Sylow subgroup of a symmetric group Consider the symmetric group of$S_{20}$ and it's subgroup $A_{20}$ consisting of all even permutations. Let $H$ be a $7$-Sylow subgroup of$A_{20}$. Is $H$  cyclic? And is correct the statement which says that any $7$-Sylow subgroup of $S_{20}$ is subset of $A_{20}$?


I know that order of H is 49 and H is not normal subgroup of $A_{20}$. But I don't understand whether it is cyclic or not.
In $|S_{20}|$, the highest power of $7$ which divides $20!$ is $7^2$. So it is clear that the Sylow-$7$ subgroup of $S_{20}$ is of order $7^2$. 

Group of order $7^2$ is either cyclic or  isomorphic to $Z_7\times Z_7$. 

If it is cyclic, then $S_{20}$ will have an element of order $49$, and it should be product of disjoint cycles. Check, whether this is possible? You will reach to your answer.
For the last option
Sn has two types of subgroups: a subgroup which contains half even and half odd permutations or it contains only even permutations.