TITLE: Subgroups of $S_4$ and homomorphisms
QUESTION [6 upvotes]: Good day,
I am given a problem: there is a homomorphism from $S_4$ onto $Z_2,$ and I am asked to find its kernel. By the first isomorphism theorem, I know that the order of the kernel must be 12. I know that $S_4$ has a subgroup of order 12, namely $A_4,$ the group of even permutations. I am tempted to say that it is the kernel. But the following concerns me:
1) How do I know that it is the subgroup of order 12 of $S_4$? How do I know there are no others? I know odd permutations do not form one, but is there a simpler/more elegant way to say it rather than going into "consider a group with some even, some odd permutation cycles. Let's show it's not closed and thus can't be a subgroup of $S_4$?"
2) A more general question, if a group has a subgroup, its order divides the group's order. But how many subgroups of such order can it maximally have (not talking about cyclic groups, who have unique subgroups for every order)? That is, let's say if I have a group of order 24, it could have 2 disjoint (besides the identity) subgroups of order 12 presumably, but could it have more? Obviously, with elements of both of the aforementioned disjoint (besides the identity) subgroups. What is the general theory behind this that could be used to answer the question (I feel like I either haven't studied it yet, or am missing something important). I know when we are dealing with cosets, everything gets broken down into equal-sized chunks, so if order of a subgroup is k and of group is x, you have exactly $\frac{x}{k}$ cosets. Thank you very very much.

REPLY [3 votes]: As it happens, less than two weeks ago I gave a talk in my local number theory seminar ("local" modifies "seminar", not "number theory"!) in which I chose to give a detailed proof of why for all $n \geq 2$, $S_n$ has a unique nontrivial homomorphism to $\{ \pm 1 \}$.  This was a strange thing to do for that audience, and I can't remember why I did it...but it will come in handy now!  
Here are two facts about $S_n$ that everyone who knows what $S_n$ is should know:
1) Two elements are conjugate in $S_n$ iff they have the same cycle type.
2) $S_n$ is generated by two-cycles.  Indeed, it is certainly generated by all of its cycles, and a $k$-cycle can be written as a product of $2$-cycles:
$(a_1,a_2,\ldots,a_k) = (a_{k-1}, a_k) \cdots (a_2, a_3)   (a_1, a_2)$.  
Now let $f$ be any homomorphism from $S_n$ into a commutative group $G$.  Since $S_n$ is generated by two-cycles, $f$ is determined by where it maps every two-cycle.  But moreover, since all two-cycles are conjugate in $S_n$, their images under $f$ will be conjugate in the abelian group $G$, and thus equal.  Thus there are, at most, as many homorphisms from $S_n$ to $G$ as there are elements $x \in G$ with $x^2 = 1$.  When $G = \{ \pm 1 \}$ there are therefore at most two homomorphisms: the trivial one, and the one which sends every two-cycle to $-1$.
One can further check that there is a group homomorphism $S_n \rightarrow \{ \pm 1\}$ sending every two-cycle to $-1$ by considering the map $\iota: S_n \rightarrow \operatorname{Aut}_{\mathbb{C}}(\mathbb{C}^n)$ which sends a permutation $\sigma$ of $\{1,\ldots,n\}$ to the unique linear transformation sending the standard basis $e_1,\ldots,e_n$ to the permuted basis $e_{\sigma(1)},\ldots,e_{\sigma(n)}$.  By easy linear algebra every such matrix has determinant $\pm 1$, and the determinant of any two-cycle is the elemtentary matrix obtained by switching two rows of $I_n$ so has determinant $-1$.  Thus $\operatorname{det} \circ \iota: S_n \rightarrow \{ \pm 1\}$ is a surjective group homomorphism, usually called the sign homomorphism.  Its kernel, usually called $A_n$, is an index $2$ subgroup of $S_n$ and is the unique such, since a different index $2$ subgroup of $S_n$ would give rise to a different surjective homomorphism from $S_n$ to $\{ \pm 1\}$.  
Remark: If I am not mistaken, the above reasoning is enough for a complete proof that $[S_n,S_n] = A_n$ for all $n \geq 2$.