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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 $G$?

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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.

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