$A_4$ has no subgroup of order $6$? Can a kind algebraist offer an improvement to this sketch of a proof?

Show that $A_4$ has no subgroup of order 6.  

Note, $|A_4|= 4!/2 =12$.
Suppose $A_4>H, |H|=6$.
Then $|A_4/H| = [A_4:H]=2$.
So $H \vartriangleleft A_4$ so consider the homomorphism
$\pi : A_4 \rightarrow A_4/H$
let $x \in A_4$ with $|x|=3$ (i.e. in a 3-cycle)
then 3 divides $|\pi(x)|$
so as $|A_4/H|=2$  we have $|\pi(x)|$ divides 2
so $\pi(x) = e_H$ so $x \in H$
so $H$ contains all 3-cycles
but $A_4$ has $8$ $3$-cycles
$8>6$, $A_4$ has no subgroup of order 6.
 A: I'm not sure if this is a combination of what people did above or not, but here's an approach that should work.
For the purpose of contradiction, assume $H\subset A_4$ is a subgroup of $A_4$ of order 6. Then, for any $a\notin H$ by properties of left cosets, then $aH\cap H=\emptyset$. Again, by properties of cosets, since $|aH|=|eH|=|H|$ (all cosets have the same number of elements), this implies that $|aH|=6$. Then, as cosets form a partition of the group $A_4$, and $|A_4|=12$, then $$A_4=H\cup aH$$ Now suppose that $a$ is a 3-cycle in $A_4$, then either $a^2\in H$ or $a^2\in aH$. If $a^2\in H$, then this implies that $a^4=a^2\cdot a^2\in H$ but, since the order of $a$ is 3 (it is a 3-cycle), then $a^4=a$ and since $a\notin H$ this is a contradiction.
Similarly, if $a^2\in aH$ by properties of cosets this implies that $(a^2)\cdot a^{-1}\in H$ which implies $a\in H$ and again this is a contradiction.
As such, H cannot be a subgroup of $A_4$ of order 6.
A: Consider the group $A_4/H$. Let $x$ be a $3$-cycle, not in $H$, and consider the cosets $H$, $xH$, and $x^2H$ in $A_4/H$. Since this is a group of order $2$, two of the cosets must be equal. But $H$ and $xH$ are distinct, so $x^2H$ must be equal to one of them. 
If $H=x^2H$, then $x^2=x^{-1}\in H$, so $x\in H$, contradiction. If $xH=x^2H$, then $x\in H$, same problem. So $H$ doesn't exist.  
A: By looking at the possible cycle types, we see that $A_4$ consists of the identity element (order $1$), $3$ double transpositions (order $2$) and $8$ $3$-cycles (order $3$).
Assume that $A_4$ has a subgroup $H$ of order $6$.
Since $A_4$ does not contain elements of order $6$, $H$ cannot be cyclic.
Therefore $H \cong S_3$, implying that $H$ contains $3$ elements of order $2$.
So $H$ contains the identity element and the $3$ double transpositions.
Since those $4$ elements form a subgroup of $A_4$, $H$ contains a subgroup of order $4$. Contradiction.
A: I am a novice, just start to preview the content in the college. The following is my solution. If I make any mistakes, please point it out.
First, as a group of order 6 is either $C_6$ or $D_6$, we can merely discuss these two cases. Moreover, an even permutation of 4 elements can only be of the form (abc)(d) or (ab)(cd)
Case1: $C_6$. This is, of course, impossible as the maximum order is 4.
Case2: $D_6$. It is of the form {$e,r,r^2,s,sr,sr^2$}. r must of the form (abc)(d), and w.l.o.g., we can assume s of the form (ad)(bc).
Then it us easy to calculate sr=(acd), $sr^2$=(abd). However, in this case, (abd)^2=(adb), which is a new element outside of the six. So contradiction.
A: I know this post is old, but there's another elegant way to prove this - a subgroup of order 6 has index 2. We prove the following statement:
Any subgroup of index 2 of a finite group must contain all elements of odd order.
Let $G$ be finite and $H\subseteq G$ a subgroup of index 2. Any subgroup of index 2 is normal, so $G/H$ is a group and we write  $\bar g:=gH$. Let $g$ be an element of odd order. Now we have $$g^{\operatorname{ord}g}=e\ \Rightarrow\ \bar g ^{\operatorname{ord}g}=\bar e\ \Rightarrow\ \operatorname{ord}\bar g\mid\operatorname{ord}g.$$ On the other hand, by Lagrange's theorem, we know that $\operatorname{ord}G/H =2$ so $$\bar g^2=\bar e\ \Rightarrow\ \operatorname{ord}\bar g\mid2.$$ Since $ \operatorname{ord}g $ is odd, it follows that $$\operatorname{ord}\bar g=1\ \Rightarrow\ \bar g=\bar e=e_{G/H}=H\ \Rightarrow\ g\in H.$$
Now since $A_4$ contains 9 elements of odd order, a subgroup of index 2 would, by the above statement have at least 9 elements, but by Lagrange's theorem has exactly 6 elements, which is a contradiction.
A: Based on reflections, $A_4$ is isometric to the rotation group of the tetrahedron. The tetrahedron has 4 vertices, so 4 subgroups of order 3. There are also 3 pairs of nonadjacent edges. So 3 subgroups of order 2. This exhausts all 12 elements of the group.
A: Here's another way to look at this.  It is well known that A4 has a normal subgroup of order 4, isomorphic to the Klein-4 group, and consisting of the identity permutation along with the 3 products of disjoint 2-cycles, (12)(34), (13)(24), and (14)(23).
As others have noted, any alleged subgroup H of order 6 must be isomorphic to S3, so H contains a unique subgroup T of order 3 which is thus characteristic in H.  Thus T is normal in A4, being a characteristic subgroup of the normal subgroup H (of index 2) of A4.
Thus A4 is the direct product of two abelian subgroups of orders 4 and 3 and must therefore be abelian itself, which it is not.  So the alleged subgroup H of order 6 does not exist.
