Let be a group and let . Prove that . Deduce that for all .
First we prove a technical lemma: for all and , . The statement is clear for . We prove the case by induction; the base case is clear. Now suppose for some ; then . By induction the statement holds for all positive . Now suppose ; we have . Hence, the statement holds for all integers .
Now to the main result.
Suppose first that is infinity and that for some positive integer . Then we have , and multiplying on the left by and on the right by gives us that , a contradiction. Thus if is infinity, so is . Similarly, if is infinite and , we have , a contradiction. Hence if is infinte, so is .
Suppose now that and for some positive integers and . We have , so that , and , so that and . Thus .
Now let and be arbitrary group elements. Letting and , we see that .