Alternative Creative Proofs that $A_4$ has no subgroups of order 6 Since I've been so immersed in group theory this semester, I have decided to focus on a certain curious fact: $A_4$ has no subgroups of order $6$.
While I know how to prove this statement, I am interested in seeing what you guys can offer in terms of unique and creative proofs of this statement!
Proofs without words would be interesting too, although I'm not sure that is possible.

So we have a group $G$ of order $12$ with a normal subgroup $H$ of order $4$ and a normal subgroup $K$ of order $6$. Then $H\cap K$ is a normal subgroup. It has order $2$ (since if $H\cap K=1$, then $|G|=24$). Thus, $|G/(H\cap K)|=6$, and it has a unique Sylow 3-subgroup, so it is not $S_3$. The only alternative is that it is cyclic. Because a quotient of $G$ contains an element of order $6$, so does $G$.
