Commutativity of a finite group

Question

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?

Answer 1

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.$\square$

Answer 2

It's certainly false for infinite groups, since there are groups with $2$ conjugacy classes, so they can't be commutative. Otherwise, take the group generated by your representatives. It is an abelian group intersecting every conjugacy class, so by Jordan's theorem, it is the whole group, which is, therefore, abelian.