Join-closedness is left residual-preserved

Property-theoretic statement
The fact about::left residual operator for composition of an intersection-closed subgroup property by any subgroup property is again intersection-closed.

Similar facts about being left residual-preserved
Some similar results about being left residual-preserved:


 * Finite-join-closedness is left residual-preserved
 * Finite-intersection-closedness is left residual-preserved
 * Intersection-closedness is left residual-preserved
 * Normalizing join-closedness is left residual-preserved
 * Conjugate-join-closedness is left residual-preserved
 * Finite-conjugate-join-closedness is left residual-preserved

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


 * Upper join-closedness is right residual-preserved
 * Intermediate subgroup condition is right residual-preserved

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).