# Center of group ring is generated by characters

Let $$\mathbb{C}[G]$$ be the group ring of $$G$$ over $$\mathbb{C}$$. Show that $$B =\left\{\sum_{g\in G}\chi_{\pi}(g)g : \pi:G\to GL(V)\text{ a representation}\right\}$$ (where $$\chi_\pi$$ is the character of $$\pi$$) is a linear basis for $$Z(\mathbb{C}[G])$$.

If $$f:G\to\mathbb{C}$$ is a class function, then it is constant on conjugacy classes and so for all $$h\in G$$ we have $$h\sum_{g\in G} f(g)g=\sum_{g\in G}f(g)hg=\sum_{g\in G} f(h^{-1}gh)h(h^{-1}gh)=\sum_{g\in G} f(g)gh=\left(\sum_{g\in G} f(g)g\right)h$$ so in particular $$B\subseteq Z(\mathbb{C}[G])$$. However, I am not sure how to show that it forms a basis. I tried to write an element $$v\in Z(\mathbb{C}[G])$$ as a linear combination of elements in $$B$$ but I couldn't make it work. Thank you in advance!

• What do you know about characters and class functions? If you know that characters are orthogonal for the appropriate hermitian structure, and if you know the dimensions involved, it should not be too hard. Oct 20, 2021 at 8:44
• What are the $\chi_{\pi}$? Irreducible characters or class functions in general? You should know that $Z(\mathbb{C}[G])$ is spanned by the different conjugacy class sums $\sum_{x \in Cl(g)}x$. If the $\chi_{\pi}$ are class functions than take one that is $1$ on a particular conjugacy class and $0$ elsewhere. Do this for each conjugacy class and you are done. Oct 20, 2021 at 10:09
• @CaptainLama I know that the characters are orthogonal, but I haven't learnt about hermitian structures yet! Oct 20, 2021 at 10:27
• @NickyHekster the $\chi_\pi$ are characters. I edited the question to make it more clear Oct 20, 2021 at 10:28
• But then, what does it mean for you that they are orthogonal? You can only be orthogonal with respect to a certain euclidean/hermitian (or similar) structure. In any case, if they are orthogonal they are linearly independent. Then you just have to check that there is the correct number of irreducible characters. Oct 20, 2021 at 14:17

It can be shown directly without any representation theory that $$f = \sum_{g\in G} c_g g\in Z(\mathbb C[G])$$ iff $$f$$ is a class function. Indeed, as $$f$$ commutes with $$h$$, we have $$f = h^{-1} f h = \sum c_g h^{-1}gh$$, and comparing the coefficients of $$f$$ and $$h^{-1}fh$$, we get $$c_{h^{-1}gh} = c_g$$.