TITLE: Commutativity of a finite group
QUESTION [5 upvotes]: In a finite group a representative can be chosen from each conjugacy class such that they all commutate. Prove that the group is commutative. Does this still hold true if the group is infinite?

REPLY [2 votes]: You can prove it for finite groups with the following lemma.

Lemma Let $G$ be a finite group, and let $H$ be a proper subgroup. Then the union of all conjugates $\bigcup_{g \in G}H^g$ is a proper subset of $G$.
Proof Let $|H|=m \gt 1$. Let $N_G(H)$ be the normalizer of $H$ in $G$, and note that it contains $H$. Hence , $[G\colon N_G(H)]\leq[G\colon H]$. $G$ acts transitively on all conjugates of $H$ by conjugation. The stabilizer of $H$ is precisely the subgroup $N_G(H)$, so by the Orbit-Stabilizer Theorem, the number of different conjugate subgroups is equal to $[G\colon N_G(H)]$. Now each of the conjugate subgroups has cardinality equal to $|H|$, and each contains the identity element $1$, so there are most $1+[G\colon N_G(H)](\vert H\vert-1)$ elements in the union. But,
$$
1+[G\colon N_G(H)](\vert H\vert-1)\leq 1+[G\colon H](\vert H\vert-1)=1+\vert G\vert-m=\vert G\vert+(1-m)<\vert G\vert
$$
since $m>1$. So the union of the conjugate subgroups is a proper subset.$\square$Now put $H= \langle x_1, \dots, x_k \rangle$, the group generated by the different representatives of the conjugacy classes of $G$. Note that this subgroup $H$ is abelian by the assumption that each of the $x_i$ commute with each other. Because the union of all conjugacy classes equals $G$, we must have that $\bigcup_{g \in G}H^g=G$. By the lemma we conclude that $H=G$ and hence $G$ is abelian.$\square$