Let $G$ be a finite simple group and $M$ and $M'$ be two maximal subgroups of $G$. Also let $m\_{M}$ be the set of minimal subgroups of $M$ and similarly $m\_{M'}$ be the set of minimal subgroups of $M'$. Is it possible that $m\_{M}\subseteq ‎m\_{M'}$.



Maximal subgroups of a finite simple group which have some minimal subgroups in common
Yes. Let $G = {\rm PSL}(3,4)$, and let $M$ be a maximal subgroup with the structure $3^2:Q\_8$.


The minimal subgroups (i.e. the subgroups of prime order) of $M$ have order $2$ or $3$ and generate a subgroup of $M$ with structure $3^2:2$. This subgroup is also contained in a maximal subgroup $M'$ of $G$ with $M' \cong A\_6$.


Another example is $G = M\_{22}$ with $M = M\_{10} = A\_6 \cdot 2$ and $M' = 2^4:A\_6$.


Yet another is $G = {\rm PSL}(3,7)$ again with $M = 3^2:Q\_8$ and $M' = (3 \times A\_4):2$.


It seems likely that there will be infinitely many examples of type $G = {\rm PSL}(3,q)$ and $M = 3^2:Q\_8$, , and this should not be hard to prove, since the maximal subgroups of ${\rm PSL}(3,q)$ are well understood.


