Let $H, K$ and $L$ be subgroups of order $2$ in some group $G$, and observe that the set $\{[h, k, l] : h \in H, k \in K, l \in L\}$ contains at most one nonidentity element, and so it generates a cyclic group. Now let $G = A_5$, the alternating group of degree $5$. Show that it is possible to choose subgroups $H, K$ and $L$ of order $2$ such that $[H, K, L] = G$.
I'm supposed to use the following hint:
Let $P \subseteq G$ have order $5$, and choose $H,K \subseteq N_G(P)$ with $H \neq K$, and choose $L \nsubseteq N_G(P)$. Show that $[H,K] = P$ and observe that $\langle P,L \rangle = G$.
I know that a subgroup of order $5$ is going to be cyclic, and can be generated by any element of order $5$, so finding $P$ isn't a problem. But, I've no idea what (or how big) $N_G(P)$ will be, so I'm unable to choose $H,K$. Even if I did, how do I show that $\langle P,L \rangle = G$?
Edit: I can manually find the solution with this hint, but, why does this hint work, and is there an easier way to compute such subgroups without manual checking?