A faster way to show that a subgroup is normal

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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If you show there are only two cosets, then the coset space has to be a group, then the quotient map has to be an isomorphism, then its kernel has to be a normal subgroup. – Asaf Karagila Nov 24 '12 at 23:23
You will note that all nonidentity elements of $H$ take the same form, whence you just have to check that all elements of $S_4$ of that form $(wx)(yz)$ are in $H$. Do you know why this has anything to do with normality? – peoplepower Nov 24 '12 at 23:25
An easier way to see that $<a,b> = A_4$ 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. – Peter Nov 25 '12 at 15:07

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.