$A_4$ has no subgroup of order $6$?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.