Let $G$ be a finite group.Let $(π,V)$ and $(γ,W)$ be irreducible representation over $\Bbb C$ Let $G$ be a finite group.Let $(π,V)$ and $(γ,W)$ be irreducible representation over $\Bbb C$.
Let $χ_π$ and $χ_γ$ be character of $(π,V)$ and $(γ,W)$.
Then, for arbitrary $h∈G$, how to prove the following ?

For arbitrary $h∈G$,
$$\sum_{g∈G} χ_π(1)χ_π(g)χ_γ(g^{-1} h) = \sum_{g∈G} χ_π(h)χ_π(g)χ_γ(g^{-1} )$$

$χ_π(1)χ_π(g)χ_γ(g^{-1}) h＝χ_π(h)χ_π(g)χ_γ(g^{-1})$ does not hold in general, but after summation, we can say the equality, but I can't grasp the mechanism.
 A: I am sorry in advance for my poor disposition and pedagogical skills, this is my first post and i am a bit intoxicated in the moment, I'll clean up my writing a bit tomorow or on sunday.
First I just want to remind you that $\pi (gh)=\pi (g) \circ \pi(h)$ for $h,g\in G$. And of the fact that $\underset{g\in G}{\sum}\overline{\chi_v(g)}\pi_v(g)=\dfrac{|G|}{dim(V)}Id_v$ which follows from using Schur's lemma, also remember the orthogonality relation of characters i.e $\underset{g\in G}{\sum}\overline{\chi_v(g)}\chi_w(g)=\left\{\begin{matrix}
|G|, \text{if }V\simeq W \\ 
0, \text{else}
\end{matrix}\right.$
I want to reform the problem a bit, just for my own convience: LHS=$dim(V)\underset{g\in G}{\sum}\chi_w(gh)\chi_v(g^{-1})$, as $\chi_v(1)=dim(V)$. Now the RHS=$\chi_v(h)\underset{g\in G}{\sum}\chi_w(g)\overline{\chi_v(g)}$, where we get the complex conjugate from $\chi(g^{-1})=\overline{\chi(g)}$. Reformulated we want to prove $dim(V)\underset{g\in G}{\sum}\chi_w(g^{-1})\chi_v(gh)=\chi_v(h)\underset{g\in G}{\sum}\chi_w(g)\overline{\chi_v(g)}$(here we also changed $g$ in the LHS)
In the case that $V$ and $W$ are not isomorphic we have $LHS=RHS=0$(try using the orthogonality relation!). For the case that $V\simeq W$ we can conider the representation $\Phi=\dfrac{dim(V)}{|G|}\underset{g\in G}{\sum}\overline{\chi_w(g)}\pi_v(gh)=
\dfrac{dim(V)}{|G|}(\underset{g\in G}{\sum}\overline{\chi_w(g)}\pi_v(g))\pi_v(h)=\dfrac{dim(V)}{|G|}\dfrac{|G|}{dim(V)}Id_v\pi_v(h)=\pi_v(h)$
and thus $Tr(\Phi)=\chi_v(h)$ which is what we wanted to show :)
 $\textbf{EDIT:}$
  first the RHS=$\chi_v(h)\underset{g\in G}{\sum}\chi_v(g)\overline{\chi_w(g)}=$
$\left\{\begin{matrix}
|G|\chi_v(h), \text{ if }V\simeq W \\ 
0, \text{else}
\end{matrix}\right.$
Now for the LHS, first remember that(by the orthogonality of charactrs and Schur's lemma) $\underset{g\in G}{\sum}\chi_v(g^{-1})\pi_w(g)= \left\{\begin{matrix}
\dfrac{|G|}{dim(V)}id_v, \text{ if }V\simeq W\\ 
0, \text{ else}
\end{matrix}\right.$
First change the variables in the sum of the LHS to $g\mapsto g^{-1}h$, $\chi_v(g)\mapsto \chi_v(g^{-1}h)$ and $\chi_w(g^{-1}h)\mapsto \chi_w(h^{-1}gh)=\chi_w(g)$. With this change of variables
realize that the left hand side is the trace of $\Phi=\chi_v(1)\underset{g\in G}{\sum}\chi_w(g)\pi_v(g^{-1}h)=dim(V)(\underset{g\in G}{\sum}\chi_w(g)\pi_v(g^{-1}))\pi_v(h)$ which we now can see is $Tr(\Phi)= \left\{\begin{matrix}
|G|\chi_v(h), \text{ if }V\simeq W\\ 
0, \text{ else}
\end{matrix}\right.$ $\square$
