Minimal normal subgroup and core-free maximal subgroup need not be permutable complements

Statement
Suppose $$G$$ is a primitive group, $$M$$ is a core-free maximal subgroup of $$G$$, and $$N$$ is a minimal normal subgroup of $$G$$. Then, $$M$$ and $$N$$ need not be permutable complements. Specifically, either of these cases is possible:


 * $$G$$ itself is a simple group and $$N = G$$.
 * $$G$$ is not a simple group, and $$N$$ is a proper subgroup of $$G$$, but $$M \cap N$$ is trivial.

Example of situation where the whole group is simple
Let $$G$$ be the alternating group on $$\{ 1,2,3,4,5 \}$$. Let $$N = G$$. Let $$M$$ be the subgroup of $$G$$ comprising those permutations that fix $$\{ 5 \}$$.

Note that:


 * $$M$$ is maximal, since it has index five.
 * Since $$G$$ is simple and $$M$$ is a proper subgroup, $$M$$ is a core-free subgroup.
 * Since $$G$$ is simple, $$N = G$$ is a minimal normal subgroup.

Example of situation where the whole group is not simple
Let $$G$$ be the symmetric group on $$\{ 1,2,3,4,5 \}$$. Let $$M$$ be the subgroup comprising those permutations that fix $$\{ 5 \}$$. Let $$N$$ be the subgroup comprising the even permutations, i.e., the alternating group on $$\{ 1,2,3,4,5 \}$$.


 * $$M$$ is maximal, since it has index five, which is prime.
 * $$M$$ is a core-free subgroup, since its conjugates are precisely the subgroups that fix the elements $$\{ 1 \}, \{ 2 \}, \{ 3 \}, \{ 4 \}, \{ 5 \}$$, and the intersection of all these is the identity element.
 * $$N$$ is a minimal normal subgroup, since $$N$$ is a simple normal subgroup.