In Aluffi's Chapter 0 there is an exercise to show that the product in the concrete construction of a free group is associative by showing that the reduction of a word is independent of the order in which cancellations are performed. I wrote a proof, and am wondering what the "standard" proof is for this fact.
I proceed by contradiction:
Fix a word $W \in F(A) $ the free group on an alphabet $A$. Suppose that the order of iterative cancellations matters. Then there exists algorithms $r_1(W,-), r_2(W,-)$, where the second argument denotes the step that the algorithm is on, such that:
- The reductions yield different words: $r_1(W,|W|) \neq r_2(W,|W|)$
- The point of final divergence is maximal: $$\forall r_i, r_j, \, \arg\min_k\{l = 1,2,\dots: r_i(w,k+l)\neq r_j(w,k+l)\} \leq \\ \hat{k} \equiv \arg\min_k\{l = 1,2,\dots: r_i(w,k+l)\neq r_j(w,k+l)\}$$
Then $r_1(w,\hat{k}) = r_2(w,\hat{k}) = w_1 a_i^{\pm 1}a_i^{\mp 1}w_2a_j^{\pm 1}a_j^{\mp 1}w_3$ for some words $w_1,w_2,w_3 \in F(A)$ and letters $a_i,a_j \in A$. But then we can construct $\hat{r_1},\hat{r_2}$ that agree with $r_1,r_2$ for $k \leq \hat{k}+1$, then remove the other pair of letters to yield $\hat{r_1}(w,\hat{k}) = \hat{r_2}(w,\hat{k}) = w_1 w_2w_3$. This contradicts that $\hat{k}$ is the maximal point of final divergence, so the reduced word is independent of the order of cancellations.