Residually operator preserves subgroup-closedness

Statement

Suppose $\alpha$ is a Subgroup-closed group property (?). Suppose $\beta$ is the group property obtained by applying the 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. Second isomorphism theorem (note that this contains and includes the statement that 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 $\alpha$: [SHOW MORE]
   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 $G/N$: [SHOW MORE]
   This follows from fact (1) (the second isomorphism theorem) and the fact that $N$ is normal in $G$.
3. $H/L$ satisfies property $\alpha$: [SHOW MORE]
   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 $\alpha$.