Free groups are residually of rank 2

Question

Let $F_X$ denote the free group on the set $X$, and $F_n$ the free group of rank $n$.

I have read that any free group is residually $F_2$, and I was trying to understand this.

For any free group $F$, it is residually free of finite rank: given $g \in F$ with $g \neq 1$, let $S$ be a set of generators of $F$ that we can use to express $g$, then $g$ is not sent to the identity under the projection $F \to F_S$.

So the question reduces to why $F_n$ is residually $F_2$, for $n \geq 2$.

Edit: I realized that this is trivial from the embedding of $F_n$ in $F_2$, for any $n \geq 1$. So the interesting question is whether the morphism $F_n \to F_2$ can be chosen to be surjective. More explicitly: let $n \geq 2$, and $1 \neq g \in F_n$. Is there an epimorphism $\phi : F_n \to F_2$ such that $\phi(g) \neq 1$?

Answer 1

The argument I had in mind uses the "super-strong approximation property".

Every finitely generated free group $F$ embeds in $SL(2, {\mathbb Z})$ which is a 2-generated group. I will need two facts about $SL(2, {\mathbb Z})$, one of which is elementary and the other is hard:

a. For every finite subset $A\subset SL(2, {\mathbb Z})$, for all but finitely many primes $p$, the projection of $A$ to $SL(2, {\mathbb Z}/p {\mathbb Z})$ is 1-1.

b. If $B\subset SL(2, {\mathbb Z})$ consists of matrices generating a free subgroup $G$ of rank 2, then for all but finitely many primes $p$, the projection of $B$ to $SL(2, {\mathbb Z}/p {\mathbb Z})$ generates $SL(2, {\mathbb Z}/p {\mathbb Z})$. This is a nontrivial fact, see

R. Matthews, L. N. Vaserstein, B. Weisfeiler, "Congruence Properties of Zariski‐Dense Subgroups, I", Proc. London Math. Soc., Series 3, 48 (3) 1984, p. 514-532.

(They proved a much more general result, I am using only a special case needed here.)

I will take $G$ to be a rank 2 free factor of the free subgroup $F< SL(2, {\mathbb Z})$ and let $A\subset F$ be an arbitrary finite subset. By taking a suitable prime $p$, we get that:

i) The restriction of the projection $$ \phi: SL(2, {\mathbb Z})\to SL(2, {\mathbb Z}/p {\mathbb Z}) $$ to $A$ is 1-to-1.

ii) $\phi(G)= SL(2, {\mathbb Z}/p {\mathbb Z})$.

Since $SL(2, {\mathbb Z}/p {\mathbb Z})$ is a 2-generated group, there exists an epimorphism $$ \eta: F_2\to SL(2, {\mathbb Z}/p {\mathbb Z}). $$ Since the group $F$ is free and $G$ is its free factor, there is a lift of the homomorphism $\phi$ to a homomorphism $$ \psi: F\to F_2, \phi=\eta\circ \psi, $$ such that $\psi(G)=F_2$; hence, $\psi$ is an epimorphism. At the same time, the restriction of $\psi$ to $A$ is 1-to-1 since $\phi$ already has this property. Thus, we proved:

Theorem. For every free group $F$ of finite rank and for every finite subset $A\subset F$, there exists an epimorphism $$ \psi: F\to F_2 $$ whose restriction to $A$ is 1-to-1.

Answer 2

Yes, it's true: for every finite subset $S$ of a free group $F$, there exists a quotient $F'$ of $F$, such that $F'$ is free of rank 2, and such that $S$ is mapped injectively into $F$.

See for instance (d) p11 in Champetier, Guirardel, Limit groups as limits of free groups: compactifying the set of free groups. (arxiv link)