# A bijection for counting commuting pairs in finite group

Let $$G$$ be a finite group and $$C_G$$ the set of conjugacy classes in $$G$$. Using the orbit-stabilizer theorem for the conjugation action, it can be found that the number of commuting pairs of elements in $$G$$, i.e. the size of the set $$A_G=\{(x,y)\in G\times G, xy=yx\}$$ equals the size of $$G$$ times the size of $$C_G$$, $$|A_G|=|G||C_G|.$$

This is surprisingly simple, and of course the right hand side counts the number of pairs containing one element from $$G$$ and one from $$C_G$$, i.e. the size of the set $$B_G=\{(a,b), a\in G, b\in C_G\}.$$

Is there a nice bijection between sets $$A_G$$ and $$B_G$$ that works for all $$G$$?

The answer is no in the following sense. Both the set of commuting pairs $$\text{Hom}(\mathbb{Z}^2, G)$$ and the set $$G \times C_G$$ are naturally acted on by the automorphism group of $$G$$, and in general it's not possible to find a bijection which is $$\text{Aut}(G)$$-equivariant.
For example, take $$G = S_3$$, the smallest nonabelian group. Then $$\text{Aut}(G) \cong S_3$$ acting by conjugation. $$G$$ has $$3$$ conjugacy classes, all of which are fixed by $$\text{Aut}(G)$$, so $$G \times C_G$$ splits up under the action of $$\text{Aut}(G)$$ into $$3$$ sets of $$3$$ orbits, of sizes $$1, 1, 1, 2, 2, 2, 3, 3, 3$$.
On the other hand, there's only one commuting pair of elements which is fixed by $$\text{Aut}(G)$$, namely the identity and the identity. So the two don't have the same fixed points.