Proof verification that $A_4$ has no subgroup of order $6$. Assume that $A_4$ has a subgroup of order $6$, and let $H$ be a subgroup of order $6$.
Since the order of $A_4$ is $12$ and the order of $H$ is $6$, we know that there are two left cosets.
Saying $A_4= H\cup gH$ for $g \in A_4$.
But, a subgroup of index $2$ in a finite group has the same number of left cosets as right cosets.
So, $gH=Hg$ for $g \in A_4$.
But, $S_4$ is not an abelian group and thus $A_4$ is not abelian group.
This implies that $gH$ is not equal to $Hg$ for $g \in A_4$ and $h \in H$.
Contradiction.
I don't know whether this is right and seems like there must be a mistake.
Anyone can check?   
 A: I see two problems in this proposed proof.  One is that a subgroup cannot be abelian if the whole group is not abelian.  That is wrong.  The other is that if a normal subgroup is not abelian, then the quotient group is not abelian.  That is also wrong.  See if you can find some counterexamples to both of those statements.
A: Maybe you want to consider the following proof. Since index$[A_4:H]=2$, $H \unlhd A_4$. $A_4/H \cong C_2$ is abelian, so $  V_4\cong [A_4,A_4] \subseteq H$. But then $|V_4|=4 \mid |H|=6$, a contradiction.
A: A subgroup of index $2$ contains all squares, in particular all elements of odd order. But in $A_4$ we have eight $3$-cycles. Way too many...
A: Mistake is here: how did you prove that bH is not Hb?? You are assuming that bh=hb for all h in H is equivalent to bH=Hb and that is not true. You need equality of sets so that there is no such c in G, that holds b*h = h * c. So, not that way, unless you want to use that the only normal subgroup of $A_4$ has 4 elements, and you clearly don't want.
So, a classical proof is like this: you take $A_4$. It contains: one identity(takes x to x, where x is a point, coincidence), 8 tri-cycles((1,2,3),....) , 3 double transpositions(so (1,2)(3,4),...). Now, let's suppose contrary: it is a subgroup H of order 6. H must contain identity, since it is a subgroup. From the cardinality(number of elements) it is clear that H must contain tri-cycle. Let it contain (1,2,3) without loss of generality. Then it has the inverse (1,3,2) of (1,2,3). So,three elements so far. We now need another three-cycle. Why? Why we cannot have the subgroup having three double transpositions, two permutations already given and identity?? It's not a subgroup! So, we introduce (1,2,4). We must have its inverse (1,4,2). So, five elements so far. But, we must have (1,2,3)(1,2,4) = (1,3)(2,4) in H, because we have both (1,2,3) and (1,2,4) in H and as well (1,2,3)(1,3)(2,4)=(2,4,3) in H, and (2,4,3) is a seventh element in H,a contradiction. 
Hope that helped!
