Is the join of two solvable subgroups solvable? 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! ^_^
 A: 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 $n>4$.
A: If you don't assume that $S\trianglelefteq G$, then the product $ST$ need not be solvable.  As the product is a subgroup of the join, and solvable groups are closed under taking subgroups, the join need not be solvable either.  Below is a simple example.
We will take $G=S_5$ the symmetric group on the set $\{1,2,3,4,5\}$, which is more or less the simplest example of a non-solvable group.  Then we take $S=\langle (1 2), (1 2 3)\rangle$ and $T=\langle (4 5), (3 4 5)\rangle$.  As each of $S,T$ is contained in a copy of $S_3$ living inside $S_5$ and is generated by a transposition and a $3$-cycle, both are isomorphic to $S_3$, which is solvable.  Now $ST$ contains $(1 2 3)(3 4 5)$ which you can easily check is a $5$-cycle.  $ST$ also contains a transposition, so $ST=G$, which is not solvable.
