Residual finiteness is subgroup-closed

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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]
  2. Let L = N \cap H. Then, L is normal in H, has finite index in H, and does not contain h: [SHOW MORE]

This completes the proof.