Transversals that are closed under multiplication in a group Let $H \le G$ be a group with subgroup $H$. A right (or left) transversal is a set of element which contains exactly one element from each right (or left) coset. Now for example for $S_3$ and $H = \{ (), (1 ~ 2) \}$ we have
$$
 H, \quad H\cdot (1 ~ 2 ~ 3) = \{ (1 ~ 2 ~ 3), (1 ~ 3) \}, \quad
 H \cdot (1 ~ 3 ~ 2) = \{ (1 ~ 3 ~ 2), (3 ~ 2) \}
$$
and the right transversal $\{ (), (1 ~ 2 ~ 3), (1 ~ 3 ~ 2) \}$ even forms a group. But what I am interested in is the case when do they are closed under multiplication? Is this always the case, i.e. can we always find a right (or left) transversal $T$ such that $TT \subseteq T$?

If you don't require your group to be finite, then it is easy to find a counter-example : if $G=\mathbb{Z}$, then any subgroup $H$ will be of the form $n\mathbb{Z}$. Then if $A\subset G$ is transversal, it must have $n$ elements exactly, and thus it cannot be closed under addition since it is finite (and non reduced to $0$).
Now that I think of it it is not hard to find a finite counterexample. Take $G=\mathbb{Z}/4\mathbb{Z}$ and $H=\mathbb{Z}/2\mathbb{Z}$. Then the cosets in $G$ are$$\{\bar{1},\bar{3}\},\ \{\bar{0},\bar{2}\}.$$So a transversal subset would have to contain exactly $2$ elements, with one of them being of order $4$; so it couldn't be closed under addition in $\mathbb{Z}/4\mathbb{Z}$.
