Why is it that every nontrivial word in a free group (it's easy to reduce to the case of, say, two generators) has a nontrivial image in some finite group? Equivalently, why is the natural map from a group to its profinite completion injective if the group is free?


Apparently, this follows from a result of Malcev's that finitely generated matrix groups over an arbitrary commutative ring are residually finite, but is there a more easily accessible proof if we only want the result for free groups?



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 $e\_i\in\{\pm 1\}$ 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 $f\_1,\dots,f\_m\in S\_{n+1}$ 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 $\gamma\gamma\alpha^{-1}\beta\gamma^{-1}\alpha(1) = 7$.


