Take the group $S_4$ and the set $X$ of all the subgroups of $S_4$. $S_4$ acts on $X$ by conjugation, given $\pi \in S_4$ and a subgroup $H$ of $S_4$ $\pi$ acts on $H$ by $\pi(H) = \pi H \pi^{-1}$ find all the subgroups of order two and which ones are conjugate to which ones, how may orbits are there which consist of subgroups of order 2?

the subgroups of order 2 are $$ \langle(12)\rangle,\langle(13)\rangle,\langle(14)\rangle,\langle(23)\rangle,\langle(24)\rangle,\langle(34)\rangle,\langle(12)(34)\rangle, \langle(13)(24)\rangle, \langle(14)(23)\rangle$$
for a group to be conjugate to another $\sigma \pi \sigma^{-1} = \pi$
if i take the group $\langle(12)\rangle$ for example $\sigma (12) \sigma ^{-1} = (12)$
$\sigma = (\sigma(1) \sigma(2))=(12)$ so would all the groups be conjugate to themselves?
also $(ab)(cd)(ab)=(cd)$ so disjoint subgroups are also conjugate?
I don't know how to approach the orbit aspect of the problem.


Two subgroups are conjugate if and only if their decompositions as a product of disjoints cycles have the same structure.

Thus all transpositions are conjugate. For instance $(12)$ and $(13)$ are conjugate, and indeed $$(23)(12)(23)=(13)$$. Similarty, the product of two disjoint transpositions: $$\{ (12)(34),(13)(24), (14)(23)$$ are conjugate to each other. As an example, $$(23)[(12)(34)](23)=(13)(24).$$


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