Let be a group acting on a set . Show that the following sets are subgroups of .
- The set (Called the kernel of the action
- For a fixed , the set (Called the stabilizer of )
By a previous exercise, for each of these sets we need to show that the identity belongs to the set and that each is closed under multiplication and inversion.
- Note that since for all . Now suppose , and let . Then , so that . Now let and ; then , so that . Thus is a subgroup of .
- We have since . Now suppose ; then we have , so that . Now let ; we have , so that . Thus is a subgroup of .