# Subnormality is permuting upper join-closed in finite

From Groupprops

This article gives the statement, and possibly proof, of a subgroup property (i.e., subnormal subgroup of finite group) satisfying a subgroup metaproperty (i.e., permuting upper join-closed subgroup property)

View all subgroup metaproperty satisfactions | View all subgroup metaproperty dissatisfactions |Get help on looking up metaproperty (dis)satisfactions for subgroup properties

Get more facts about subnormal subgroup of finite group |Get facts that use property satisfaction of subnormal subgroup of finite group | Get facts that use property satisfaction of subnormal subgroup of finite group|Get more facts about permuting upper join-closed subgroup property

## Contents

## History

The result was proved both by Maier and by Wielandt.

## Statement

### When the whole group is finite

Suppose is a finite group and is a subgroup of . Suppose are intermediate subgroups of such that (i.e., they are Permuting subgroups (?)) and is a subnormal subgroup in both and . Then, is also subnormal in the product of subgroups .

### Equivalent formulation when the whole group is not finite

Suppose is a group and is a finite subgroup of . Suppose are intermediate finite subgroups of such that (i.e., they are Permuting subgroups (?)) and is a subnormal subgroup in both and . Then, is also subnormal in the product of subgroups .

Note that these two formulations are equivalent because even if is not finite, the product is still finite since both and are finite.

## Related facts

- Subnormality is not permuting upper join-closed
- Subnormality is not finite-upper join-closed (we can get counterexamples even in finite groups)