# Center of pronormal subgroup is subset-conjugacy-determined in normalizer

From Groupprops

CONVENTION WARNING:This article/section uses the right-action convention. The left and right action conventions are equally powerful and statements/reasoning here can be converted to the alternative convention (the main reason being that every group is naturally isomorphic to its opposite group via the inverse map). For more on the action conventions and switching between them, refer to switching between the left and right action conventions.

## Contents

## Statement

Suppose is a group and is a Pronormal subgroup (?). Suppose are subsets of (the Center (?) of ) and is such that . Then, there exists such that, for all , .

(Here, denotes conjugation by , following the right-action convention. The statement and proof remain the same for the left-action convention).

## Definitions used

### Pronormal subgroup

`Further information: Pronormal subgroup`

A subgroup of a group is termed a **pronormal subgroup** if for any , there exists such that .

## Related facts

### Stronger facts

### Corollaries

- Center of Sylow subgroup is subset-conjugacy-determined in normalizer
- Abelian pronormal subgroup is subset-conjugacy-determined in normalizer
- Abelian and abnormal implies subset-conjugacy-closed

## Proof

**Given**: is a pronormal subgroup, for and .

**To prove**: There exists such that for all .

**Proof**: Since , we have . Now, since , we have , so . Also, , so .

Thus, . By pronormality, there exists such that . In particular, there exists such that .

Now consider . We check the two required conditions:

- : Clearly, . So, , and hence .
- for all : Pick . We have . Now, and , so fixes . Thus, .