# Pronormal implies weakly closed in intermediate nilpotent

## Statement

### Statement with symbols

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

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

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

## 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).

### Pronormal subgroup

Further information: Pronormal subgroup

A subgroup $H$ of a group $G$ is termed paranormal in $G$ if for any $g \in G$, there exists $x \in \langle H, H^g \rangle$ such that $H^x = H^g$.

### 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 $K$.

Now suppose $g \in G$ is such that $H^g \le K$. Then, $\langle H, H^g \rangle \le K$. By fact (3), $H$ is normal in $\langle H, H^g \rangle$. Thus, for any $x \in \langle H, H^g \rangle$, we have $H^x \le H$.

By the definition of pronormality, we also have $x \in \langle H, H^g \rangle$ such that $H^x = H^g \rangle$. Since $H^x \le H$, we get $H^g \le H$, completing the proof.