Residual finiteness is subgroup-closed

Verbal statement
Any subgroup of a residually finite group is itself a residually finite group.

Statement with symbols
Suppose $$G$$ is a residually finite group and $$H$$ is a subgroup of $$G$$. Then, $$H$$ is also a residually finite group.

Facts used

 * 1) uses::Finiteness is subgroup-closed: Any subgroup of a finite group is finite.
 * 2) uses::Residually operator preserves subgroup-closedness: If $$\alpha$$ is a subgroup-closed group property, so is the property obtained by applying the residually operator to $$\alpha$$.
 * 3) uses::Normality satisfies transfer condition
 * 4) uses::Index satisfies transfer inequality

Proof from given facts
The proof follows directly by combining facts (1) and (2).

Hands-on proof
Given: A residually finite group $$G$$, a subgroup $$H$$ of $$G$$.

To prove: For any non-identity element $$h \in H$$, there exists a normal subgroup $$L$$ of $$H$$ of finite index not containing $$h$$.

Proof:


 * 1) There exists a normal subgroup $$N$$ of finite index in $$G$$ not containing $$h$$: Since $$H \le G$$, we in particular have $$h \in G$$ is a non-identity element. The existence of $$N$$ is now guaranteed by the assumption that $$G$$ is residually finite.
 * 2) Let $$L = N \cap H$$. Then, $$L$$ is normal in $$H$$, has finite index in $$H$$, and does not contain $$h$$: Normality of $$L$$ in $$H$$ follows from fact (3), and finite index follows from fact (4). $$h \notin L$$ follows from the fact that $$h \notin N$$.

This completes the proof.