Re: clarity on groups
Well, fortunately, there is only ONE subgroup of $S_3$ of order 3, which is generated by any 3-cycle.
The USUAL notation for such a 3-cycle is:
$(1\ 2\ 3)$ which is shorthand for the mapping:
$1 \to 2$
$2 \to 3$
$3 \to 1$.
The subgroup generated by this is:
$H = \{e, (1\ 2\ 3), (1\ 3\ 2)\}$.
Lagrange's Theorem tells us there will be exactly TWO cosets, $H$ and $Ha$ where $a$ is any permutation not in $H$. The 2-cycle (transposition) $(1\ 2)$ will do, and we find:
$H(1\ 2) = \{(1\ 2), (1\ 2\ 3)(1\ 2), (1\ 3\ 2)(1\ 2)\}$
$= \{(1\ 2), (1\ 3), (2\ 3)\}$