Strongly intersection-closed not implies union-closed

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup metaproperties. That is, it states that every subgroup satisfying the first subgroup metaproperty (i.e., strongly intersection-closed subgroup property) need not satisfy the second subgroup metaproperty (i.e., union-closed subgroup property)
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Statement

General statement

It is possible to have a strongly intersection-closed subgroup property $p$ (a subgroup property closed under arbitrary intersections, including the empty intersection, and therefore true for every group as a subgroup of itself) that is not a union-closed subgroup property.

Statement with symbols using a specific group

It is possible to have a strongly intersection-closed subgroup property $p$ and a group $G$ such that there is a collection $H_{i}, i \in I$ of subgroups of $G$ all satisfying $p$ such that the union $⋃_{i \in I} H_{i}$ is a subgroup of $G$ not satisfying $p$ .

Proof

Proof idea

Closure under intersections can be achieved by making all the subgroups small, but we also need to throw in the whole group. Therefore, we need to impose a condition of either being equal to the whole group or the group order being bounded by some small value.

We can then try to take unions of such subgroups to obtain a bigger subgroup.

We have three layers of nontrivial groups here: the subgroups, their union (which can't be the whole group because the whole group satisfies the property), and the whole group. The size should at least double each time. Therefore the smallest counterexample should have order at least $2^{3} = 8$ .

These insights suffice to come up with the proof.

Proof details

The construction is as follows:

$p$ is the property defined as follows: a subgroup satisfies $p$ if it is either equal to the whole group or has order at most two.
$G$ is elementary abelian group:E8, i.e., $Z_{2} \times Z_{2} Z_{2}$ .
$H_{1}, H_{2}, H_{3}$ are two-element subgroups with non-identity elements respectively $(1, 0, 0)$ , $(0, 1, 0)$ , and $(1, 1, 0)$ .

We now prove the individual pieces:

$p$ is strongly intersection-closed: By definition, every group satisfies $p$ as a subgroup of itself. For an intersection of one or more subgroups satisfying $p$ , if they are all equal to the whole group, the intersection is also the whole group. If any of them is proper, then that subgroup has size at most two, so the intersection (which is contained in that subgroup) also has size at most two, and therefore satisfies $p$ .
$H_{1}, H_{2}, H_{3}$ all satisfy $p$ by construction.
$H_{1} \cup H_{2} \cup H_{3}$ is a Klein four-group ${(0, 0, 0), (1, 0, 0), (0, 1, 0), (1, 1, 0)}$ so it is a subgroup of order four, and therefore is a subgroup not satisfying $p$ .