Finite-relative-intersection-closed implies transitive

This article gives the statement and possibly, proof, of an implication relation between two subgroup metaproperties. That is, it states that every subgroup satisfying the first subgroup metaproperty (i.e., Finite-relative-intersection-closed subgroup property (?)) must also satisfy the second subgroup metaproperty (i.e., Transitive subgroup property (?))
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Statement

Suppose $p$ is a subgroup property that is a finite-relative-intersection-closed subgroup property. Explicitly, this means that whenever $H,K,L \le G$ are such that $H,K$ are both contained in $L$, $H$ satisfies $p$ in $G$, and $K$ satisfies $p$ in $L$, then $H \cap K$ satisfies $p$ in $G$.

Then, $p$ is a transitive subgroup property: if $K \le H \le G$ are groups such that $K$ satisfies $p$ in $H$ and $H$ satisfies $p$ in $G$, then $K$ satisfies $p$ in $G$.

Related facts

• Characteristicity is transitive, characteristicity is not finite-relative-intersection-closed

Proof

We can set $L = G$ with the notation used in the definitions to complete the proof.