# Difference between revisions of "Paranormal implies polynormal"

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., paranormal subgroup) must also satisfy the second subgroup property (i.e., polynormal subgroup)
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## Statement

Any paranormal subgroup of a group is also a polynormal subgroup.

## Definitions used

For these definitions, $H^g = g^{-1}Hg$ denotes the conjugate of $H$ by $g$, using the right-action convention (the action convention doesn't really matter). For subgroups $H,K \le G$, $H^K$ is the smallest subgroup of $G$ containing $H$ and invariant under the action of $K$ by conjugation.

### Paranormal subgroup

Further information: Paranormal subgroup

A subgroup $H$ of a group $G$ is termed paranormal in $G$ if $H$ is a contranormal subgroup of $\langle H, H^g \rangle$.

### Polynormal subgroup

Further information: Polynormal subgroup

A subgroup $H$ of a group $G$ is termed polynormal in $G$ if $H$ is a contranormal subgroup of $H^{\langle g \rangle}$.

## Facts used

1. Contranormality is upper join-closed

## Proof

Given: A paranormal subgroup $H$ of a group $G$.

To prove: For any $g \in G$, $H$ is contranormal in $H^{\langle g \rangle}$.

Proof: Clearly, $H^{\langle g \rangle}$ is generated by $H^k$ for all $k \in \langle g \rangle$, which in turn means that it is generated by the subgroups $\langle H, H^k \rangle$. $H$ is contranormal in each of these by definition of paranormality, so by fact (1), $H$ is contranormal in $H^{\langle g \rangle}$.