Residual finiteness is subgroup-closed

This article gives the statement, and possibly proof, of a group property (i.e., residually finite group) satisfying a group metaproperty (i.e., subgroup-closed group property)
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Statement

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. Finiteness is subgroup-closed: Any subgroup of a finite group is finite.
2. 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. Normality satisfies transfer condition
4. Index satisfies transfer inequality

Proof

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 $h$: [SHOW MORE]
   Since $H\leq G$, 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 $h$: [SHOW MORE]
   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.