Cyclic subgroup and left coset question

Hi, I have a question that I am not sure how to work out:

Let H be the cyclic subgroup generated by g= ( 1 2 3)

( 1 3 2)

Find all left cosets of S3 modulo H.

Am i correct in that there will be two distinct left cosets, and if so how do I figure out what they are?

Re: Cyclic subgroup and left coset question

First, S3 has order 6 while $\displaystyle \begin{pmatrix}1 & 2 & 3 \\ 1 & 3 & 2\end{pmatrix}$ has order 2 (g just swaps 2 and 3 so doing g twice swaps back and gives the identity) so that there are 6/2= 3 left cosets. Further, those cosets **partition** S3 so there are, in fact three left cosets. One, the one containing the identity, is just H itself. $\displaystyle \begin{pmatrix}1 & 2 & 3 \\ 2 & 1 & 3\end{pmatrix}$ is not in H and so generates another coset: taking it with the identity gives itself and taking it with $\displaystyle \begin{pmatrix}1 & 2 & 3 \\ 1 & 3 & 2\end{pmatrix}$ gives $\displaystyle \begin{pmatrix}1 & 2 & 3 \\ 3 & 1 & 2\end{pmatrix}$. Can you find the third coset?