# Weakly normal implies weakly closed in intermediate nilpotent

## Statement

### Statement with symbols

Suppose $H \le K \le G$ are groups such that:

• $H$ is a Weakly normal subgroup (?) of $G$.
• $K$ is a Nilpotent group (?).

Then, $H$ is a Weakly closed subgroup (?) of $K$.

## Definitions used

For these definitions, $H^g = g^{-1}Hg$ denotes the conjugate subgroup by $g \in G$. (This is the right-action convention; however, adopting a left-action convention does not alter any of the proof details).

### Paranormal subgroup

Further information: Weakly normal subgroup

A subgroup $H$ of a group $G$ is termed paranormal in $G$ if for any $g \in G$, $H^g \le N_G(H)$ implies $H^g \le H$. In other words, it is weakly closed in its normalizer.

### Weakly closed subgroup

Further information: Weakly closed subgroup

Suppose $H \le K \le G$ are groups. We say $H$ is weakly closed in $K$ with respect to $G$ if, for any $g \in G$ such that $H^g \le K$, we have $H^g \le H$.

## Proof

Given: $H \le K \le G$ with $H$ a paranormal subgroup of $G$ and $K$ a nilpotent group.

To prove: For any $g \in G$ such that $H^g \le K$, we have $H^g \le H$.

Proof: By fact (2), $H$ is a subnormal subgroup of $K$. By fact (1), $H$ is therefore normal in $K$. Thus, $K \le N_G(H)$.

Since $H$ is weakly closed in $N_G(H)$ by definition, whenever $H^g \le N_G(H)$, we get $H^g \le H$. In particular, whenever $H^g \le K$, we get $H^g \le H$, so $H$ is weakly closed in $K$.