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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?

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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 ST. 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).

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Hey Jack, what a GREAT answer and solution! Thanks very much, this problem kept me awake for a while. Quite surprising that it is true indeed. I also kept calculating in small groups. That's why I had the suspicion that it should be true. Something to go in textbooks in the future, because I have not seen this anywhere, did you? –  Nicky Hekster Nov 17 '11 at 21:31
    
@Nicky: No, I had never seen it. I'm pretty sure I would have remembered it, since st conjugate to ts has always been useful to me, and this argument is so simple. Your set generalization sounded false to me, since you needed different conjugators to get one element of ST to one element TS, so I was most certainly looking for counterexamples. –  Jack Schmidt Nov 17 '11 at 22:33
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