Solvable $T$-groups Let $G$ be a finite solvable $T$-group, then it is easy to show that $G$ is supersolvable. I want to find out if the finiteness condition is necessary as I have not been able to find a reference for a proof or either a counterexample. 
 A: A $T$-group is a group in which every subgroup is subnormal.
Proposition: A solvable $T$-group is supersolvable if and only if it is finitely generated.
Proof: All supersolvable groups are finitely generated - in fact finitely presented - so the "only if" direction is clear.
So let $G$ be a finitely generated solvable $T$-group. We prove that $G$ is supersolvable by induction on the derived length. The result is clear for derived length $0$ (and also for $1$).
Let $N$ be the last nontrivial term in the derived series for $G$. So $N$ is abelian, and all subgroups of $N$ are subnormal and hence normal in $G$. By inductive hypothesis $G/N$ is supersolvable, so it is enough to prove that $N$ is finitely generated.
Let $X$ be a finite generating set of $G/N$, and $R$ a finite set of defining relators for the group $G/N$ on the generating set $\{xN : x \in X \}$. Then $N$ is the normal closure in $G$ of the set $\hat{R}$ of evaluations in $N$ of the relators in $R$. But since all subgroups of $N$ are normal in $G$, we have $N = \langle \hat{R} \rangle$ is finitely generated, and we are done.
