Prove that every nonidentity element of a free group has infinite order.
We begin with a lemma.
Lemma: Let . Denote by the reduced length of . Then . Proof: We proceed by induction on . If has length 1, then for some letter . Then , and this word is reduced; hence . Suppose . There are two cases; if , then . If , where , then there are again two cases. If , then , a contradiction. If , then is reduced, and we again have . Suppose now that every word of length at most satisfies for some . Let be a word of length , where and are letters and is a word. Now . If , then . Since is a reduced word of length at least 1, . Now . If , then is reduced, and we have .
Let be a reduced word with . Then the length of is positive, so that by the lemma, . Moreover, the words have arbitrarily large reduced length. But if is finite, there exists a maximal reduced length among the reduced lengths of powers of .