gr.group theory - Why are free groups residually finite?

Here is a direct proof for free groups.

Let $x_1,dots,x_m$ be the generators of our group. Consider a word $x_{i_n}^{e_n}dots x_{i_2}^{e_2}x_{i_1}^{e_1}$ where $e_iin{pm 1}$ and there are no cancellations (that is, $e_k=e_{k+1}$ if $i_k=i_{k+1}$).

I'm going to map this word to a nontrivial element of $S_{n+1}$, the group of permutations of $M:={1,dots,n+1}$. It suffices to construct permutations $f_1,dots,f_min S_{n+1}$ such that $f_{i_n}^{e_n}dots f_{i_2}^{e_2}f_{i_1}^{e_1}ne id_M$. For each $k=1,dots,n$, assign $f_{i_k}(k)=k+1$ if $e_k=1$, or $f_{i_k}(k+1)=k$ if $e_k=-1$. This gives us injective maps $f_1,dots,f_m$ defined on subsets of $M$. Assign yet unassigned values of $f_i$'s arbitrarily (the only requirement is that they are bijections). The resulting permutations satisfy $f_{i_n}^{e_n}dots f_{i_2}^{e_2}f_{i_1}^{e_1}(1)=n+1$.

Edit: As pointed out by Steve D in comments, this proof can be found in a book by Daniel E. Cohen, "Combinatorial group theory: a topological approach" (1989). The book can be found on the net if you are determined; the proof in on page 7 and in Proposition 5 on page 11.

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Edit [DZ]: I have a hard time reading multiple subscripts, so here is an example of Sergei Ivanov's construction.

Take the word $cca^{-1}bc^{-1}a$. This has length $6$, so we will find a homomorphism to $S_7$ whose image of this word is nontrivial because it sends $1$ to $7$. We'll choose values of permutations so that the $k$th suffix sends $1$ to $k+1$:

Suffix 1: $a$

$alpha=bigg(begin{array}{} 1&2 &3 &4& 5& 6& 7 \ 2&?&?&?&?&?&? end{array} bigg)$

Suffix 2: $c^{-1}a$

$gamma=bigg(begin{array}{} 1&2 &3 &4& 5& 6& 7 \ ?&?&2&?&?&?&? end{array} bigg)$

Suffix 3: $bc^{-1}a$

$beta=bigg(begin{array}{} 1&2 &3 &4& 5& 6& 7 \ ?&?&4&?&?&?&? end{array} bigg)$

Suffix 4: $a^{-1}bc^{-1}a$

$alpha=bigg(begin{array}{} 1&2 &3 &4& 5& 6& 7 \ 2&?&?&?&4&?&? end{array} bigg)$

Suffix 5: $ca^{-1}bc^{-1}a$

$gamma=bigg(begin{array}{} 1&2 &3 &4& 5& 6& 7 \ ?&?&2&?&6&?&? end{array} bigg)$

Suffix 6: $cca^{-1}bc^{-1}a$

$gamma=bigg(begin{array}{} 1&2 &3 &4& 5& 6& 7 \ ?&?&2&?&6&7&? end{array} bigg)$

These conditions on $alpha, beta,$ and $gamma$ don't conflict, they can be extended to permutations, and then $gammagammaalpha^{-1}betagamma^{-1}alpha(1) = 7$.