Residually operator preserves subgroup-closedness

Statement
Suppose $$\alpha$$ is a fact about::subgroup-closed group property. Suppose $$\beta$$ is the group property obtained by applying the fact about::residually operator to $$\alpha$$, i.e., a group $$G$$ satisfies $$\beta$$ if every non-identity element is outside some normal subgroup for which the quotient group has property $$\alpha$$. Then, $$\beta$$ is also a subgroup-closed group property.

Facts used

 * 1) uses::Second isomorphism theorem (note that this contains and includes the statement that uses::normality satisfies transfer condition)

Proof
Given: A subgroup-closed group property $$\alpha$$. A group $$G$$ with the property that for every non-identity element $$g \in G$$, there exists a normal subgroup $$N$$ of $$G$$ not containing $$g$$ such that $$G/N$$ satisfies $$\alpha$$. A subgroup $$H$$ of $$G$$.

To prove: For any non-identity element $$h \in H$$, there exists a normal subgroup $$L$$ of $$H$$ such that $$h \notin L$$ and $$H/L$$ satisfies property $$\alpha$$.

Proof:


 * 1) There exists a normal subgroup $$N$$ of $$G$$ such that $$h \notin N$$ and $$G/N$$ satisfies $$\alpha$$: Since $$h \in H$$, we have $$h \in G$$, and we can apply the given condition on $$G$$ (namely, that $$G$$ is residually $$\alpha$$).
 * 2) $$L = N \cap H$$ is normal in $$H$$ and $$H/L$$ is isomorphic to a subgroup $$HN/N$$ of $$G/N$$: This follows from fact (1) (the second isomorphism theorem) and the fact that $$N$$ is normal in $$G$$.
 * 3) $$H/L$$ satisfies property $$\alpha$$: From the previous step, $$H/L$$ is isomorphic to a subgroup of $$G/N$$, which satisfies property $$\alpha$$. Since $$\alpha$$ is subgroup-closed, this forces $$H/L$$ to also satisfy $$\alpha$$.