Let $p$ be a prime number and let $q$ be relatively prime to $p$.
Prove that if $G$ is a $p$-group and $g \in G$,then there exists $x \in G$ with $qx=g$.
My question:I was able to prove this when $G$ is finite,but how do we prove it when $G$ is infinite?Any help would be appreciated.
Say that $g$ is of order $p^n$. Then by Bezout's theorem, there are $u,v$ with $p^nu + qv = 1$
In particular, $g = (p^nu + qv)g = up^nv + q(vg) = q(vg)$. So $x=vg$ works.
Note that I used additive notation to fit your question, but this works just as well over nonabelian groups