TITLE: Product of class sums
QUESTION [6 upvotes]: Let $C_i$ be the conjugacy classes of a finite group $G$. Consider the class sums $z_i=\sum_{g\in C_i} g$. It is well known that ${z_i}$ form a basis of the center of the group algebra $\mathbb{C}G$. How can we show that $z_iz_j$ is a $\mathbb{N}$ linear combinations of the $z_i$

REPLY [4 votes]: Consider the proof that the class sums form a $\Bbb C$-basis for the group algebra $\Bbb C[G]$. Say that
$$x\in Z(\Bbb C[G]), \qquad x=\sum_{g\in G}a(g)g.$$
(Think of the coefficients $a(g)$ as a function $a:G\to\Bbb C$.) Then for any $\sigma\in G$ we have
$$\sigma x\sigma^{-1}=\sum_{g\in G}a(g)\sigma g\sigma^{-1}=\sum_{g\in G}a(\sigma^{-1}g\sigma) g.$$
But since $x$ is central, we know $x=\sigma x\sigma^{-1}$. Comparing coefficients below,
$$\sum_{g\in G}a(g)g=\sum_{g\in G}a(\sigma^{-1} g\sigma)g$$
we find $a(g)=a(\sigma^{-1}g\sigma)$ for all $g,\sigma\in G$, i.e. $a$ is a class function. Thus $a(C)$ makes sense, where we denote by $C$ a conjugacy class of $G$, and so as $C$ ranges over classes we have
$$x=\sum_C a(C)\sum_{g\in C}g.$$
Conversely, class sums are invariant under conjugation by $G$, so any class sum or combination of class sums commutes with any $g\in G$, and hence commutes with every element of $\Bbb C[G]$.
Now if $x,y\in Z(\Bbb C[G])$ then so is $xy$. Therefore if $x$ and $y$ are two class sums, $xy$ must be a $\Bbb C$-linear combination of class sums. But how do we get that it's an $\Bbb N$-linear combination?
Consider instead the semiring $\Bbb N[G]$ (note that our $\Bbb N$ must contain $0$; think about abelian groups $G$ to see this). It is comprised of $\Bbb N$-linear combinations of elements of $G$, with the same obvious rules for addition and multiplication, but this time there aren't additive inverses. The same argument that is written above shows every element of the center $Z(\Bbb N[G])$ (the elements that commute with everything else) are $\Bbb N$-linear combinations of class sums. Clearly the product of any two central elements is central, so the product of class sums must be an $\Bbb N$-linear combination of class sums!
This shows that all of our already-present logic and understanding is perfectly sufficient for the problem at hand, and no new thinking is necessary. One can also give a counting interpretation to the coefficients. If $z_\Gamma=\sum_{g\in \Gamma}g$ denote class sums, and $[g]$ denotes the conjugacy class of an element $g\in G$, then for any three elements $a,b,c\in G$ the coefficient of $z_{[c]}$ in the expansion of the product $z_{[a]}z_{[b]}$ equals $\#\{(x,y)\in G^2:x\sim a,y\sim b,xy=c\}$. Indeed in order to count such a thing with generatingfunctionological thinking, one would naturally look for the coefficient of $c$ in the expansion of $(\sum_{x\sim a}x)(\sum_{y\sim b}y)$!