Let $G$ be a finite group. Let $N$ be the number of conjugacy classes of $G$. Prove that the number of all pairs $\left(a,b\right)\in G \times G$ satisfying $ab = ba$ is $N \cdot \left|G\right|$.
The first part of this problem asks to describe $\operatorname{Hom}(\mathbb{Z}^2,G)$ as a subset of $G \times G$ which turned out that $\operatorname{Hom}(\mathbb{Z}^2,G)$ is the set of pairs of elements which commute in G.
Now, I am trying to understand, and later show the fact that
$\#\operatorname{Hom}(\mathbb{Z}^2,G) = \#G \cdot N$
where $N$ is the number of orbits of the group action of conjugation by $G$ on itself.
My first guess was to try constructing a bijection between the two sides of the equality but I am having a hard time understanding $\#G \cdot N$ as a set. Is this a correct approach?
Write $\text{Cl}(a)$ for the conjugacy class of $a\in G$, and $C(a)$ for the centralizer of $a$ in $G$.
Note that $(a,b)$ is a commuting pair if and only if $b\in C(a)$. Thus the number of commuting pairs is (counting by the first element in the pair) $$\text{# pairs} = \sum_{a\in G} |C(a)|.$$
Now, choose representatives $a_1,\dotsc, a_N$ of the $N$ conjugacy classes. If $g,h\in G$ are conjugate, then $|C(g)| = |C(h)|$, so we may rewrite the equation above as $$\text{# pairs} = \sum_{i=1}^N |\text{Cl}(a_i)|\cdot |C(a_i)|.$$ By the orbit-stabilizer theorem, $|\text{Cl}(a_i)| = |G : C(a_i)|$, so the foregoing simplifies to $$\text{# pairs} = \sum_{i=1}^N |G:C(a_i)|\cdot |C(a_i)| = \sum_{i=1}^N |G| = \text{#}G\cdot N.$$