Q:

Let $G$ be a group and we suppose that the commutator of $G$ has order $m$

Let $G$ be a group and we suppose that the commutator of $G$ has order $m$. I want to show that each element of $G$ has at most $m$ different conjugates. That means that $\left|x^G\right|\leq m$ for all $x\in G$ where x^G={z ∈ G : z ∼ x} = {z ∈ G : z = gxg−1 for some g ∈ G}. I have no idea to prove this, I do not know how can I start.

A:

Hint:
$$gxg^{-1}=[g,x]x\ \in [G,G]x\,.$$

