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.
 A: A group of order $6$ is isomorphic to $C_6$ or $S_3$. We can rule out $C_6$ because $A_4$ has no element of order 6; therefore the subgroup must be isomorphic to $S_3$. But $S_3$ has three elements of order $2$. In $A_4$ the three elements of order $2$ generate a group of order $4$, so they can't be contained in a group of order $6$.
A: 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 $G$.
A: Well if you have the time and you really want to convince yourself you could always list down all the elements of $A_4$.
$A_4=\{(1), (12)(34), (13)(24), (14)(23),(123), (132), (124),(142),(134), (143), (234), (243)\}$
Multiply the elements with each other to see which elements should go together inside a subgroup. You will see that these are the following subgroups you could make:
the trivial subgroup of order $1$:
$<(1)>=\{(1)\}$
three (cyclic) subgroups of order $2$:
$<(12)(34)>=\{(1),(12)(34)\}$
$<(13)(24)>=\{(1),(13)(24)\}$
$<(14)(23)>=\{(1),(14)(23)\}$
four (cyclic) subgroups of order $3$:
$<(123)>=\{(1),(123),(132)\}$
$<(124)>=\{(1),(124),(142)\}$
$<(134)>=\{(1),(134),(143)\}$
$<(234)>=\{(1),(234),(243)\}$
the subgroup of double transpositions of order $4$ (isomorphic to the Klein-4 group):
$\{(1),(12)(34),(13)(24),(14)(23)\}$
and the group itself of order $12$:
$A_4=\{(1), (12)(34), (13)(24), (14)(23),(123), (132), (124),(142),(134), (143), (234), (243)\}$
