Intersection of Maximal subgroups

Question

Let G be a group. Show that the intersection of all maximal subgroups of G is a normal subgroup. I proved that the normalizer of a Maximal subgroup is either the subgroup itself of the Maximal subgroup is normal. Also I showed that every subgroup is present in a Maximal subgroup

Answer 1

Hint: if $M$ is maximal, then for any $g \in G$ also $M^ g=g^{-1}Mg$ is also maximal.

Answer 2

Let $\mathcal{M}$ be the set of maximal subgroups of $G$. Then $G$ acts on $\mathcal{M}$ by conjugation:

Suppose $M\in\mathcal{M}$ and $g\in G$, and that $gMg^{-1}\leq N\leq G$. Then $M\leq g^{-1}Ng\leq G$ so either $g^{-1}Ng=M$ or $g^{-1}Ng=g^{-1}Gg=G$. This implies either $N=gMg^{-1}$ or $N=G$. Hence $gMg^{-1}\in \mathcal{M}$ as required.

Now, as the map $M\mapsto gMg^{-1}$ is a bijection $\mathcal{M}\to \mathcal{M}$ we have $$\bigcap_{M\in \mathcal{M}}M=\bigcap_{M\in\mathcal{M}}gMg^{-1}=g\left(\bigcap_{M\in \mathcal{M}}M\right)g^{-1}.$$