Let be a set, let be a subgroup and let and be the sets of fixed and moved points as defined in this previous exercise. Prove that if then stabilizes and .
We begin with some lemmas.
Lemma 1: If is injective and is a collection of subsets of , then . Proof: If , then for some . Thus for all , hence . If , then for every there exists such that . Since is injective, we in fact have for all . Hence , so that .
Lemma 2: If is injective and , then . Proof: Let . Then for some with . If , then since is injective, we have , a contradiction. We thus have . Let . Now for some . If , then , a contradiction. Thus , so that .
If , then (using the lemmas to the previous exercise) we have
Thus fixes both and .