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

Statement
Suppose $$G$$ is a nilpotent 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.

Then, suppose the collection of subgroups of $$G$$ satisfying $$\alpha$$ has a maximal element $$M$$, i.e., there is a subgroup $$M$$ satisfying $$\alpha$$ and $$M$$ is not contained in any bigger subgroup of $$G$$ satisfying $$\alpha$$.

Then the following are true:


 * 1) $$M$$ contains every subgroup of $$G$$ satisfying $$\alpha$$.
 * 2) $$M$$ is the unique maximal element among subgroups of $$G$$ satisfying $$\alpha$$.
 * 3) $$M$$ is a normal subgroup of $$G$$.
 * 4) $$M$$ is a characteristic subgroup of $$G$$.

Note that, in general, a maximal element need not exist. However, for a finite group (in our case a finite nilpotent group) and more generally for a Noetherian group (in our case a finitely generated nilpotent group) a maximal element must always exist, so the above conditions hold for it.

Facts used

 * 1) uses::Nilpotent implies every subgroup is subnormal
 * 2) uses::Normalizing join-closed subgroup property implies every maximal element is intermediately subnormal-to-normal

Proof of (3)
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 $$\alpha$$.

To prove: $$M$$ is normal in $$G$$.

Proof:

Proof of (1) using (3)
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 $$\alpha$$. $$A$$ is a subgroup of $$G$$ satisfying $$\alpha$$.

To prove: $$A \le M$$.

Proof:

Proof of (2) using (1)
This is immediate.

Proof of (4) using (1)
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 $$\alpha$$. An automorphism $$\sigma$$ of $$G$$.

To prove: $$\sigma(M) \le M$$

Proof: This follows directly from (1) and the observation that because $$\alpha$$ is a subgroup property and $$M$$ satisfies it, $$\sigma(M)$$ also satisfies $$\alpha$$ on account of subgroup properties being automorphism-invariant.