Consider the regular representation of a finite group $G$ and let $X_{reg}$ be its character. Let $(\pi, V)$ be any finite dimensional representation of $G$ with character $X$.
Show that $<X_{reg}, X>=dimV$
The regular representation of $G$ is the permutation representation of the action "acting on itself by left multiplication."
$<X_{reg}, X>=\frac{1}{|G|}\sum_{g \in G}X_{reg}(g)X(g^{-1})$
I have the fomrula, but am not sure how to use this in practice. Would appreciate your help, thanks
Best Answer
Let $V_{\rm reg}=\bigoplus_{g\in G}\Bbb Cg$ be the vector space with basis the elements of $G$. Then the regular representation associates to each $g\in G$ the matrix $M_g$ that permutes the basis according to the multiplication table of $G$. Since $gx\neq x$ for all $x\in G$ and for all $1\neq g\in G$, the matrices $M_g$ have trace zero for all $g\neq 1$.
Thus $X_{\rm reg}(1)=|G|$ and $X_{\rm reg}(g)=0$ for $g\neq1$.
At this point the formula you want should be fairly obvius.