# Pronormality is not finite-upper join-closed

From Groupprops

This article gives the statement, and possibly proof, of a subgroup property (i.e., pronormal subgroup)notsatisfying a subgroup metaproperty (i.e., finite-upper join-closed subgroup property).This also implies that it doesnotsatisfy the subgroup metaproperty/metaproperties: Upper join-closed subgroup property (?), .

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## Contents

## Statement

We can have a subgroup such that is a pronormal subgroup in two intermediate subgroups and , but is *not* pronormal in the join .

## Related facts

### Related facts about pronormality

- Pronormality satisfies intermediate subgroup condition: If and is pronormal in , then is also pronormal in .

### Related facts about upper join-closedness

- 2-subnormality is not upper join-closed
- Subnormality is not upper join-closed
- Normality is upper join-closed
- Characteristicity is not upper join-closed

## Proof

### Example

`Further information: symmetric group:S3, symmetric group:S4`

Let be the symmetric group on the set , be the subgroup comprising permutations on , be the subgroup comprising permutations on , and be the two-element subgroup generated by . Observe that:

- is a pronormal subgroup in both and . In fact, it is a maximal subgroup in both, and maximal implies pronormal.
- .
- is not pronormal in . That's because the subgroup generated by is a conjugate of , and these two subgroups are not conjugate in the subgroup they generate.