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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Get more facts about normal subgroup |Get facts that use property satisfaction of normal subgroup | Get facts that use property satisfaction of normal subgroup|Get more facts about upper join-closed subgroup property
Statement with symbols
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
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 .