** Surface subgrops of graphs of free groups with cyclic edge groups **

**1. Gromov’s questions **

**Tits’ Ping-pong Lemma**. *Let be a non-elementary hyperbolic group. If do not commute, then some high powers of and generate a free subgroup.*

Hence there exist heaps of free subgroups in hyperbolic groups. What about one-ended subgroups? Known: they willcome out in at most finitely many conjugacy classes. Need to rule out free groups, of course.

**Question (Gromov)**. Does every one-ended hyperbolic group contain a surface subgroup ?

The most important development is Kahn-Markovic’s positive result for 3-manifold groups. In 2012, Calegari-Walker gave a positive answer for random groups.

**Question (Gromov)**. Does every one-ended hyperbolic group which is not a surface group, contain a finitely generated one-ended subgroup of infinite index?

**Question**. Does every one-ended finitely presented group contain either a surface group or a Baumslag-Solitar subgroup ?

**2. Free groups and relative questions **

Relative means relative to a given set of relators. I.e. study group pairs where is e finitely generated free group and a finite collection of words. A surface pair is where is a surface with boundary and represents its boundary components.

Note that a pair can be doubled into

Bestvina-Feighn: is hyperbolic iff is not a proper power.

Shentzer: is one-ended iff is not contained in a proper free factor. In this case, we say that the group pair is *irreducible*.

Theorem 1 (Wilton)Every irreducible group pair admits an essential admissible map of a surface with boundary .

*Essential* is a bit more restrictive than injective. *Admissible* means that along the boundary, the gluing map has the same degree on all components.

**3. Consequences **

** 3.1. Graphs of free groups **

Corollary 2For any one-ended graph of free groups with cyclic edge groups ,

- either containes a Baumslag-Solitar subgroup,
- or is hyperbolic and contains a surface subgroup.

There were earlier results with C. Gordon and Kim. In 2008, Calegari proved this under condition that . He reduced the problem to linear equations, and the homology assumptions provided a solution. In 2010, Kim and Oun solved the case of doubles of rank 2 free groups, where Calegari’s equation could be scrutinized.

** 3.2. Residually free groups **

A group is *fully residually free* if every finite subset is mapped injectively by some homomorphism to some free group.

Corollary 3If is a finitely generated non-free fully residually free group, then contains a surface subgroup and for any free group .

This follows from a structure theorem for such groups. This answers Remeslennikov’s question that arised in Alan Reid’s course.

** 3.3. Rigid groups **

Say that a group is *rigid* if it does not split over 1 or .

Theorem 4 (Louder-Touikan’s Strong Accessibility)If is a one-ended hyperbolic group, then cntains a quasi-convex subgroup such that

- either is rigid,
- or s a graph of free groups with cyclic edge-groups.

Hence Gromov’s question is reduced to the rigid case.

**4. Ideas in the proof **

- Study all essential admissible maps with irreducible. There is at least one, itslef.
- Look at
Argue that there is a map that maximizes .

- Show that this is a surface pair.

One can maybe implement the algorithm, but it does not seem to be efficient.

**5. Comment **

Arzhantseva: one cannot fix the genus of surfaces (1997).