TITLE: Abstract Algebra Cosets and Lagrange's Theorem
QUESTION [7 upvotes]: 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.

REPLY [8 votes]: 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.