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.

Related facts

The Noetherian group requirement is almost tight for this statement. Specifically, if a group $G$ contains a proper subgroup $H$ that is not contained in any finitely generated subgroup of $G$ , then we can find a strongly intersection-closed subgroup property that is not a multi-invariance property. The trick is to define the property as follows: a subgroup of $G$ has the property if it is either equal to all of $G$ or it is contained in a finitely generated subgroup of $G$ . The lack of multi-invariance follows from the fact that any collection of functions that preserves all such subgroups must preserve $H$ , because the $n$ -ary operation is only looking at finitely many elements at a time so these live inside a finitely generated subgroup of $H$ .

As a concrete example, let $G$ be the finitary symmetric group on the set of integers and $H$ be the subgroup that fixes all the negative integers. $G$ is locally finite, so finitely generated subgroups are the same as finite subgroups in this case. $H$ is infinite and therefore not contained in any finitely generated subgroup, and the result follows.

Facts used

Multi-invariance implies strongly intersection-closed

Proof

Forward direction (multi-invariance implies strongly intersection-closed)

This follows directly from Fact (1).

Reverse direction (strongly intersection-closed implies multi-invariance)

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.