Let be a finite group and let .
- Prove that if then for some integer .
- Show conversely that if for some integer , then . [Hint: Show first that for any integer , so that . If has order , show that the elements are distinct for , so that and conclude that .]
- Let . By definition, we have , so that for some integer .
- We prove some lemmas.
Lemma 1: Let be a group and let . Then for all integers , . Proof: First we prove the conclusion for nonnegative by induction on . If , we have . Now suppose the conclusion holds for some ; then . By induction, the conclusion holds for all nonnegative . Now suppose ; then . Thus the conclusion holds for all integers .
Lemma 2: Let be a group and let such that for some integer . Then is a subgroup of . Proof: Let ; by Lemma 1 we have , so that . Thus . Now let . Then . By the Subgroup Criterion, then, .
Lemma 3: Let be a group and let such that for some integer and such that , . Then are distinct for . Proof: Choose distinct . By a previous exercise, . Suppose now that ; by cancellation we have , a contradiction. Thus the are distinct.
Now to the main result; suppose for some integer . Since has finite order, for some . By Lemma 2, , and by Lemma 3 we have . Since is finite, then, we have . Thus .