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I'm working with $\mathbb S_4$, and I have a subgroup of $\mathbb S_4$ called $G$.

$G$ is generated by $a=(12)(34)$ and $b=(123)$, which I've actually found to be $A_4$ by multiplying elements by $a$ and $b$ until I can't find any newer elements (actually , is there a simpler way to do this as well?)

Then I have a subgroup of $G$ called $H$, and $H$ is generated by $a=(12)(34)$ and $c=(13)(24)$. I know I can just test whether each conjugate ($ghg^{-1}$) is an element of $H$, but this is long and tedious.

Is there a trick to show this without having to calculate all the conjugates?

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    An easier way to see that $ = A_4$ is first to see that $a$ and $b$ are both even permutations, so the group generated by them consists of even permutations only. Next, show that all 3-cycles are generated by them, and $A_4$ is generated by the 3-cycles.2012-11-25

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Conjugating a permutation preserves its cycle structure. $H$ just contains the identity and the only three elements of cycle structure $(2,2)$, so any conjugate of $H$ has to be $H$, which is exactly normality.