# Transitive normality is not finite-join-closed

From Groupprops

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

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## Contents

## Statement

### Verbal statement

The join of two transitively normal subgroups of a group need not be transitively normal.

### Statement with symbols

Suppose is a group and are two transitively normal subgroups of . Then, the join , which in this case equals the product , need not be a transitively normal subgroup.

## Related facts

### Related metaproperty dissatisfactions for transitively normal subgroups

- Transitive normality is not finite-intersection-closed
- Transitive normality is not centralizer-closed

### Related properties that are join-closed

- Normality is strongly join-closed: In particular, this implies that any join of transitively normal subgroups is a normal subgroup, even though it need not be transitively normal.

## Proof

`Further information: symmetric group:S3`

Let be the cyclic group of order three. Let be the symmetric group of degree three, and be the subgroup of order three in . Define:

.

Then, we have:

- and are both normal subgroups of prime order. In particular, the only normal subgroups they have are the whole group and the trivial subgroup, both of which are normal in . Thus, and are both transitively normal.
- is not transitively normal: . is elementary abelian of order nine. Pick an isomorphism and consider the subgroup of given by . Then, is normal in , since is abelian. However, is not normal in , because conjugation by an element in sends to , which is not in . Thus, is not transitively normal.

Note that in this example, is the center of and is also a direct factor of . This shows that taking the join of a transitively normal subgroup with the center or with a direct factor does not guarantee a transitively normal subgroup.