
Left and Right Cosets
Let G = S3, the symmetric group of degree 3 and let H = {i,f} where f(x1) = x2, f(x2) = x1, f(x3) = x3
a) find all the left cosets of H in G
b) find all the right cosets of H in G
c) Is every left coset of H a right coset of H?
Please show explicit steps I'm very confused on how to prove these coset problems. Thanks so much!

cycle notation
I find it a lot easier to use cycle notation when working with the symmetric group. example $\displaystyle (1,2,4, 7)\in S_7$ is the permutation that takes 1 to 2, 2 to 4, 4 to 7 and 7 to 1. The unmentioned numbers are left the same.
Your coset $\displaystyle H=\{(1), (1,2)\}$ it has size two and $\displaystyle S_n=n! \Rightarrow S_3=3!=6$. So you should expect 3 cosets each of size 2. Basically you just gotta multiply them out and see what happens.
Here are the left cosets of H
$\displaystyle (1)H=\{(1), (1,2)\}$
$\displaystyle (1,3)H=\{(1,3), (1,2,3)\}$
$\displaystyle (2,3)H=\{(2,3), (1,3,2)\}$
Right cosets of H found similarly
$\displaystyle H(1)=\{(1), (1,2)\}$
$\displaystyle H(1,3)=\{(1,3), (1,3,2)\}$
$\displaystyle H(2,3)=\{(2,3), (1,2,3)\}$
Compare these cosets and see that the last two do not match up, so these are not the same. In particular this tells you that H is infact not a normal subgroup of $\displaystyle S_3$ because $\displaystyle (1,3)H \not = H(1,3)$