Endo-invariance implies strongly join-closed

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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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Statement

Verbal statement

Any subgroup property that arises as an invariance property with respect to endomorphisms in the function restriction formalism is strongly join-closed, viz it is both join-closed and trivially true.

Symbolic statement

Let p be an endomorphism property. Let I be a (possibly empty) indexing set. Let H_i is a family of subgroups of G indexed by I. Assume that for every function f on G satisfying p, f(H_i)H_i (viz H_i satisfies the invariance property for p).

Then, if H denotes the join of (viz, subgroup generated by) all H_is, H also satisfies the invariance property for p. In other words, whenever f is a function on G satisfying p, f(H)H.

Examples

Definitions used

Invariance property

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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.

Facts used

  1. Homomorphisms commute with joins: This states that if \sigma:G \to K is a homomorphism, and H_i, i \in I is a collection of subgroups of G whose join is H, then \sigma(H) is the join of \sigma(H_i), i \in I.

Proof

Given: A property p of endomorphisms, a group G with subgroups H_i, i \in I whose join is a subgroup H of G. Further, each H_i is invariant under each endomorphism \sigma of G that satisfies p.

To prove: H is invariant under each endomorphism \sigma of G that satisfies property p.

Proof: We pick any endomorphism \sigma of G that satisfies p.

  1. For each i \in I, \sigma(H_i) is contained in H_i: This follows from the assumption on the H_is.
  2. \sigma(H) is the join of the \sigma(H_i)s: This follows from fact (1), with K = G.
  3. \sigma(H) is contained in H: Since each \sigma(H_i) is contained in H_i, it is in particular contained in H, hence their join, which is \sigma(H), is also contained in H.

This completes the proof.