Endo-invariance implies strongly join-closed
This article gives the statement and possibly, proof, of an implication relation between two subgroup metaproperties. That is, it states that every subgroup satisfying the first subgroup metaproperty (i.e., Endo-invariance property (?)) must also satisfy the second subgroup metaproperty (i.e., Strongly join-closed subgroup property (?))
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Let be an endomorphism property. Let be a (possibly empty) indexing set. Let is a family of subgroups of indexed by . Assume that for every function on satisfying , ⊆ (viz satisfies the invariance property for ).
Then, if denotes the join of (viz, subgroup generated by) all s, also satisfies the invariance property for . In other words, whenever is a function on satisfying , ⊆ .
- A join of normal subgroups is normal. Here, the subgroup property of being normal is the property of being invariant under inner automorphisms, and is hence the invariance property for the property of being an inner automorphism.
- A join of characteristic subgroups is characteristic. Here the subgroup property of being characteristic is the invariance property with respect to the property of being an automorphism.
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Strongly join-closed subgroup property
A subgroup property is termed strongly join-closed if given any family of subgroups having the property, their join (viz the subgroup generated by them) also has the property. Note that just saying that a subgroup property is join-closed simply means that given any nonempty family of subgroups with the property, the join also has the property.
Thus, the property of being strongly intersection-closed is the conjunction of the properties of being intersection-closed and trivially true, viz satisfied by the trivial subgroup.
- Homomorphisms commute with joins: This states that if is a homomorphism, and is a collection of subgroups of whose join is , then is the join of .
Given: A property of endomorphisms, a group with subgroups whose join is a subgroup of . Further, each is invariant under each endomorphism of that satisfies .
To prove: is invariant under each endomorphism of that satisfies property .
Proof: We pick any endomorphism of that satisfies .
- For each , is contained in : This follows from the assumption on the s.
- is the join of the s: This follows from fact (1), with .
- is contained in : Since each is contained in , it is in particular contained in , hence their join, which is , is also contained in .
This completes the proof.