I was reading an article. There is written that if $G$ is an $FC$-group which is also a soluble $\overline T$-group, then the locally finite group $G/Z(G)$ is metabelian.
- $G$ is an $FC$-group if every element has only finitely many conjugates.
- $G$ is a $\overline T$-group if every subgroup of $G$ is a $T$-group.
- $G$ is a $T$-group if every subnormal subgroup is normal.
I think the other definitions are well known.
It is clear for me that $G/Z(G)$ is locally finite, but I don't understand why $G/Z(G)$ is metabelian.
I think (so I am not sure) I can use that
- if $H$ is a locally finite $\overline T$-group, then $H$ is locally metabelian,
- locally metabelian groups are metabelian,
but I can't prove none of them.
