Finding cosets of a subgroup of $S_{3}$ I'm trying to figure out how to get all the left cosets of subgroup $S_{3}$ generated by $(1,2)$. I know what $\langle (1,2) \rangle = \{ (0,0),(1,2) \}$. I'm just not sure how the answer key got $(1,3)\langle(1,2) \rangle = \{ (1,3), (1,2,3) \}$ and $(2,3) \langle (1,2) \rangle = \{ (2,3), (1,3,2) \}$
 A: $S_3=\{(), (12), (13), (23), (123),(132) \}$ is of order $6.$
The subgroup $\langle (12) \rangle= \{(),(12) \}$ is of order $2.$ By Lagrange's theorem, the number of elements of every subgroup divides the order of the group. All cosets have the same number of elements, and partition the group. Hence, there are $3$ cosets of $2$ elements each.
We can look for elements of $S_3$ outside of the subgroup $\langle (12) \rangle$ and multiply them on the left by all the elements of $\langle (12) \rangle.$ Since subgroups are closed under multiplication, there is no point in multiplying by elements of the subgroup.
In the OP the first line of the solution involves multiplying on the left the subgroup by $(13)$ with the permutations applied from right to left:
$$(13)(12)=(123)$$
because $1$ goes to $2$ in the transposition on the right, i.e. $(12),$ and it is not moved anywhere else by the transposition on the left, i.e. $(13).$ The next element to consider is $2,$ which goes to $1$ by $(12)$ and then goes to $3$ by $(13).$ Finally $3$ is not moved on the right transposition, but goes to $1$ on the left transposition.
Evidently, multiplying $(13)$ times the identity element will not change it. Therefore the first coset found is
$$(13)\langle (12) \rangle=\{(13), (123) \}$$
And we have the other remaining elements constituting the last coset: $\{(23),(132)\}.$
