TITLE: About alternating group $A_4$
QUESTION [7 upvotes]: This is a simple exercise telling that $A_4$ cannot have a subgroup of order $6$. Here in my way:
Obviously, for any group $G$ and a subgroup $H$ of it with index $2$; we have $∀$$ g\in G$ ,$g^2\in H$. I suppose that $A_4$ has such this subgroup, named $H$, of order 6. Then for any $\sigma\in A_4$; $\sigma^2\in H$. I think maybe the contradiction happens when we enumerate all $\sigma^2$. May I ask if there is another approach for this problem? Thanks.

REPLY [5 votes]: Your approach and many more can be found in this article, where $11$ different proofs are given.

Michael Brennan. Des Machale. Variations on a Theme: $A_4$ Definitely Has no Subgroup of Order Six!,  Mathematics Magazine, Vol. $73$, No. $1$ (2000) JSTOR