Aluffi IV.1.11 suggests proving that if a group $G$ is finite and there exist $g_1, \dots, g_r$ — representatives of all the $r$ conjugacy classes such that they commute pairwise, then the group is commutative. He also gives a hint to think about the sizes of the classes.
Working backwards, $G$ is commutative if $Z(G) = G$, that is, if each conjugacy class is trivial, that is, its size is $1$ and $[g_i] = \{ g_i \}$. But how do I prove this?
I know this problem is solvable by resorting to the fact that for any $H$ a proper subgroup of $G$, $G$ is never the union of conjugates of $H$, but this fact is given as an exercise that follows this one, so I guess the author intended a different proof.
Best Answer
If $G$ is the trivial group, we are done. Let $|G|=n$. If $r=n$, we are done. Suppose $r<n$. Without loss of generality, let $g_{1}$ be the identity element. For $i\geq 2$, $|[g_{i}]|=|G:C_{G}(g_{i})|$. Since $g_{i}$ commutes with all $g_{j}$'s, $|C_{G}(g_{i})|\geq r$. So $|[g_{i}]|=n/|C_{G}(g_{i})|\leq n/r$. Then $$ n=|G|=|[g_{1}]|+\sum_{i=2}^{r}|[g_{i}]| \leq 1+\frac{n}{r}(r-1) =1+n-\frac{n}{r}. $$ So $n\leq r$, contradiction.