# Normality is upper join-closed

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

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## Contents

## Statement

### Statement with symbols

Suppose is a subgroup of , is a nonempty indexing set, and are subgroups of containing , such that (i.e., is a normal subgroup of ) for each . Then, is normal in the join of the s.

## Related facts

### Related facts about normality

- Join lemma for normal subgroup of subgroup with normal subgroup of whole group
- Normality is not UL-join-closed
- Normality is strongly join-closed
- Normality is strongly intersection-closed
- Normality is UL-intersection-closed
- Normality satisfies intermediate subgroup condition
- Normality satisfies transfer condition

### Related facts about upper join-closedness

The fact about normality generalizes to the following:

Left-inner right-monoidal implies upper join-closed: A subgroup property that has a function restriction expression with the left property being inner automorphisms and the right property being monoidal (closed under composition) is upper join-closed.

Other manifestations of the general fact include:

Here are some related properties that are *not* upper join-closed:

- Characteristicity is not upper join-closed
- Conjugacy-closedness is not upper join-closed
- Subnormality is not finite-upper join-closed, subnormality is not permuting upper join-closed
- 2-subnormality is not finite-upper join-closed, 2-subnormality is not permuting upper join-closed

### Analogues and breakdowns of analogues in other algebraic structures

- Ideal property is upper join-closed for Lie rings: If is a subring of a Lie ring such that is an ideal in two subrings , where , an indexing set, then is also an ideal in the Lie subring generated by all the s.
- Ideal property is not upper join-closed for alternating rings
- Normality is not upper join-closed for algebra loops

## Proof

**Given**: A group , a subgroup , a nonempty indexing set , and a collection of subgroups , such that is normal in for each .

**To prove**: is normal in the join of the s.

**Proof**: Let be the join of the s. For , we can write:

where for some index element . Thus, if denotes conjugation by , we have:

Now, since is normal in , each acts as an automorphism of . Thus, their composite, namely , is also an automorphism of . In other words, for every , showing that is normal in .