Proving an identity regarding character of irreducible representation

Let $$\frak{X}$$ be an irreducible representation of a finite group $$G$$ affording the character $$\chi$$. Prove that for every $$x,y\in G$$: $$\chi(x)\chi(y)=\frac{\chi(1)}{|G|}\sum_{z\in G}\chi(yzxz^{-1}).$$

My attempt: I know that I can replace $$zxz^{-1}$$ with $$w$$, and then the sum will be only on the conjugacy class of $$x$$, we'll denote it by $$[x]$$, but I don't know how many times every summand will appear after the change of variables, and to be honest I don't know how to take it from here. Any help would be appreciated.

1 Answer

Theorem Let $$\chi \in Irr(G)$$, and $$x,y \in G$$, then $$\chi(x)\chi(y)=\frac{\chi(1)}{|G|}\sum_{z \in G}\chi(xy^z)$$

Before proving this theorem we need to set notation and an observation. Write $$\mathfrak{X}$$ for any representation affording the character $$\chi$$. Let $$x \in G$$, then the $$\color{blue}{conjugacy \ class}$$ of $$x$$ in $$G$$ is denoted by $$\color{blue}{K_x}$$ and the $$\color{darkgreen}{sum \ of \ its \ elements}$$ as central element of the group algebra $$\mathbb{C}[G]$$ is denoted by $$\color{darkgreen}{\hat{K}_x}$$. As is well-known, the Schur Lemma implies the formula $$\mathfrak{X}(\hat{K}_x)=\omega_{\chi}(\hat{K}_x)I$$ where $$I$$ is the identity matrix of dimension $$\chi(1)$$, and taking traces yields $$\omega_{\chi}(\hat{K}_x)=\frac{\chi(x)|K_x|}{\chi(1)}$$ Observe that if we let $$y$$ run over $$G$$, and look at $$x^y$$, each element of $$K_x$$ appears $$|C_G(x)|$$ times. Hence we have the useful equality in $$\mathbb{C}[G]$$:

$$|C_G(x)|\hat{K}_x=\sum_{y \in G} x^y$$

Now let us proceed proving the theorem. Working in $$\mathbb{C}[G]$$ and applying the previous formula, we have $$\sum_{z \in G}xy^z=\sum_{z \in G}xz^{-1}yz=x\sum_{z \in G}y^z=x|C_G(y)|\hat{K}_y$$ This gives $$\mathfrak{X}(\sum_{z \in G}xy^z)=\sum_{z \in G}\mathfrak{X}(xy^z)=\mathfrak{X}(x)|C_G(y)|\omega_{\chi}(\hat{K}_y)I$$ Taking traces at both sides in the formula above gives $$\sum_{z \in G}\chi(xy^z)=\chi(x)|C_G(y)|\omega_{\chi}(\hat{K}_y)=\chi(x)|C_G(y)|\frac{\chi(y)|\hat{K}_y|}{\chi(1)}=\chi(x)\chi(y)\frac{|G|}{\chi(1)}$$ which proves the theorem.$$\square$$

Corollary Let $$\chi \in Irr(G)$$. Then

$$\sum_{x,y \in G}\chi([x,y])=\frac{|G|^2}{\chi(1)}.$$

Proof In the Theorem, put $$y=x^{-1}$$. Then the formula yields $$|\chi(x)|^2=\chi(x)\chi(x^{-1})=\frac{\chi(1)}{|G|}\sum_{z \in G}\chi([x^{-1},z]).$$ Since $$\chi$$ is irreducible we have $$\sum_{x^{-1} \in G}|\chi(x^{-1})|^2=\sum_{x \in G}|\chi(x)|^2=|G|$$, so summing over all $$x \in G$$ in the formula above gives the desired result.$$\square$$

Corollary Let $$\chi \in Irr(G)$$, and $$x,y \in G$$, then $$\chi(x)\chi(y)=\frac{\chi(1)}{|G|^2}\sum_{g,h \in G}\chi(x^gy^h)$$

Proof The theorem tells us that $$\sum_{g,h \in G}\chi(x^gy^h)=\sum_{g\in G}(\sum_{h \in G}\chi(x^gy^h))=\sum_{g\in G}\frac{|G|}{\chi(1)}\chi(x^g)\chi(y)=\sum_{g\in G}\frac{|G|}{\chi(1)}\chi(x)\chi(y)=\frac{|G|^2}{\chi(1)}\chi(x)\chi(y).$$