Inner product in Maschke's Theorem I am working through Maschke's Theorem on page 16 in Bruce Sagan's The Symmetric Group:
In order to prove the theorem the author constructs an inner product $\langle v, w \rangle' = \sum_{g \in G} \langle gv , gw \rangle$.
However, I fail to see why $\langle v, w \rangle'$ satisfies the definiteness property of the inner product. In other words, I do not know how to prove $\langle v,v \rangle = 0$ if and only if $v = 0$.
The "if"-implication is obvious, but I cannot see why $v = 0$ if $\langle v, w \rangle' = 0$.
Thank you very much for your help!

As $g.V=V$ for all $g \in G$ (because this is an action of $G$ on $V$), and each of these is a linear map, we have that $g:V \rightarrow V$  is a linear isomorphism. So for all $g \in G$, $g.v=0$ if and only if $v=0$. So in order to show $\langle v,v \rangle'=0 \Rightarrow v=0$, observe that $\langle v, v \rangle' = \sum_{g \in G} \langle gv , gv \rangle = 0$ if and only if $v =0$ by the remark above and by positive-definiteness of $\langle .,.\rangle$.
