# Subnormality is not finite-upper join-closed

From Groupprops

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

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## Statement

Suppose is a group, is a subgroup and are subgroups containing . Then, it can happen that is a subnormal subgroup of and of , but is *not* a subnormal subgroup of the join of subgroups .

## Related facts

- 2-subnormality is not upper join-closed: Note that the same example works.
- Normality is upper join-closed
- Subnormality is permuting upper join-closed in finite: This result is also known as the Maier-Wielandt theorem.
- Subnormality is not permuting upper join-closed

## Proof

### Example of the symmetric group

`Further information: symmetric group:S5`

Let be the symmetric group on the set . Let and be the dihedral groups given as follows:

Define . Then, is a two-element subgroup comprising and the identity permutation.

Observe that:

- is a 2-subnormal subgroup in both and . In particular, is subnormal in both and .
- The join of and is . This follows from some straightforward computation.
- is not a subnormal subgroup of . In fact, is a contranormal subgroup of : the normal closure of in is the whole of . This follows from the fact that transpositions generate the finitary symmetric group.