Show that $S_5$ has a normal subgroup $H$ such that $S_5/H$ is of order $20$
I think if there exist such H then the order of $H$ should be $6$ . And order of $S_5$ is $5!$ Ie $120$ and so $S_5$ has sylow $2$ subgroup of order $8$ and a sylow $3$ subgroup and a sylow $5$ subgroup so how can I prove that there exist a normal subgroup of order $6$?