Join-closedness is left residual-preserved

This article gives the statement, and possibly proof, of a subgroup metaproperty (i.e., Intersection-closed subgroup property (?)) satisfying a subgroup metametaproperty (i.e., Left residual-preserved subgroup metaproperty (?))
View all subgroup metametaproperty satisfactions $|$ View all subgroup metametaproperty dissatisfactions

Statement

Property-theoretic statement

The Left residual operator for composition (?) of an intersection-closed subgroup property by any subgroup property is again intersection-closed.

Related facts

Similar facts about being left residual-preserved

Some similar results about being left residual-preserved:

Similar facts about being right residual-preserved

Any fixed-subgroup-expressible subgroup metaproperty is right residual-preserved. Instances of this are:

Proof

Hands-on proof

Given: A join-closed subgroup property $p$ , a subgroup property $q$ . Let $r$ be the left residual of $p$ by $q$ .

To prove: $r$ is join-closed.

Proof: Suppose $r$ is not join-closed. Then, there exists a group $G$ with a nonempty collection of subgroups $H_{i},i\in I$ such that each $H_{i}$ satisfies property $r$ but the join of the $H_{i}$ s does not satisfy property $r$ .

Let $H$ be the join of the $H_{i}$ s. Then, by the definition of left residual, there exists a group $K$ containing $G$ such that $G$ has property $q$ in $K$ and $H$ does not have property $p$ in $K$ .

Now, since each of the $H_{i}$ has property $r$ in $G$ , and $G$ has property $q$ in $K$ , we obtain that the $H_{i}$ all have property $p$ in $K$ . Thus, we have a collection of subgroups of $K$ that have property $p$ in $K$ but whose join does not have property $p$ in $K$ .

Property-theoretic proof

This proof directly follows from facts (1) and (2).