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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 $|G|$?

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You should be able to check that if $[\cdot, \cdot]$ is any $G$-invariant inner product on $V$, then for any $t > 0$ the map sending $(x,y)$ to $t [x,y]$ will also be a $G$-invariant inner product on $V$. So the appearance of the scalar $\frac{1}{|G|}$ in this definition is not necessary to make the map an inner product, or to make it $G$-invariant. It is just there for convenience. Note for example that if the inner product $\langle \cdot, \cdot \rangle$ you started with were already $G$-invariant, the $\frac{1}{|G|}$ factor makes $(\cdot,\cdot) = \langle \cdot, \cdot \rangle$. –  leslie townes Feb 17 '12 at 2:24
    
@leslietownes Great, thanks a lot. –  Matt Feb 17 '12 at 2:41
    
This specific choice also has some nice properties, e.g. with respect to this inner product the characters of iso classes of irreps form an orthonormal basis for functions on your group. –  Eric O. Korman Feb 20 '12 at 22:55
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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 $|G|$.

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