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 $A,B\leq G$ 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. Nilpotent implies every subgroup is subnormal
2. Normalizing join-closed subgroup property implies every maximal element is intermediately subnormal-to-normal

Proof

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:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$M$ is subnormal in $G$.	Fact (1)	$G$ is nilpotent		Fact-direct
2	$M$ is intermediately subnormal-to-normal in $G$	Fact (2)	$\alpha$ is normalizing join-closed, $M$ is maximal with respect to satisfying $\alpha$.		Fact-given direct
3	$M$ is normal in $G$.			Steps (1), (2)	Step-combination direct, along with definition of subnormal-to-normal.

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: $A\leq M$.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$M$ is normal in $G$.			previous part of proof, see #Proof of (3) above
2	$A$ normalizes $M$.			Step (1)	Follows directly from Step (1) and definition of normal subgroup
3	$AM$ satisfies $\alpha$.		$A,M$ both satisfy $\alpha$ $\alpha$ is normalizing join-closed	Step (2)	Step-given combination direct
4	$AM=M$		$M$ is maximal among subgroups of $G$ satisfying $\alpha$	Step (3)	Step-given combination direct
5	$A\leq M$			Step (4)	Step-direct

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: $\sigma (M)\leq M$

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.