# 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.

## Proof

### Example of situation where the whole group is simple

Further information: alternating group:A5, A5 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

Further information: symmetric group:S5

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.