Can a kind algebraist offer an improvement to this sketch of a proof?

Show that A4 has no subgroup of order 6.

Note, |A4|=4!/2=12.

Suppose A4>H,|H|=6.

Then |A4/H|=[A4:H]=2.

So H⊲ 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**

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.

**Attribution***Source : Link , Question Author : Stephen Cox , Answer Author : Ben West*