# 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 $H_i$s, $H$ 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

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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 $H_i$s.
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.