Induction of characters

Question

This is a very basic question on induction of characters.

Let $G$ be a finite group and $H$ a subgroup. Let $\mathcal{C}_G$ and $\mathcal{C}_H$ denote the spaces of class functions for $G$ and $H$ respectively. There is a linear map $$\text{Ind}^G_H: \mathcal C_H \rightarrow \mathcal C_G$$ defined by sending $f\in \mathcal C_H$ to the function $$(\text{Ind}^G_Hf)(g):= \sum_{xH,x^{-1}gx\in H}f(x^{-1}gx).$$

I'm having a bit of trouble showing that Ind$^G_H(f)$ is a class function for $G$ if $f$ is a class function for $H$.

I tried to show directly that Ind$^G_H(f)(y^{-1}gy)=$Ind$^G_H(f)(g) \; \forall g, y \in G$ but wasn't sure how to simplify the sum appearing on the left hand side.

I'd be grateful if someone could help me with this.

Answer 1

One way of seeing this is to make the definition a bit more uniform. If $h\in H$, then $x^{-1}gx\in H\Leftrightarrow (hx)^{-1}g(hx)=h^{-1}(x^{-1}gx)h\in H$. But $f$ is a class function of $H$, so this implies that $$ f((hx)^{-1}g(hx))=f(x^{-1}gx), $$ whenever either of the arguments of $f$ is an element of $H$.

As $H$ is finite, this means that an equivalent definition for the induced class function is $$ (\operatorname{Ind}_H^G f)(g)=\frac1{|H|}\sum_{x\in G, x^{-1}gx\in H}f(x^{-1}gx). $$ Does this removal of the constraint of $x$ belonging to a fixed set of representatives of left cosets help you complete the proof?

Answer 2

If $F$ is the induced character: $$F(g)=\frac{1}{|H|}\sum_{x\in G,x^{-1}gx\in H}f(x^{-1}gx)$$ so that $$F(y^{-1}gy)=\frac{1}{|H|}\sum_{x\in G,(yx)^{-1}gyx\in H}f((yx)^{-1}gyx),$$ so $F(y^{-1}gy)=F(g)$ simply by summing over $yx$ in the second sum.