Transitive and transfer condition implies finite-relative-intersection-closed

Statement
Suppose $$p$$ is a fact about::transitive subgroup property satisfying the fact about::transfer condition. Then, $$p$$ is a fact about::finite-relative-intersection-closed subgroup property.

Transitive subgroup property
A subgroup property $$p$$ is termed transitive if whenever $$L \le H \le G$$ are groups such that $$L$$ has property $$p$$ in $$H$$, and $$H$$ has property $$p$$ in $$G$$, we have that $$L$$ has property $$p$$ in $$G$$.

Transfer condition
A subgroup property $$p$$ is said to satisfy the transfer condition if whenever $$H, K \le G$$ are subgroups such that $$H$$ satisfies property $$p$$ in $$G$$, we have that $$H \cap K$$ satisfies property $$p$$ in $$K$$.

Finite-relative-intersection-closed subgroup property
A subgroup property $$p$$ is termed finite-relative-intersection-closed if whenever $$H, K \le G$$ are subgroups such that $$H$$ satisfies property $$p$$ in $$G$$ and $$K$$ satisfies property $$p$$ in some subgroup containing both $$H$$ and $$K$$, then $$H \cap K$$ satisfies property $$p$$ in $$G$$.

Examples

 * Subnormal subgroup: Subnormality is finite-relative-intersection-closed follows from the facts that subnormality is transitive and subnormality satisfies transfer condition.

Proof
Given: A group $$G$$ with subgroups $$H, K$$ such that $$H$$ satisfies property $$p$$ in $$G$$ and $$K$$ satisfies property $$p$$ in some subgroup containing both $$H$$ and $$K$$. $$p$$ is both transitive and satisfies the transfer condition.

To prove: $$H \cap K$$ satisfies property $$p$$ in $$G$$.

Proof:


 * 1) (Given data used: $$K$$ satisfies $$p$$ in $$L$$, $$p$$ satisfies transfer condition): $$H \cap K$$ satisfies property $$p$$ in $$H$$: $$K$$ satisfies property $$p$$ in $$L$$, and $$H$$ is a subgroup of $$L$$. Since $$p$$ satisfies the transfer condition, we have that $$K \cap H$$ satisfies property $$p$$ in $$H$$.
 * 2) (Given data used: $$p$$ is transitive, $$H$$ satisfies $$p$$ in $$G$$): By the previous step, we have $$H \cap K$$ satisfies property $$p$$ in $$H$$, and $$H$$ satisfies property $$p$$ in $$G$$. Since $$p$$ is transitive, we obtain that $$H \cap K$$ satisfies $$p$$ in $$G$$.