# $F_2$ is residually finite, but what are the trivially-intersecting subgroups?

A group $G$ is residually finite if for all $g\in G$ with $g\not=1$ there exists a normal subgroup of finite index, $N_g\lhd_f G$ such that $g\not\in N_g$. Note that $\cap_{g\in G} N_g=1$.

It is well-known that $F_2$ is residually finite. To prove this, simply recall that linear groups are residually finite, and $F_2$ is linear because the matrices, $$\left( \begin{array}{ccc} 1 & 2 \\ 0 & 1 \end{array} \right) \text{ & } \left( \begin{array}{cc} 1 & 0 \\ 2 & 1 \end{array} \right)$$ generate a free group. However, this proof is somewhat unsavoury. I would quite like to know what the subgroups $N_g$ are.

Does there exist a "nice" set of finite-index subgroups of $F_2$ which intersect trivially?

Nice is, of course, a subjective term. By "nice" I could mean "take the same form". However, I doubt this can happen (if they "take the same form" then presumably some rule dictates this form, and this rule is defined by a word, or a collection of words, and so the intersection of the subgroups is non-trivial). Alternatively, I could mean characteristic, which is nice in a different sense. I suppose if you can given a reason why I might think your set is nice that would be...nice.

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You can take congruence subgroups; that is, $\Gamma_2(n)=\{g\in F_2\ |\ g\equiv I\pmod{n}\}$, where $I$ is the 2x2 identity matrix and $n$ is any integer. Of course, I am talking about the linear rep. of $F_2$ here. – Steve D Sep 26 '11 at 12:58
I have just realised that the characteristic property always happens. As in, for a given finite index subgroup $H$ of $G$ one can find a finite index subgroup of $H$ which is characteristic in $G$. This holds because there are only ever a finite number of subgroups for a given finite index, and because the intersection of finitely many finite index subgroups is again of finite index. So that nice-ness can always be made to happen... – user1729 Sep 26 '11 at 15:23
Sorry for being ignorant, but what is $F_2$? A notation for $GL(2,F)$? Or $GL(2,\mathbb Z)$? – Henning Makholm Sep 26 '11 at 18:17
@Henning: the free group on two generators. – Chris Eagle Sep 26 '11 at 18:50

Here is one possibility, among many. Fix a prime number $p$. For any group $G$, define $\gamma_{1}^{p}(G) = G$ and, for $n\geq 1$, define $$\gamma_{n+1}^{p}(G) = \left(\gamma_{n}^{p}(G)\right)^{p}[G,\gamma_{n}^{p}(G)],$$ where $[A,B]$ denotes the subgroup generated by commutators of the form $[a,b]$, with $a\in A$ and $b\in B$. If $G$ is finitely generated, then $G/\gamma_{n}^{p}(G)$ is a finite $p$-group, for all $n$. In particular, for the free group $F_{2}$ of rank two, the groups $F_{2}/\gamma_{n}^{p}(F_{2})$ are all finite. Moreover, since free groups are residually $p$-finite, we have $\bigcap_{n\geq 1}\gamma_{n}^{p}(F_{2}) = 1$.