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)
View all group metaproperty satisfactions | View all group metaproperty dissatisfactions |Get help on looking up metaproperty (dis)satisfactions for group properties
Get more facts about residually finite group |Get facts that use property satisfaction of residually finite group | Get facts that use property satisfaction of residually finite group|Get more facts about subgroup-closed group property
Statement with symbols
Suppose is a residually finite group and is a subgroup of . Then, is also a residually finite group.
- Finiteness is subgroup-closed: Any subgroup of a finite group is finite.
- Residually operator preserves subgroup-closedness: If is a subgroup-closed group property, so is the property obtained by applying the residually operator to .
- Normality satisfies transfer condition
- Index satisfies transfer inequality
Proof from given facts
The proof follows directly by combining facts (1) and (2).
Given: A residually finite group , a subgroup of .
To prove: For any non-identity element , there exists a normal subgroup of of finite index not containing .
- There exists a normal subgroup of finite index in not containing : [SHOW MORE]
- Let . Then, is normal in , has finite index in , and does not contain : [SHOW MORE]
This completes the proof.