Finite 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., finite subgroup) 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 finite subgroup|Get more facts about powering-invariant subgroup
- Periodic implies powering-invariant
- Finite index implies powering-invariant
- Finite and normal implies quotient-powering-invariant
- Normal of finite index implies quotient-powering-invariant
Given: A group , a finite subgroup , a prime number such that the map is bijective from to itself.
To prove: The map , restricted to , is bijective from to itself.
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation|
|1||The map sends to within itself.||is a subgroup.|
|2||The map (restricted to ) is injective from to itself.||The map is bijective from to itself.||Step (1)||The map is bijective on , hence injective on , hence its restriction to must also be injective.|
|3||the map (restricted to ) is injective from to itself.||is finite.||Step (2)||Since is finite, any injective map from to itself must be bijective.|