Let $H$ and $K$ be two proper non-trivial subgroups of the alternating group $A_5$ and $\langle H,K\rangle < A_5$. We can show that there exists a maximal subgroup $M$ of $A_5$ such that $H\not\leq M$ and $K\not\leq M$. To see this let $\Omega:=\{1,2,3,4,5\}$ and ${\rm Supp}_{\Omega}(H):= \{\omega\in\Omega\mid \omega^{h}\neq \omega,$ for some $h\in H\}$. Since $|{\rm Supp}_{\Omega}(H)|\geq 3$ and $|{\rm Supp}_{\Omega}(K)|\geq 3$, ${\rm Supp}_{\Omega}(H)\cap {\rm Supp}_{\Omega}(K)\neq\emptyset$. Suppose that $l\in {\rm Supp}_{\Omega}(H)\cap {\rm Supp}_{\Omega}(K)$. Thus $H\not\leq {\rm stab}_{A_5}(l)$ and $K\not\leq {\rm stab}_{A_5}(l)$ and ${\rm stab}_{A_5}(l)$, which denotes the stabilizer of $l$ in $A_5$, is a maximal subgroup of $A_5$.
By a GAP code I could see that this property is true for $A_6$ and $A_7$. (Also it is true for $S_5$, $S_6$ and $M_{10}$).
Could we generalize this property to $A_n; n\geq 5$?
(If one of the subgroups $H$ or $K$ is transitive this property is true, but I have no idea, when $H$ and $K$ are both intransitive).
Any answer or idea will be appreciated! Thank you.
If there exists $l$ in the supports of both $H$ and $K$, then you can choose $M={\rm Stab}_{A_5}(l)$.
Otherwise, let $l$ be in the support of $H$ but not of $K$, and let $m$ be in the support of $K$ but not of $H$, and choose $M = {\rm Stab}_{A_5}(\{l,m\})$.