# Finite-relative-intersection-closed implies transitive

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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 $G$.

## Proof

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