First the problem:

If I am correct the problem is simply asking about a specific example, one that does not necessarily generalize. I can easily show that, for a in H, aH = Ha = H for any subgroup, due to closure of H. And, as it happens I have showed (by construction) that for any element a of $\displaystyle S_3$ that is not in H, $\displaystyle aH \neq Ha$. (I note that this property is specific to the subgroup H. If I take a new cyclic subgroup K = <(123)> of $\displaystyle S_3$ that aK = Ka for all a in $\displaystyle S_3$.) (Note: please forgive me if my cyclic notation is off. I rarely use it.)Let H be the cyclic subgroup (of order 2) of $\displaystyle S_3$ generated by $\displaystyle \binom {1~2~3}{2~1~3}$. Then no left coset of H (except H itself) is also a right coset. There exist an element a in $\displaystyle S_3$ such that $\displaystyle aH \cap Ha = \{ a \}$.

The last part is bugging me. $\displaystyle aH \cap Ha = \{ a \}$ is a true statement for all a in $\displaystyle S_3$! So why are they commenting on it??

-Dan