\langle , \rangle$ be an arbitrary (Hermitian) inner product on the $\mathbb C$-vector space $V$, and define $(,) : V \times V \to \mathbb C$ by $(x,y) = \displaystyle \frac{1}{|G|} \sum_{g \in G} \langle gx, gy \rangle$. Then $(,)$ is a Hermitian inner product.  Pick $h \in G$. Then $ (hx, hy) = \displaystyle \frac{1}{|G|} \sum_{g \in G} \langle ghx, ghy \rangle = \frac{1}{|G|} \sum_{g' \in G} \langle g'x,g'y \rangle = (x,y)$.
The point of this is to show that all $\mathbb C$-representations of a finite group $G$ have a $G$-invariant Hermitian inner product.  I'm being a bit slow here: why are we dividing by $|G|$? 

A:

I believe that we divide by $|G|$ as a normalising factor.  When you calculate $\sum_{g \in G} \langle gx, gx \rangle$ because you sum over $G$ you will get $(|G|)||gx||^2$. I believe you divide through by $|G|$ to remove the $|G|$ term in the front. 
Note several things:
1) Because $G$ is finite it makes sense to divide by $|G|$. If $G$ is infinite I think there are things like Representation Theory for compact groups to deal with this (where this time one divides over some integral)
2) In characteristic 0 everything is fine (as in $\mathbb{C}$ here).  In characteristic $p$ you may have some trouble dividing by $|G|$.

