Products of sets in a group Let $S$ and $T$ be non-empty subsets of a group $G$. As usual $ST=\{st : s \in S, t\in T\}$. What can be said of the subgroups $\langle ST\rangle$ and $\langle TS\rangle$? For example if the identity $1 \in S \cap T$, then it is easy to see that $\langle ST\rangle$  =  $\langle TS\rangle$. Also, if S and T are both singletons, then $\langle ST\rangle$ and $\langle TS\rangle$ are conjugated, since $ST$  and $TS$ are conjugated sets in this case. Are $\langle ST\rangle$ and $\langle TS\rangle$ always conjugated? 
 A: 
Yes, $\langle ST \rangle^s = \langle TS \rangle$ for any $s \in S$.

Proof: Consider $(s_i t_j)^s = s^{-1} s_i t_j s = s^{-1} t^{-1} t s_i t_j s = (ts)^{-1} (ts_i) (t_js) \in \langle TS \rangle$ where t is any element of T.  Similarly, $(t_i s_j)^{s^{-1}} = s t_i s_j s^{-1} = s t_i s_j t t^{-1} s^{-1} = (st_i) (s_jt) (st)^{-1} \in \langle ST\rangle$.  The first sentence shows $\langle ST \rangle ^s \leq \langle TS \rangle$ and the second shows $\langle ST\rangle^s \geq \langle TS\rangle$, so we are done. $\square$

In case you are curious, here is how I proceeded:
I think you know already that $\langle ST \rangle$ and $\langle TS \rangle$ need not be equal, since $st$ need not be a power of $ts$, for instance taking $s=(1,2)$ and $t=(1,2,3)$.  Of course, st and ts are conjugate individually, by either s or t.  In particular, if either S or T has one element, then ST and TS are conjugate.
A quick check of the smallest non-abelian group shows that ST and TS need not be conjugate in general: take S = { 1, (1,2) } and T = { (1,2,3), (2,3) } and then ST and TS are not even the same size.  In this case the subgroups generated by ST and TS are still the same, so we have to try a little harder.
We can use a sort of universal construction for the counterexample.  If we don't want the subgroups generated by ST and TS to be conjugate, we might as well make G minimal with respect to containing S and T, that is $G = \langle S \cup T \rangle$.  We also don't want things to be conjugate that don't have to be, so we don't want G to have any extra relations: G should be the free group on S ∪ T.  Now we just need to solve a subgroup conjugacy problem in a free group.
I didn't know how to do this, so I checked finite quotients of the free group (code available on request), and discovered there were no counterexamples of small order or nilpotency class.  I then began to suspect I either needed a very complicated counterexample, or that it might even be true.
Using the GAP package called FGA by Christian Sievers, one sees that this also holds if |S| ≤ |T| ≤ 100.  In fact, one can always take the conjugating element to be an arbitrary element of S.  For example, |S| = |T| = 2 is handled with the following code:
gap> m := 2;; n := 2;;
gap> f := FreeGroup( m + n );;
gap> s := GeneratorsOfGroup( f ){[ 1 .. m ]};;
gap> t := GeneratorsOfGroup( f ){[ m + 1 .. m + n ]};;
gap> st := Subgroup( f, ListX( s, t, \* ) );;
gap> ts := Subgroup( f, ListX( t, s, \* ) );;
gap> IsConjugate( f, st, ts );
true
gap> st^f.1 = ts;
true

Obviously, one conjectures this always holds, and in fact the proof was an easy calculation (at the top of the answer).
