**1. Torsion in and applications **

Skipping train tracks and mapping tori.

** 1.1. Finite subgroups **

Theorem 1has finitely many classes of finite subgroups.

**question**: Does this holds in arbitrary finitely presented linear groups ?

** 1.2. Proof **

Prove a fixed point theorem.

Proposition 2 (Khramstov, Culler, Zimmermann independantly)Every finite subgroup fixes points in the spine .

Indeed, let be a finite subgroup, let be its inverse image in . It is virtually free, so a finite graph of finite groups (Stallings). acts cocompactly on the Bass-Serre tree . Mod out by the kernel, a copy of action. The obtained graph is a finite covering of the graph (of groups). So is a group of isometries of a graph of rank , whence a fixed point.

**Example**: The isometry group of a rose is the wreath product (it is the largest finite subgroup in ).

**Example**: The isometry group of a cage (aka multitheta) is .

Remark 1Let be a finite subgroup. acts on some finite graph of rank . If has a fixed point in , then lifts to .

Lifting is not always possible. For instance does not lift. Here is an other example.

**Example**: Consider simplicial -cycle with cycles glued at vertices. Let rotate the picture. Then the inverse image of in is torsion free.

** 1.3. Consequences **

- Natural map does not split.
- Neither does

This implies

Theorem 3Let be the standard -module. Then . However .

** 1.4. Trying to construct homomorphisms **

Start with a rose. Take a finite Galois covering . Assume the corresponding subgroup is characteristic (i.e. -invariant). Individual elements of , acting on the base rose, lift to . But this need not provide a lift of the action in general. We look for special cases where it does.

Restrict attention to abelian covers, with groups or .

Theorem 4The action of on the rose lifts to the abelian covering with Galois group iff with .

**Example**: If is even, is odd, so coprime to , so we get a homomorphism for . These examples were discovered earlier by Bogopolski and Puga, with a more algebraic argument, based on presentations.

**Proof**: The exact sequence fits in a commutative diagram with the corresponding sequence for groups of homotopy equivalences of graphs. Playing with finite subgroups, we know the former does not split, so the latter does not either.

Theorem 5 (Kielak)A homomorphism with infinite image can exist only if .

**2. **

Definition 6

Gersten’s presentation shows that is generated by .

Theorem 7 (Magnus)The subgroup generated by the and the is normal. It follows that is finitely generated.

Not very geometric. Unknown wether is finitely presented.

** 2.1. Maps from higher rank lattices **

In some sense, contains the non-linear part of .

Theorem 8 (Bridson, Wade)Let be an irreducible lattive in a higher rank semisimple Lie group. Then every homomorphism has finite image.

Theorem holds under a much weaker assumption on : say is -averse if no finite index subgroup

** 2.2. Proof **

**Case 1**: Assume image contains an iwip. Then apply

Theorem 9 (Dahmani, Guirardel, Osin)If is iwip, some power of it normally generates a free group.

Let . It is a free infinite index normal subgroup of . This contradicts Margulis’ normal subgroup theorem.

**Case 2**: Assume image contains no iwip. Then apply

Theorem 10 (Handel, Mosher)If subgroup does not contain iwip’s, then, up to finite index, leave invariant some free factor up to conjugacy. In particular, ‘s action on homology is block triangular.

Then use induction on . Prove that every finite index subgroup acts on via a finite group.

**Step 3**: Reduce to the case when maps into . For this use Lie algebra methods à la Magnus.

**Case 4**: Assume image . has a filtration by ‘s such that are free abelian and is trivial. maps trivially to all these quotients. The ‘s arise from the lower central series of . Since is residually nilpotent (Magnus), .

Proposition 11 (Andreadakis)is free abelian.

The corresponding homomorphisms to ‘s are unfortunately called Johnson homomorphisms nowadays.