Reduced words form the free group on an alphabet

For a word $w$, read its letters from left to right while maintaining a reduced stack: append a new letter unless it is the formal inverse of the stack's last letter, in which case delete that last letter. Induction on the length of $w$ gives a reduced output $\mathrm{red}(w)$, with $\mathrm{red}(r)=r$ for every reduced word $r$.

The same induction shows that reading a neighbouring pair $a{a}^{-1}$ or $a^{-1}a$ has exactly the same net effect on every preceding stack as omitting that pair. Thus $\mathrm{red}$ is unchanged by an elementary cancellation or reverse insertion; hence $w$ is freely equivalent to $\mathrm{red}(w)$, and two freely equivalent reduced words are equal.

Let $F(X)$ be the set of reduced words. For reduced $r,s$, define $r\cdot s:=\mathrm{red}(rs)$, let the empty word be $e$, and let $r^{-1}$ be the reversal of $r$ with each letter formally inverted. Step 2.1 gives $\mathrm{red}(\mathrm{red}(rs)t)=\mathrm{red}(rst)=\mathrm{red}(r\mathrm{red}(st))$.

The equality in step 3.1 makes the product associative. The empty word is a two-sided identity, and $r{r}^{-1}$ and $r^{-1}r$ reduce by successive central cancellations to the empty word; therefore every reduced word has the stated two-sided inverse. Hence $F(X)$ is a group.

Send $x\in X$ to the one-letter word $x$. For $u:X\to G$, evaluate a word by replacing $x$ with $u(x)$ and $x^{-1}$ with $u(x{)}^{-1}$ and multiplying in order. Each elementary cancellation evaluates to an adjacent inverse pair, so evaluation is unchanged by step 2.1 and defines $\widehat{u}:F(X)\to G$; it extends $u$ and preserves the product by the definition in step 3.1.

Any homomorphism $h:F(X)\to G$ extending $u$ is forced, by writing each reduced word as its ordered product of one-letter words and their inverses, to agree with the evaluation map of step 4.2. Thus $\widehat{u}$ is unique.

Steps 4.1--5.1 establish the group and the extension-and-uniqueness property of [L4], so the reduced-word group is the free group on $X$.