TITLE: Is the join of two solvable subgroups solvable?
QUESTION [5 upvotes]: It is an exercise in Rotman's "introduction to the theory of groups" to show that if $S \trianglelefteq G$ and $T \leq G$ are solvable, then so is $ST$. 
I know $ST = S \vee T$ in the lattice of subgroups when $S$ is normal, so I'm curious if $S \vee T$ is solvable more generally (assuming, of course, both $S$ and $T$ are).
My proof relied somewhat heavily on the normalcy of $S$, so I'm not sure how to adapt my proof to this more general setting, and a few attempts haven't led to anything. 
Thanks in advance! ^_^

REPLY [6 votes]: No.  Indeed, every finite group can be written as a finite join of solvable subgroups, namely its cyclic subgroups.  If a join of two solvable subgroups was always solvable, then by iterating we would conclude that every finite group was solvable.  So, every finite non-solvable group must have a counterexample among its subgroups.
For a very explicit counterexample, $S_n$ is generated by the $n$-cycle $(1\ 2\ \dots\ n)$ and the transposition $(1\ 2)$, and so is the join of the cyclic subgroups they generate, but is not solvable for $n>4$.