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Our goal is to understand the structure of subgroups of graphs of groups. This will enable us to prove things like: F \ast_{z} F is LERF.

Exercise 22. A group is coherent if every fg subgroup is fp. Prove that \pi_1 of a finite graph of groups with coherent vertex groups and cyclic edge group is coherent.

Recall, that if H \subseteq \pi_1 \mathcal{G}=G is fg then H has an induced graph of groups structure. \mathcal{H} =H \backslash T_H \subseteq H \backslash T where T is the Bass-Serre tree of \mathcal{G}.

Topologically, H defines a covering space X^H \rightarrow X_g, which inherits a graph of spaces structure with underlying graph H\backslash T.

\Gamma' =H \backslash T_H \subseteq H \backslash T is a finite connected subgraph and the image of \Gamma' in X^H is a connected subcomplex X' such that \pi_1 X' = \pi_1 X^H =H. X' has the structure of a graph of space with underlying graph \Gamma'

There is an explicit algebraic description of H\backslash T, which follows immediately from Lemmas 16,17,18.

Lemma 23. (a) The vertices of H\backslash T are in bijection with \coprod_{v \in V(\Gamma)} H \backslash G/G_v=\coprod_{v \in V(\Gamma)}\{HgG_v\mid g \in G\}.

The vertex HgG_v is labelled by: H \cap gG_vg^{-1}, well-defined up to conjugation in H.

(b) The edges of H\backslash T are in bijection with \coprod_{e \in E(\Gamma)} H \backslash G/G_e. The vertex HgG_e is labelled with H \cap g G_e g ^{-1}.

(c) The edges of H\backslash T adjoining the vertex corresponding to HgG_v  are in bijection with \coprod_{e~\mathrm{adjoining}~v} (g^{-1} H g \cap G_v ) \backslash G_v/G_e.

Now we will try to understand this topologically. In particular, we will understand the edge maps of a covering space of a graph of spaces.

Let f:X \rightarrow Y be a continuous map and Y' \rightarrow Y a covering map. If f':X \rightarrow Y' makes the diagram commutes, then f' is a lift of f.

Lemma 24. Fix basepoint x \in X, y=f(x) \in Y, and y' \in Y' mapping to y. There is a lift f':X \rightarrow Y' with f'(x)=y' if and only if f_{\ast} \pi_1(X,x) \subseteq \pi_1(Y',y').  Furthermore, if this lift exists it is unique.

It may be impossible to lift at y'. But it is possible if we pass to a covering space. Eg., if \tilde{X} \rightarrow X is the universal cover.  Intuitively, an elevation is a minimal lift.

Definition: Let X,Y,Y', x,y,y' be as above. A based connected covering space (X',x') \rightarrow (X,x) together with a based map f':(X',x')\to (Y',y') such that


commutes is an elevation (of f at y') if whenever the diagram


commutes and X' \rightarrow \bar{X} is a covering map of degree larger than 1 then the composition (\bar{X},\bar{x}) \rightarrow (X,x) \rightarrow (Y,y) does not lift to Y' at y'.

The unbased covering map X' \rightarrow X or equivalently the conjugacy class of the subgroup \pi_1(X') \subseteq \pi_1(X) is called the degree of the elevation f'.

Let \Gamma be a finitely generated group. It has a surjection F_s \rightarrow \Gamma for S finite. Elements of the kernel are called relations and the elements of S are called genereators. Suppose the kernel is generated as a normal subgroup by a subgroup K \subseteq F_S. Then we write \Gamma \cong  \langle S|R \rangle  to mean \Gamma \cong F_S/ \langle \langle R \rangle \rangle. If R is finite, the \Gamma is said to be finitely presentable.

Example: \mathbb{Z} \cong \langle b,c|bcb^{-1}c^{-1},b^2c^{-3} \rangle .

Let’s develop a topological point of view.

Let X^{(1)} be the standard rose for F_S. Each relation r \in R corresponds to a (homotopy class of a) map f_r:S^1 \rightarrow X^{(1)}. We construct a 2-complex X gluing on a 2-cell using the map f_r for each relation r \in R.

Lemma 6: \pi_1(X) \cong \Gamma
Proof: This is a simple application of the Seifert-van Kampen Theorem.

We call X a presentation complex for \Gamma, and we deduce that every finitely generated group is \pi_1 of a 2-complex.

Exercise 8: Every finitely presented group is \pi_1 of a closed 4-manifold.

Therefore, \Gamma acts freely and properly discontinuously on some 2-complex \tilde{X}, the universal cover of X.

Definition: Let S be a finite generating set for \Gamma as above. Then Cay_S(\Gamma)=\tilde{X}^{(1)} is the Cayley graph of \Gamma with respect to S.

To see that this only depends on S, let’s give the more standard definition. The vertices of Cay_S(\Gamma) are just the elements of \Gamma. For each generator g \in S, two vertices \gamma ,\delta are joined by an edge iff \gamma g=\delta

The group \Gamma acts by left translation.


  1. \mathbb{Z}\cong \langle 1\rangle.
  2. \mathbb{Z} \cong \langle 2,3 \rangle.
  3. The tree for F_S.

Definition: The Cayley graph Cay_S(\Gamma) induces a natural length metric on \Gamma, denoted d_S and called the word metric. Note that d_S(1,\gamma)=l_S(\gamma), the word length of \gamma.

The action of \Gamma on Cay_S(\gamma) is by isometries.

Given a metric space (X,d), a geodesic is just an isometric embedding of a compact interval  into X.

A metric space is geodesic if any pair of point is joined by a geodesic. Note that Cay_S(\Gamma) is a geodesic metric space.

Definition: Let \lambda ,\epsilon \geq 0.  A (\lambda,\epsilon)-quasi isometric embedding is a map of metric space f:X \rightarrow Y such that \lambda^{-1}d_X(x_1,x_2)-\epsilon \leq d_Y(f(x_1),f(x_2)) \leq \lambda d_X(x_1,x_2)+\epsilon.
If for some C, we also have that for every y \in Y, there is x \in X such that d_Y(f(x),y) \leq C, then f is a quasi-isometry.

Exercise 9: Quasi-isometry is an equivalent relation (use the Axiom of Choice).

Exercise 10: Let S and S' be two finite generating sets for \Gamma.  Then id:(\Gamma,d_S) \rightarrow (\Gamma,d_{S'}) is a (\lambda,0) quasi-isometry.

Example: (\mathbb{Z}^2,l^1) \hookrightarrow (\mathbb{R}^2,l^2) is a quasi-isometry.


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