Let be a group and let such that and commute with . Prove that .
We begin with some lemmas.
Lemma 1: Let be group elements and a positive integer. Then . Proof: We proceed by induction on . If , the conclusion is trivial. Suppose the conclusion holds for some . Then , and the conclusion holds for all .
Lemma 2: Let and be as in the problem statement and let be a positive integer. For all , we have . Proof: We proceed by induction on . For the base case , the conclusion is trivial (using Lemma 1). For the inductive step, suppose the conclusion holds for some . Then, noting that , we have the following.
Thus the conclusion holds for all .
Now the main result follows from Lemma 2 with .