# More on subgroups of R. Thompson group F

Together with Gili Golan we wrote another paper on $F$. Note that  Savchuk proved that for every number $\alpha$ in $(0, 1)$ the subgroup of all elements of $F$ that fix $\alpha$ is maximal (and its Schreier graph is amenable). He asked if $F$ has any other maximal subgroups of infinite index. That problem is soved (in the negative) in our paper. In fact we give two solutions.

First we  show that Jones’ subgroup $\vec F$ of $F$ is maximal in a certain subgroup of index $2$ of $F$ which itself is isomorphic to $F$. (Recall that $\vec F$ is isomorphic to the Thompson-Brown group $F_3$ and is 3-generated.) Moreover, we prove that there are exactly three subgroups of $F$ containing $\vec F$.  The preimage of $\vec F$ under the natural injective endomorphism of $F$ is then a maximal subgroup of $F$ and it is not difficult to show that this subgroup does not fix a point in the open unit interval. The fact that there are only three subgroups of $F$ containing Jones’ subgroup is very counter-intuitive and answers a question by Saul Schleimer.

The second solution gives many implicit examples (which may or may not be finitely generated). Note that the generator $x_0$ of $F$ does not fix any point in the open unit interval. Now take any element $g$ such $H=\langle x_0, g\rangle$ is a proper subgroup of $F$ and not inside any proper subgroup of finite index. Then any maximal subgroup  containing $H$ is an example we need. The fact that $H$ is not inside any proper subgroup of finite index is easy to establish. Indeed, every non-Abelian homomorphic image of $F$ is $F$ itself. So it is enough to show that the images of $x_0, g$ in $F/[F,F]$ generate the whole $F/[F,F]$. For example, one can take $g=x_1x_2x_1^{-1}$.

The non-trivial thing is to prove that $H$ is proper. Indeed, the only previously known (more precisely – previously published) way to show that a subgroup of $F$ is proper is to show that it is inside some proper subgroup of finite index of $F$ or fixes a point on the open unit interval. We employ a 2-dimensional analog of Stallings’ solution of the membership problem for subgroups of free groups. For every  subgroup of $F$ viewed as the diagram group over the Dunce hat, we construct what we called a Stallings 2-core which is a directed 2-complex with a distinguished 1-path $p$ and a map into the Dunce hat (which is also a directed 2-complex). In the case of the free group, the Stallings core is an automaton which accepts a reduced word if and only if it belongs to the subgroup. In the case of $F$, the Stallings 2-core is a 2-automaton which “accepts” reduced diagrams from $H$ (but not only these diagrams). In the case of $g=x_1x_2x_1^{-1}$ we construct the Stallings 2-core and show that $x_1$ is not “accepted” by it. So $x_1\not\in H$, and $H$ is indeed a proper subgroup of $F$. Although the method of Stallings 2-cores was never published before, we discussed it with Victor Guba long ago (in 1998 or even earlier). We wanted to write a paper about it but never had time to do that.

Note that the question of when the Stallings 2-core of a subgroup $H$ of a diagram group accepts only the diagrams from $H$ is very interesting because in that case, $H$ is a diagram group itself. This is how, together with Victor Guba, we represented the derived subgroup $[F,F]$ as a diagram group (see Theorem 26 in this paper).

If $U$ is a finite set of finite binary fractions, then it is easy to see that the subgroup $H_U$ of all elements of $F$ that fix each number from $U$ is just the direct product of $|U|+1$ copies of $F$. Savchuk studied these subgroups too, proving that their Schreier graphs are also amenable. We prove in our paper that for every $U$ there are only finitely many subgroups of $F$ containing $H_U$ (moreover the lattice of subgroups of $F$ containing $H_U$ is anti-isomorphic to the Boolean lattice of subsets of $U$). This implies that $F$ is quasi-residually finite (in our terminology), that is it has a decreasing sequence of finitely generated subgroups $F > P_1 > P_2 > ...$ such that the intersection of these subgroups is trivial and for each $i$ there are only finitely many subgroups of $F$ containing $P_i$.   That is also a counter-intuitive property of $F$.