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.