# Prove that the alternating group has a subgroup of order 12

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?

Consider the Alternating group $A_{4}$, it has order 12, and it's in $A_{5}$. Aletnaring group $A_{n}$ is not cyclic for $n >3$
• Why is $A_{4}$ a subgroup of $A_{5}$? Is it always the case that a smaller alternating group will be a subgroup of a larger alternating group? – EmaLee Apr 13 '15 at 0:16
As a hint to your subgroup of order $12$ question, note that $A_n \subset A_m$ for all $n < m$.