Every nonidentity element in a free group has infinite order

Prove that every nonidentity element of a free group has infinite order.

We begin with a lemma.

Lemma: Let w \in F(S). Denote by \mathsf{len}(w) the reduced length of w. Then \mathsf{len}(w^2) > \mathsf{len}(w). Proof: We proceed by induction on \mathsf{len}(w). If w has length 1, then w = a for some letter a. Then w^2 = a^2, and this word is reduced; hence \mathsf{len}(w^2) = 2 > 1 = \mathsf{len}(w). Suppose \mathsf{len}(w) = 2. There are two cases; if w = a^2, then \mathsf{len}(w^2) = 4 > 2 = \mathsf{len}(w). If w = ab, where b \neq a, then there are again two cases. If b = a^{-1}, then w = 1, a contradiction. If b \neq a^{-1}, then w^2 = abab is reduced, and we again have \mathsf{len}(w^2) > \mathsf{len}(w). Suppose now that every word of length at most k satisfies \mathsf{len}(w^2) > \mathsf{len}(w) for some k \geq 2. Let w = aub be a word of length k+1, where a and b are letters and u is a word. Now w^2 = aubaub. If b = a^{-1}, then w^2 = au^2b. Since u is a reduced word of length at least 1, \mathsf{len}(u^2) > \mathsf{len}(u). Now \mathsf{len}(w^2) \geq \mathsf{len}(u^2) + 2 > \mathsf{len}(u) + 2 = \mathsf{len}(w). If b \neq a^{-1}, then w^2 = aubaub is reduced, and we have \mathsf{len}(w^2) = 2 \mathsf{len}(w) > \mathsf{len}(w). \square

Let w \in F(S) be a reduced word with w \neq 1. Then the length of w is positive, so that by the lemma, \mathsf{len}(w^2) > \mathsf{len}(w). Moreover, the words w^{2^k} have arbitrarily large reduced length. But if |w| is finite, there exists a maximal reduced length among the reduced lengths of powers of w.

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