$\{(1), (12)(34), (13)(24), (14)(23)\}$ is the only non-cyclic proper subgroup of $A_4$? What can I use to display the following:
$\{(1), (12)(34), (13)(24), (14)(23)\}$ is the only non-cyclic proper subgroup of $A_4$.
What I've started to do: list all the elements of $A_4$ and finding their orders.
 A: You can think in following way: $A_4$ is a group of order 12. By Lagrange theorem all of its subgroups must have orders which | 12, and not be 12. So, their orders are 1, 2, 3, 4 and 6. Order one is cyclic, two and three are prime, so they are also cyclic. If group of order 4 was cyclic, then $A_4$ would have the element of order 4, which is not possible, since a permutation of order 4 in $S_4$ is of type $(a,b,c,d)$ which is not in $A_4$. So, one example is among subgroups of order 4. 
Claim: $A_4$ has no subgroup of order 6.
 Proof: Let H be a subgroup of order 6 in $A_4$. $A_4$ consists of neutral, eight three-cycles and three elements of type $(a,b)(c,d)$, called double transpositions. So, elementary cardinality gives us that in H it must be at least one three-cycle , without loss of generality let it be $(1,2,3)$. With it, H, being a subgroup, must contain $(1,3,2)$ , its inverse. So far, we had three elements in H, namely two three-cycles and neutral. So, we must have double transposition or another three-cycle.
Option one: We have one double transposition in H, let it be, again without loss of generality, (1,2)(3,4). Now, (1,2,3)(1,2)(3,4)=(1,3,4) must be in H, and so must be its inverse (1,4,3). Now H has neutral, (1,2,3),(1,3,2), (1,2)(3,4), (1,3,4), (1,4,3) which is six elements, and it is enough to show that H is not closed under multiplication, take (1,2,3)(1,3,4) = (2,3,4) which is not in H.
Similar is deal with another option. 
Now, the only possibility is group of order 4. Here we consult again Lagrange theorem: Order of group is a multiple of order of any element. So, we have a group of order 4. Now, all elements can have order 1 or 2. So, in H, of order 4, it can not be three-cycles, since they have order 3. We are left with neutral and three double transposition, which is non-cyclic group and your example and the only possibility.
