Normalizing join-closed subgroup property implies every maximal element is intermediately subnormal-to-normal

Statement
Suppose $$G$$ is a group and $$\alpha$$ is a subgroup property that can be evaluated for subgroups of $$G$$ such that $$\alpha$$ is a  normalizing join-closed subgroup property: if $$A,B \le G$$ are such that both $$A,B$$ satisfy $$\alpha$$, and $$B$$ normalizes $$A$$, then $$AB$$ also satisfies.

Suppose $$M$$ is a maximal element among the subgroups of $$G$$ satisfying $$\alpha$$. In other words, $$M$$ satisfies $$\alpha$$ and it is not contained in any bigger subgroup satisfying $$\alpha$$. Then, $$M$$ is an proves property satisfaction of::intermediately subnormal-to-normal subgroup inside $$G$$. In other words, if $$M$$ is a 2-subnormal subgroup inside some subgroup $$L$$ of $$G$$, then $$M$$ is normal inside $$L$$.

Related facts

 * Normalizing join-closed subgroup property in nilpotent group implies unique maximal element

Proof
Given: A group $$G$$, a normalizing join-closed subgroup property $$\alpha$$ of $$G$$, a subgroup $$M$$ of $$G$$ maximal with respect to inclusion among subgroups satisfying $$\alpha$$. A subgroup $$L$$ of $$G$$ such that $$M$$ is 2-subnormal in $$L$$. Let $$g \in L$$. Consider the subgroup $$N = gMg^{-1}$$ of $$G$$.

To prove: $$N \le M$$ (Note that once we show this, it also must be equal to $$M$$ by the equivalence of definitions of normal subgroup).

Proof: