TITLE: Prove $S_4$ has only 1 subgroup of order 12
QUESTION [5 upvotes]: The subgroup in $S_4$ that I know has order 12 is the subgroup of all even permutations, otherwise known as the alternating group $A_4$. However, I know this from a fact and not because I am able to show a subgroup of order 12 exists in $S_4$ in the first place. If I had not been told there existed a subgroup of order 12 in $S_4$, I would not have known there was one. So I guess this is actually a two-parter for me: 1) How do I go about showing that a subgroup of order 12 does indeed exist in $S_4$? I know by Lagrange that if there exists a subgroup, then the order of that subgroup must divide the order of the group. However, this doesn't say anything about the existence of the subgroup. And Sylow only verifies subgroups of a prime to some power order, which 12 is not. I also know that there might be a cyclic subgroup of order 12, but without manually multiplying every possible permutation in $S_4$, I don't know any other way to check the existence of this or if this subgroup is isomorphic to $A_4$ because I don't know how to write these permutations as functions in order to check if there is a homomorphism (I know that a permutation is bijective by definition). And 2) How do I show that this group of order 12 is the only group of order 12 in $S_4$?

REPLY [3 votes]: Here is a link which will help you with the question and the comment of Derek:
$A_n$ is the only subgroup of $S_n$ of index $2$. .