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.