Given a group $G$, let $G_m$ be the group generated by the set $S=\{g^m|g\in G\}$.

Prove that if $G_m$ and $G_n$ are both abelian, then $G_{\gcd(m,n)}$ is also abelian.


1 Answer 1


I will try and generalize Nicky Hekster's solution to the coprime case. Let $\gcd(m,n)=r$, with $m=ra$, $n=rb$, and $a,b$ coprime

Let $S = \{g^r \mid g \in G \}$, and let $M = G_m = \langle k^a \mid k \in S \rangle$, and $N = G_n = \langle k^b \mid k \in S \rangle$. So $M$ and $N$ are abelian normal subgroups of $G$, with $[M,N] \le M \cap N \le Z(\langle M,N \rangle)$.

It is enough to prove that any two elements $g^r$ and $h^r$ of $S$ commute. We have $r = \lambda m + \mu n$ for some $\lambda,\mu \in {\mathbb Z}$, and since $g^m$ commutes with $h^m$ and $g^n$ commutes with $h^n$, it is enough to show that $u=g^m= (g^r)^a$ commutes with $v = h^n = (g^r)^b$.

Now $u^{-1}v^{-1}uv=z \in M \cap N$, so $v^{-1}uv = uz$. Since $z$ centralizes $u$ and $v$, $v^{-1}u^bv = u^bz^b$. But $u^b,v \in G_n$, which is abelian, so $v^{-1}u^bv=u^b$ and hence $z^b=1$. Similarly $z^a=1$ and, since $a$ and $b$ are coprime, $z=1$, and we are done.

  • $\begingroup$ Very nice argument! $\endgroup$ May 10, 2015 at 18:52
  • $\begingroup$ Thanks! But it's really just an adaptation of the proof for the coprime case. $\endgroup$
    – Derek Holt
    May 10, 2015 at 19:19
  • $\begingroup$ I see, $(u^{-1}h^{-1}u)^n=u^{-1}h^{-n}u=u^{-1}v^{-1}u\in G_n.$ $\endgroup$ May 11, 2015 at 6:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .