Find the subgroups of A4 I had a question I was hoping for some help on:

Find all of the subgroups of $A_4$

Here is what I know:
$A_4$ is the alternating group on 4 letters. That is it is the set of all even permutations. The elements are:
$(1), (12)(34), (13)(24), (14)(23), (123), (132), (124), (142), (134), (143), (234), (243)$ 
which totals to 12 elements. Which means, the subgroups should have order 1,2,3,4,6 and 12. The groups of order 1 and order 12 are trivial. I also know the groups of order 2 which are:
$[1,(12)(34)], [1, (13)(24)], [1, (14)(23)]$
but would someone be able to help me find the subgroups of order 3, 4 and 6 and be able to provide a good, easy explanation of how you got to them for me? I'm pretty sure there is no such groups of order 6, but I don't know how to explain why.
Thank you so much for your help in advance, I really appreciate it!
 A: If you know that the only groups of order 4 (up to isomorphism) are the cyclic group and the Klein 4-group, and that the only groups of order 6 (up to isomorphism) are the cyclic group and $S_3$, then you can just look for copies of those groups in $A_4$. 
For a cyclic group of order 4, you need an element of order 4. Are there any in $A_4$? 
For a Klein 4-group, you need three elements of order 2, each of which commutes with the other two. Are there such elements in $A_4$? 
For a cyclic group of order 6, you need an element of order 6. Are there any in $A_4$? 
For a copy of $S_3$, you need an element $a$ of order 2, and an element $b$ of order 3, and you need $ba=ab^2$. Now you have a bit of work to do here, since there are several elements of order 2, and several of order 3; but by the inherent symmetries of the set-up, you can assume the element of order 3 is $(123)$, and then there aren't so many cases you have to look at. 
EDIT: but here's an easier way to handle $S_3$. $S_3$ has 3 elements of order 2, none of which commute with the others. Does $A_4$ have that?
A: If you know Sylow's 1st theorem, then you can easily find the subgroup(s) of order $4$.
By Sylow's 1st theorem, $A_4$ must have at least one subgroup of order $4$. By Lagrange's theorem, the order of every element must divide the order of the group, so the elements of a group of order $4$ can only have orders $1$, $2$ or $4$. Now you can form the Sylow $2$-subgroup(s) by looking at your list of elements!
There are many ways to show that $A_4$ has no subgroup of order $6$. Counting elements and making some multiplications may be the most elementary way.
If $H \leq A_4$ and $|H| = 6$, then $H$ has at least one element of order $2$ and one of order $3$ (by Cauchy's theorem). Let $(a \ b \ c) \in H$. Then also $(a \ b \ c)^2 \in H$. You know that one of the $3$ elements of order $2$ is in $H$. You can check all three individually by multiplying them from both sides with the $3$-cycle and its square, and you should end up with a contradiction in all three cases.
