For a positive integer $n\ge 4$ and a prime number $p\le n$,let $U_{p,n}$ denote the union of all Sylow $p$ subgroups of $A_n$ .Also let $K_{p,n}$ denote the subgroup of $A_n$ generated by $U_{p,n}$ and let $|K_{p,n}|$ denote the order of $K_{p,n}$.
Show that $|K_{2,4}|=4$ and $|K_{2,5}|=60.$
My try:Number of Sylow $2$ subgroups of $A_4=1+2k=n_2|3 \implies n_2=1,3$ .$A_4$ has only one Sylow $2$ subgroup of order $4$ i.e. $\{e,(12)(34),(13)(24),(14)(23)\}\implies |U_{2,4}|=4\implies |K_{2,4}|=4$.
But I am unable to solve the problem in case of $A_5$ .Please give some hints