# Locally graded group with all proper subgroups abelian

A group $G$ is said to be locally graded if every finitely generated nontrivial subgroup of $G$ contains a proper subgroup of finite index.

I have to prove that a locally graded group with all proper subgroups being Abelian is itself Abelian or finite. I can suppose that a group, $G$, is not finite and prove that it is Abelian, but I don't know how to proceed.

Thanks.

• Perhaps you could define "locally graded?" – Matt Samuel May 14 '15 at 16:36
• Edit your question to add the definition of locally graded, don't put it in an answer. – Todd Wilcox May 14 '15 at 16:53

If $G\neq 1$, then $G$ being locally graded, it has a proper subgroup of finite index $H$. Hence $H$ is abelian; in particular $G$ is virtually abelian.
Assume that $G$ is not abelian and infinite. Then any two non-commuting elements of $G$ generate $G$; in particular $G$ is finitely generated.
Being finitely generated and virtually abelian, $G$ has a normal subgroup of finite index $N$ that is free abelian of finite rank. Pick a prime $p$ greater than $|G/N|$, and let $q>p$ be another larger prime. Then the finite group $H=G/N^{pq}$ also has all its proper subgroups abelian; it's an extension of $N/N^{pq}$ by $G/N$, which have coprime order; hence this extension is split, so $H=(N/N^{pq})\rtimes K$ with $K\simeq G/N$. Since $N/N^{pq}$ splits as the direct product of its Sylows $N/N^p$ and $N/N^q$, $H$ has the proper subgroup $N/N^q\rtimes K$. Hence $N/N^q\rtimes K\simeq G/N^q$ is abelian. Since this holds for every prime $q>p$ and $\bigcap_{q>p}N^q=1$, we deduce that $G$ is abelian, contradiction.