For a positive integer $n\ge 4$ and a prime $p\le n$, let $U_{p,n}$ denote the union of all $p$-Sylow subgroups of the alternating group $A_n$ on $n$ letters. Also let $K_{p,n}$ denote the subgroup of $A_n$ generated by $U_{p,n}$. Then the order of $K_{2,4}=5$ and order of $K_{2,5}=60$.
It is given in a question paper of an examination for $\mathbf{NET}$. I can't do it. If someone knows, please help me.
For $A_4$: You can just list down all 12 elements and you will see that there is a unique 2-Sylow subgroup $H = \{e, (12)(34), (13)(24), (14)(23)\}$. Hence, $U_{2,4} = K_{2,4} = H$ so $|K_{2,4}| = 4$.
For $A_5$ : You are looking at subgroups of order 4. Since any group of order 2 is contained a 2-sylow subgroup, you can just count the elements of order 2. $A_5$ has 15 elements of order 2 of the form $(12)(34), (12)(35)$, etc. Hence, $|U_{2,5}| \geq 15+1 = 16$ and so $|K_{2,5}| \in \{20, 30, 60\}$ (since it must divide 60)
Assume $|K_{2,5}| = 20$, then let $A_5$ act on $A_5/K_{2,5}$ by left translation, and this will give you a homomorphism $A_5 \mapsto S_3$. Since $A_5$ is simple, this homomorphism must be injective, which means $$ 60 = |A_5| \leq |S_3| = 6 $$ Hence, $|K_{2,5}| \neq 20$.
Again, if $|K_{2,5}| = 30$, then it has index 2, and so has to be normal in $A_5$, which is impossible. Thus, $|K_{2,5}| = 60$.