## 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 $w$.