Commensurator of subgroup is subgroup
Suppose is a subgroup of a group . Consider the commensurator of in , defined as the set of all such that is a subgroup of finite index in both and , i.e., and are Commensurable subgroups (?). Then, is a subgroup of .
Given: A group , a subgroup of . is the set of all such that has finite index in both and .
To prove: is a subgroup of .
|1||The identity element of is in||, so the intersection also equals . This has index in both and , which is finite.|
|2||If , then||Consider the map given by . This is an inner automorphism of , and hence preserves intersections and the index of subgroups. We have and . Thus, we get . Since the index of in both and is finite, so is the index of in both and .|
|3||If , then .||PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]|