# Subnormality is finite-relative-intersection-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., subnormal subgroup) satisfying a subgroup metaproperty (i.e., finite-relative-intersection-closed subgroup property)
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## Statement

### 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 $G$.

## Related facts

### 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. 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. 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 $G$.# Transitive and transfer condition implies finite-relative-intersection-closed
3. 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 $G$.