Free implies residually finite

Statement
Any free group is a residually finite group, i.e., for every non-identity element of a free group, there is a normal subgroup of finite index in the whole group not containing that element.

Related facts

 * Free implies residually nilpotent
 * Free implies residually solvable
 * Free abelian implies residually finite
 * Finitely generated abelian implies residually finite

Proof idea
The idea is to use the fact that finite groups are big enough to accommodate a particular word evaluating to a non-identity element.

Proof details
Given: A free group $$F$$ with freely generating set $$T$$. A non-identity element $$a \in F$$.

To prove: There exists a normal subgroup $$N$$ of $$F$$ such that $$F/N$$ is a finite group and $$a \notin N$$.

Proof: We write:

$$a = a_na_{n-1} \dots a_2a_1$$

as a reduced form expression for $$a$$ in terms of $$T$$. Thus, for each $$a_i$$, either $$a_i \in T$$ and $$a_i^{-1} \in T$$. We now define a function $$f:T \to S_{n+1}$$ where $$S_{n+1}$$ is the symmetric group on the set $$\{1,2,3,\dots,n+1\}$$:


 * $$f(t)$$ is the identity element if $$t$$ is not equal to any of the $$a_i$$s or their inverses.
 * Suppose $$A$$ is the set of $$i$$s such that $$a_i = t$$ and $$B$$ is the set of $$j$$s such that $$a_j^{-1} = t$$. Then, set $$f(t)$$ as any permutation $$\sigma$$ that sends each $$i \in A$$ to $$i + 1$$, and for each $$j \in B$$, sends $$j + 1$$ to $$j$$. This is well-defined since an element and its inverse cannot occur adjacently in the reduced form expression for a word.

We have thus obtained a function $$f:T \to S_{n+1}$$. This extends uniquely to a homomorphism from $$F$$ to $$S_{n + 1}$$, because $$F$$ is free. Moreover, under this homomorphism, we see that the image of $$a$$ sends $$1$$ to $$n + 1$$, and is not the identity element. The kernel of this homomorphism is thus a normal subgroup of $$F$$ with finite quotient group (a subgroup of $$S_{n+1}$$) and such that the kernel does not contain $$a$$.