Suppose that $H$ is a subgroup of $S_4$ and that $H$ contains $(12)$ and $(234)$. Prove that $H = S_4$.
Since $(234) \in H$ and $(12) \in H$, this means $(234)(12) \in H$ to get $(1342) \in H$, and the orders of the three cycle and the four cycle is $|(1342)|=4$ and $|(234)|=3$ and pretty much get stuck there.
The order of $H$ must be divisible by both $3$ and $4$. So $H$ has order $12$ or $24$. The only subgroup of $S_4$ of order $12$ is $A_4$, and $(12) \notin A_4$. So the order of $H$ is $24$, which implies $H = S_4$.
Remember if G is a finite group and H is a subgroup of G, then the |H| divides |G|. So, |H| divides 24. Now 3/|H|, 4/||H| and |H|/24 gives you either |H|=12 or |H|=24. Take |H|=12. Since the only subgroup of 12 in S4 is A4. So H=A4. now, since (1432)=(13)(14)(12), which is an odd permutation, contradicts the previous statement. So, |H|=24 and since |S4|=24, H=S4.
