Existence of subgroup of order six in $A_4$

Solution 1:

Assume $H \le A_4$ is a subgroup of order 6. Then $H$ contains a unique subgroup $C$ of order 3. So $C$ is characteristic in $H.$ And as $H$ is normal in $A_4,$ we obtain $C$ is normal in $A_4.$ On the other hand, if $(a b c)$ is a generator of $C,$ conjugating $(abc)$ by $(ab)(cd) \in A_4$ we obtain $(bad) \not\in C,$ from which we obtain not only a contradiction, but a potential pun - not too shabby.

Solution 2:

There is a proof in the QUESTION here.

Note that There are twelve elements in $A_4$ : $(1),\ (12)(34),\ (13)(24),\ (14)(23),\ (123),\ (132),\ (124),\ (142),\ (134),\ (143),\ (234),\ (243)$.