Finding a subgroup $H \leq S_4$ such that a set $X$ is a coset by $H$.

Question

Let $X=\{(1432), (13)\} \subseteq S_4$. How can I go about finding a subgroup $H$ of $S_4$ such that $X$ is a left coset of $G$ by $H$, i.e. $X=\sigma H$ for some $\sigma \in S_4$?

In particular, is there some systematic way to do this? I've been experimenting by creating different subgroups, but haven't really noticed anything that helps. It would be nice if I could say $H$ consists of permutations of even order, but this doesn't have to be true since we can just take $\sigma$ to have even order (maybe even this is wrong).

In any case, I'm looking for a hint in how one could go about this systematically. Thanks.

Answer 1

$X$ and $H$ must be of the same order, so $H$ has size 2. One of those elements of $H$ must be the identity, so call $H=\{1, \alpha\}$. You know that $(13)H = X$, so get to work on the algebra and work out who $\alpha$ is.

Answer 2

One has $|H| = |\sigma H| = |X| = 2$. Then $H=\{e,\alpha\}$, so $\sigma H = \{\sigma, \sigma\alpha\} = X = \{(1432),(13)\}$.

If $\sigma=(13)$, then $H=\{e,(13)(1432)\}=\{e,(14)(23)\}.$

If $\sigma=(1432)$, then $H=\{e,(1234)(13)\}=\{e,(14)(23)\}.$