Subnormality is finite-relative-intersection-closed

Property-theoretic statement
The property of being a subnormal subgroup is finite-relative-intersection-closed.

Statement with symbols
Suppose $$H, K$$ are subgroups of $$G$$ such that $$H$$ is a subnormal subgroup of $$G$$, and $$K$$ is a subnormal subgroup inside some subgroup $$L$$ containing both $$H$$ and $$K$$. Then, $$H \cap K$$ is subnormal in $$G$$.

More specifically, if $$H$$ is $$h$$-subnormal in $$G$$ and $$K$$ is $$k$$-subnormal in $$L$$, then $$H \cap K$$ is $$(h + k)$$-subnormal in $$G$$.

Applications

 * Subnormality is permuting join-closed: If $$H$$ and $$K$$ are both subnormal subgroups and they permute, i.e., $$HK = KH$$, then $$HK$$ is subnormal. The proof of this uses certain subgroups in intermediate steps that are subnormal as a consequence of the fact that subnormality is finite-relative-intersection-closed.

Facts used

 * 1) uses::Subnormality satisfies transfer condition: If $$A \le G$$ is $$a$$-subnormal and $$B \le G$$, then $$A \cap B$$ is $$a$$-subnormal in $$B$$.
 * 2) uses::Subnormality is transitive: If $$A \le B \le G$$ are groups, and $$A$$ is $$a$$-subnormal in $$B$$ and $$B$$ is $$b$$-subnormal in $$G$$, then $$A$$ is $$(a + b)$$-subnormal in $$G$$.# uses::Transitive and transfer condition implies finite-relative-intersection-closed
 * 3) uses::Transitive and transfer condition implies finite-relative-intersection-closed

Proof
Given: $$H, K \le G$$. $$H$$ is $$h$$-subnormal in $$G$$ and $$K$$ is $$k$$-subnormal in a subgroup $$L$$ of $$G$$ containing both $$H$$ and $$K$$.

To prove: $$H \cap K$$ is $$(h + k)$$-subnormal in $$G$$.

Proof:


 * 1) (Facts used: fact (1); Given data used: $$H \le L$$, $$K$$ is $$k$$-subnormal in $$L$$): $$H \cap K$$ is $$k$$-subnormal in $$H$$: $$K$$ is $$k$$-subnormal in $$L$$, and $$H \le L$$, so by fact (1), $$K \cap H$$ is $$k$$-subnormal in $$H$$.
 * 2) (Facts used: fact (2); Given data used: $$H$$ is $$h$$-subnormal in $$G$$): $$H \cap K$$ is $$k$$-subnormal in $$H$$ and $$H$$ is $$h$$-subnormal in $$G$$, so by fact (2), $$H \cap K$$ is $$(h + k)$$-subnormal in $$G$$.