Question

Prove, by contradiction, that A₄ has no subgroup of order 6.

Answer 1

Suppose $A_4$ has a subgroup $H$ of order 6. There is only groups of order 6 (up to an isomorphism) : $S_3$ and $\mathbb{Z}/6\mathbb{Z}$. As there is no element of order $6$ in $A_4$, we should have $H\simeq S_3$ and thus there is $3$ distinct elements of order 2 in $H$. There is only $3$ elements of order 2 in $A_4$ : these are $(12)(34),(13)(24),(14)(23)$, so they should all be included in $H$. However, $\{id, (12)(34),(13)(24),(14)(23) \}$ is a group of order 4, and as $4 \nmid 6$ it is a contradiction, so there is no $H$ of order 6.