# Normalizer of pronormal implies abnormal

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., pronormal subgroup) must also satisfy the second subgroup property (i.e., subgroup with abnormal normalizer)
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## Statement

### Verbal statement

The Normalizer (?) of a pronormal subgroup of a group is an Abnormal subgroup (?).

### Statement with symbols

Suppose $H$ is a pronormal subgroup of $G$. Then, the normalizer $N_G(H)$ is an abnormal subgroup of $G$.

## Definitions used

### Pronormal subgroup

Further information: Pronormal subgroup

A subgroup $H$ of a group $G$ is termed pronormal if for any $g \in G$, there exists $x \in \langle H, H^g \rangle$ such that $H^x = H^g$. Here, $H^g := g^{-1}Hg$.

### Abnormal subgroup

Further information: Abnormal subgroup

A subgroup $H$ of a group $G$ is termed abnormal if for any $g \in G$, $g \in \langle H, H^g \rangle$.

## Related facts

### Corollaries

• Sylow-normalizer implies abnormal: The normalizer of a Sylow subgroup is an abnormal subgroup. In particular, it is self-normalizing, and any subgroup containing it is also self-normalizing.

## Proof

Given: Group $G$, pronormal subgroup $H$, $K = N_G(H)$.

To prove: For any $g \in G$, $g \in \langle K, K^g \rangle$.

Proof: By the definition of pronormality, there exists $x \in \langle H, H^g \rangle$ such that $H^g = H^x$. Thus, $H^{gx^{-1}} = H$, so $gx^{-1} \in N_G(H)$, hence $g \in N_G(H)x = Kx$. We know that $x \in \langle H, H^g \rangle \le \langle K, K^g \rangle$, so $g \in \langle K, K^g \rangle$.