# Commutativity of a finite group

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?

• Please clarify which subgroup or normal subgroup is used here. – Stefan Hamcke Dec 16 '13 at 0:37
• No subgroup is used. We simply take elements from each conjugacy glass. – Vaprus Dec 16 '13 at 12:02
• Oops! I confused conjugacy class with coset. But the setting here is that you are given a group $G$ and a fixed subgroup $H$, right? – Stefan Hamcke Dec 16 '13 at 16:34

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.$\square$

• The Lemma is also known as "Jordan's theorem". – Igor Rivin Dec 16 '13 at 15:12
• I wasn't one of them (since it is a perfectly good answer, albeit identical to part of mine in content). As you probably have noticed long ago, this venue is more than a little frustrating. – Igor Rivin Dec 16 '13 at 21:32
• OK, Igor, thanks for mentioning. I did not know that it was Camille Jordan who proved this for the first time (apparently in 1872!). – Nicky Hekster Dec 16 '13 at 21:33
• Igor, fully agree! The OP asked for more explanation, that's why I wrote it down. – Nicky Hekster Dec 16 '13 at 21:34
• Actually, I did not realize, for some reason, that Jordan's theorem had a one line proof (Serre's paper has harder arguments, I think), so it may have been partly a case of knowing too much (and also, trying to reverse engineer the question -- I had assumed that the OP knew about Jordan's theorem, since they were probably not expected to find the proof you wrote down). – Igor Rivin Dec 16 '13 at 21:38

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.

• Could you please be more specific on how you apply Jordan's theorem? – Vaprus Dec 16 '13 at 12:21
• @Vaprus Jordan's theorem says that if you have a subgroup that intersects every conjugacy class, it is the whole thing. There is not much more to say. – Igor Rivin Dec 16 '13 at 13:05
• I'm sorry, I can't find this specific Jordan's theorem. Might you have a source for this formulation? – Vaprus Dec 16 '13 at 15:09
• @Vaprus yes, there is a very nice paper by J. P. Serre: ams.org/journals/bull/2003-40-04/S0273-0979-03-00992-3/… – Igor Rivin Dec 16 '13 at 15:11