# Why do elements of coprime orders commute in nilpotent groups?

I want to show the following statement:

Let $G$ be a nilpotent group and $a,b\in G$ such that there exist $m,n\in\mathbf{N}_{>0}$ such that $\text{gcd}(m,n)=1$ and $a^m=b^n=1$. Then $ab=ba$.

If $G$ is finite, this is clear to me, since I know that then $G$ is the direct product of its Sylow subgroups.

I have found a sketch of a proof in Hall’s Theory of groups: If $G=G_1,G_2,\ldots$ is the lower central series of $G$, show that $[a,b]\in G_i$ for any $i$, which then implies $[a,b]=e$. Hall even gives another hint: If $[a,b]\in G_i$, show that $[a,b]^m\in G_{i+1}$ and $[a,b]^n\in G_{i+1}$. Unfortunately I can’t do that. I managed the case $m=2$:
$$a^{-1}b^{-1}aba^{-1}b^{-1}ab=a^{-1}(b^{-1}a^{-1}ba)a(a^{-1}b^{-1}ab)=[a,[a,b]],$$
but I don’t see how to generalize this.

Another hint would be nice.

#### Solutions Collecting From Web of "Why do elements of coprime orders commute in nilpotent groups?"

You might want to note that the finite groups $G$ having the property that for all $a, b \in G$ with $gcd(order(a), order(b))=1$, one has $order(ab)=order(a).order(b)$, are the nilpotent groups.

In other words, for a finite group $G$ the order function is “multiplicative” iff $G$ is nilpotent.

Since you are okay in the finite case, can you show that the subgroup generated by $a$ and $b$ is finite?