6
$\begingroup$

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 $\phi(g) \neq 1$?

$\endgroup$
12
  • 1
    $\begingroup$ @Shaun I am using the formal construction with words and concatenation. I do not remember in which paper I read it, but it was used to show that all free groups are residually finite (the residual finiteness of $F_2$ follows from it being embeddable in $GL_2(\mathbb{Z})$). $\endgroup$
    – frafour
    Dec 27, 2018 at 18:21
  • 1
    $\begingroup$ Related. $\endgroup$
    – Shaun
    Dec 27, 2018 at 19:08
  • 3
    $\begingroup$ Hint: Every finite rank free group embeds in $F_2$. $\endgroup$ Dec 28, 2018 at 1:06
  • 1
    $\begingroup$ One can prove more: Given a finite rank free group $F$ and a nontrivial element $g\in F$, there exists an epimorphism $\phi: F\to F_2$ such that $\phi(g)\ne 1$. But this is a more difficult result. $\endgroup$ Dec 28, 2018 at 17:31
  • 4
    $\begingroup$ If you modify the question I will write this as an answer: It requires a bit of an argument. $\endgroup$ Dec 28, 2018 at 21:06

2 Answers 2

5
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Are you surprised this result is hard to prove? Because I am. As you've shown, it's enough to get a surjection $F\rightarrow G$, where $1<rk(G)<rk(F)$, and $g$ isn't in the kernel. I tried doing it for $G$ a finite $p$-group, knowing I could play with which prime $p$, but couldn't make it work. $\endgroup$ Dec 30, 2018 at 3:18
5
$\begingroup$

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)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .