# Baer norm not is hereditarily normal

This article gives the statement, and possibly proof, of the fact that for a group, the subgroup obtained by applying a given subgroup-defining function (i.e., Baer norm) doesnotalways satisfy a particular subgroup property (i.e., hereditarily normal subgroup)

View subgroup property satisfactions for subgroup-defining functions View subgroup property dissatisfactions for subgroup-defining functions

## Contents

## Statement

The Baer norm of a group need not be a hereditarily normal subgroup: not every subgroup of the Baer norm need be normal in the whole group.

## Definitions used

### Baer norm

`Further information: Baer norm`

The Baer norm of a group is defined as the intersection of the normalizers of all its subgroups.

## Related facts

- Baer norm is hereditarily permutable
- Baer norm is Dedekind: Every subgroup of the Baer norm is normal within the Baer norm.
- Normality is not transitive: Combining the fact that the Baer norm is normal, every subgroup of the Baer norm is normal in it, and not every subgroup of the Baer norm is normal in the whole group, gives a proof that normal subgroups of normal subgroups need not be normal.
- Permutable not implies normal: Combining the fact that every subgroup of the Baer norm is permutable in the whole group, but not every such subgroup is normal in the whole group, we get examples of permutable subgroups that are not normal.

## Facts used

## Proof

### Example of a group of prime power order

`Further information: prime-cube order group:p2byp`

Let be an odd prime. Let be the semidirect product of a cyclic group of order and a cyclic group of order acting nontrivially. In other words: is a group generated by , with the relations and :

Thus, has order .

Let . is a subgroup of order in . Moreover, is *not* normal in , because it is the acting group in a semidirect product for a nontrivial action.

First, note that (the subgroup generated by elements of order ) is a subgroup of exponent (fact (1)). It cannot equal , since . Also, , so has order at least . Hence, its order is exactly , and it is elementary Abelian on and .

Note that the normalizer of any subgroup of order or is the whole group. The normalizer of any subgroup of order must contain that subgroup, and also the center, which is generated by . Hence, either the subgroup is normal, or its normalizer is a subgroup of order containing two permuting subgroups of order , forcing it to be (again by fact (1)). Thus, every normalizer is either the whole group or . Moreover, for the subgroup , the normalizer must be precisely , since is not normal in . Thus, the Baer norm is precisely .

But the subgroup is not normal in . Thus, the Baer norm is not hereditarily normal.