Prove that $A_{5}$ has a subgroup of order $12$.
What I have so far: The order of an alternating subgroup $A_{n}$ = $n!/2$. So the order of $A_{5}$ is $60$. We know that the order of a subgroup must be divide the order of the group, and this is the case since $12$ divides $60$.
Also, I have another question. Are all alternating groups cyclic?