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

Statement

Suppose $p$ is a Transitive subgroup property (?) satisfying the Transfer condition (?). Then, $p$ is a Finite-relative-intersection-closed subgroup property (?).

Definitions used

Transitive subgroup property

Further information: Transitive subgroup property

A subgroup property $p$ is termed transitive if whenever $L\leq H\leq G$ 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

Further information: Transfer condition

A subgroup property $p$ is said to satisfy the transfer condition if whenever $H,K\leq G$ 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

Further information: Finite-relative-intersection-closed subgroup property

A subgroup property $p$ is termed finite-relative-intersection-closed if whenever $H,K\leq G$ 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 $G$.