Finite index implies powering-invariant
This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., subgroup of finite index) must also satisfy the second subgroup property (i.e., powering-invariant subgroup)
View all subgroup property implications | View all subgroup property non-implications
Get more facts about subgroup of finite index|Get more facts about powering-invariant subgroup
- Poincare's theorem states that any subgroup of finite index contains a normal subgroup of finite index
- Normal of finite index implies quotient-powering-invariant
- Finite implies powering-invariant
- Powering-invariant over quotient-powering-invariant implies powering-invariant
Given: A group , a subgroup of finite index in .
To prove: is powering-invariant in .
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation|
|1||Let be the normal core of in . Then, is a normal subgroup of finite index in .||Fact (1)||has finite index in .||Fact-given direct.|
|2||is a quotient-powering-invariant of .||Fact (2)||Step (1)||Step-fact direct.|
|3||is a powering-invariant subgroup of .||Fact (3)||Step (1)||By Step (1), is finite, hence , as a subgroup of it, is finite. Thus, by Fact (3), it is powering-invariant in .|
|4||is powering-invariant in .||Fact (4)||Steps (2) and (3)||Step-combination direct.|