Let $G$ be a finite group. We can define a relation $R$ on $G$ by setting $aRb \iff b = g^{-1} a g$ for some $g \in G$. The conjugacy class of an element $a \in G$ is defined as its equivalence class: $$[a] := \{b \in G: \hspace{0.1cm} aRb\} = \{g^{-1}ag \in G: \hspace{0.1cm} g \in G\}$$
Let $[G,G]$ be the commutator subgroup of $G$, defined to be the subgroup generated by elements of the form $aba^{-1}b^{-1}$, $a,b \in G$. I want to prove that the number of conjugacy classes of $G/[G,G]$ is $|G/[G,G]| = |G|/|[G,G]|$ but I have no idea how to prove it. I believe I have to use the fact that $G/[G,G]$ is abelian, but that's all I know. Any help would be appreciated!
The conjugacy classes of an abelian group all have exactly one element, therefore the number of conjugacy classes of a finite abelian group $H$ is equal to the number elements of $H$.