Multi-invariance equals strongly intersection-closed in Noetherian group

Statement

Let $p$ be a subgroup property and $G$ be a Noetherian group (i.e., a group in which every subgroup is finitely generated). Then, the following are equivalent:

$p$ is a multi-invariance property when restricted to $G$ , there is a collection ${F_{n}}_{n \in N}$ of $n$ -ary functions on $G$ such that the subgroups satisfying $p$ are precisely the ones invariant under all these functions.
$p$ is a strongly intersection-closed subgroup property when restricted to $G$ , and every (possibly empty) collection ${H_{i}}_{i \in I}$ of subgroups of $G$ all satisfying $p$ , the intersection $⋂_{i \in I} H_{i}$ also satisfies $p$ .

Since both subgroup properties keep the ambient group fixed, we have framed the statement in terms of a single Noetherian group, but we can also frame it broadly in terms of subgroup properties over the collection of all Noetherian groups.

Facts used

Multi-invariance implies strongly intersection-closed

Proof

Forward direction

This follows directly from Fact (1).

Reverse direction

Construction of functions

The idea is to define one $n$ -ary function $f_{n} : G^{n} \to G$ for each $n$ as follows:

For $(g_{1}, g_{2}, \dots, g_{n}) \in G^{n}$ , let $H$ be the subgroup of $G$ generated by all the $g_{i}$ .
If $H$ satisfies $p$ , define $f_{n} (g_{1}, g_{2}, \dots, g_{n})$ to be the identity element.
Otherwise, let $K$ be the intersection in $G$ of all subgroups of $G$ satisfying $p$ and containing $H$ . So, $H < K$ , and $K$ satisfies $p$ by the "strongly intersection-closed" assumption. Pick $f_{n} (g_{1}, g_{2}, \dots, g_{n})$ to be any element of $K$ that is outside $H$ (the choice of element is arbitrary).

We now need to prove that the subgroups of $G$ invariant under all the $f_{n}, n \in N$ are precisely the subgroups satisfying $p$ .

Proof that subgroups that don't satisfy the property are not multi-invariant

Given: A group $G$ , a subgroup $H$ of $G$ that doesn't satisfy $p$ .

To prove: There exists $n \in N$ and $(h_{1}, h_{2}, \dots, h_{n}) \in H^{n}$ such that $f_{n} (h_{1}, h_{2}, \dots, h_{n}) \notin H$ .

Proof: Since $G$ is Noetherian, $H$ is finitely generated. Take $h_{1}, h_{2}, \dots, h_{n}$ to be a generating set for $H$ .

Since $H$ does not satisfy $p$ , the function construction shows that $f_{n} (h_{1}, h_{2}, \dots, h_{n})$ is in $K$ but not in $H$ , where $K$ is the intersection of all subgroups of $G$ satisfying $p$ and containing $H$ . Therefore, $f_{n} (h_{1}, h_{2}, \dots, h_{n}) \notin H$ .

Proof that subgroups that do satisfy the property are multi-invariant

Given: A group $G$ , a subgroup $L$ of $G$ that satisfies $p$ , $n \in N$ and $(l_{1}, l_{2}, \dots, l_{n}) \in L^{n}$ .

To prove: $f_{n} (l_{1}, l_{2}, \dots, l_{n}) \in L$ .

Proof: Let $H$ be the subgroup of $G$ generated by $l_{1}, l_{2}, \dots, l_{n}$ . Since all the generators are in $L$ , $H \leq L$ .

If $H$ satisfies $p$ , then $f_{n} (l_{1}, l_{2}, \dots, l_{n})$ is the identity element, and therefore $f_{n} (l_{1}, l_{2}, \dots, l_{n}) \in L$ as desired.

If $H$ does not satisfy $p$ , then $f_{n} (l_{1}, l_{2}, \dots, l_{n})$ is in $K$ but not in $H$ , where $K$ is the intersection of all subgroups of $G$ satisfying $p$ and containing $H$ . By the definition of $K$ being the intersection of all subgroups of $G$ satisfying $p$ and containing $H$ , and the fact that $L$ is such a subgroup (it satisfies $p$ and contains $H$ ), we have that $K \leq L$ . Therefore, $f_{n} (l_{1}, l_{2}, \dots, l_{n}) \in L$ as desired.