Endo-invariance implies strongly join-closed

From Groupprops
Jump to: navigation, search
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 (?))
View all subgroup metaproperty implications | View all subgroup metaproperty non-implications


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.


Definitions used

Invariance property


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.


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.