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In this lecture, we will use the Normal Form Theorem to understand and construct groups.  In particular, we will construct a group that is not residually finite (RF).  Note that in the case when \mathcal{G} is an HNN-extension, then the Normal Formal Theorem is called Britton’s Lemma.

Example: The (p,q)-Baumslag-Solitar group has the following presentation:


For example \mathrm{BS} (1,1) \cong \mathbb{Z}^{2}.

Consider \mathrm{BS}(2,3).  We will show that this group is non-Hopfian (i.e., that there exists a homomorphism \varphi: \mathrm{BS}(2,3) \to \mathrm{BS}(2,3) that is a surjection with nontrivial kernel), hence is not RF.  Consider \psi: \mathrm{BS}(2,3) \to \mathrm{BS}(2,3) defined by \psi(t) = t and \psi(a) = a^{2}.  First, we show this is a homomorphism, then a surjection, then that it has nontrivial kernel.

Homomorphism:  \psi(ta^{2}t^{-1}) = ta^{4}t^{-1} = ta^{2}t^{-1}ta^{2}t^{-1} = a^{3}a^{3} = a^{6} = \psi(a^{3}).  Hence, the map is a homomorphism.

Epimorphism: To see that \psi is a surjection, we check that a, t \in \mathrm{Im}(\psi).  This is obvious for t.  Further a^{2} = \psi(a) \in \mathrm{Im}(\psi) and so ta^{2}t^{-1} = a^{3} \in \mathrm{Im}(\psi).  But then a = a^{3}a^{-2} \in \mathrm{Im}(\psi).

Nontrivial kernel: Consider c = ata^{-1}t^{-1}a^{-2}tat^{-1}a.  Then \psi(c) = a^{2}ta^{-2}t^{-1}a^{-4}ta^{2}t^{-1}a^{2} = a^{2}a^{-3}a^{-4}a^{3}a^{2} = 1.  So we just need to show that c is nontrivial.  We use Britton’s Lemma (Normal Form Theorem).  What Britton’s Lemma says is that there will be t's that can be removed from our word using the relation if it is trivial.  But this can’t be done with c, so we must have c \neq 1 as required.

Remark: \mathrm{BS}(2,3) can never embed in a word-hyperbolic group by a previous Lemma.

Exercise 20: Show that \mathrm{BS}(1,2) \cong \mathbb{Q}_{2} \rtimes_{2} \mathbb{Z}, where \mathbb{Q}_{2} is the dyadic rationals and we have \mathbb{Z} acting via multiplication by 2.  One can deduce that \mathrm{BS}(1,2) is linear, hence RF by Selberg’s Lemma.

Theorem 15 (Higman-Neumann-Neumann): Every countable group embeds in a 3-generator group.

Proof: Let G = \{ 1 = g_{0}, g_{1}, \ldots\}.  Consider G \ast \mathbb{Z} = G \ast \langle s \rangle.  This has some nice free subgroups, unlike G:  Let s_{n} = g_{n}s^{n}.  Let \Sigma_{1} = \{ s_{n} \vert n \geq 1\}.  By the Normal Form Theorem, \langle \Sigma_{1} \rangle \cong F_{\Sigma_{1}} (basically since there will always be some g’s between the s’s whenever you multiply any two elements together).  Let \Sigma_{2} = \{ s_{n} \vert n \geq 2\}.  Again, \langle \Sigma_{2} \rangle \cong F_{\Sigma_{2}}.  Since $latex\Sigma_{1}$ and \Sigma_{2} are both countable, F_{\Sigma_{1}} \cong F_{\Sigma_{2}}.

Consider the HNN-extension $\Gamma = (G \ast \langle s \rangle)\ast_{F_{\Sigma_{1}} \tilde F_{\Sigma_{2}}}$ and let t be the stable letter.  For every n \geq 1, ts_{n}t^{-1} = s_{n+1}.  But \Gamma is generated by g_{1}, s, and t: s_{n+1} = ts_{n}t^{-1}, so g_{n+1}s^{n+1} = tg_{n}s^{n}t^{-1} or g_{n+1} = tg_{n}s^{n}t^{-1}s^{-n-1}.  So by induction g_{n} \in \langle g_{1}, s, t \rangle. for all n.  But by construction \Gamma is generated by \Sigma_{1} \cup \{ s\} \cup \{ t \}. \square

Remark: In fact 3 can be replaced by 2.

Isometries of Trees

Before we saw that graphs of groups correspond to groups acting on trees.  As such, we now turn to isometries of trees.  Let T be a tree and let \gamma \in \mathrm{Isom}(T).  The translation length of \gamma is defined to be |\gamma| := \inf_{x \in T} d(x, \gamma x).  Let \mathrm{Min}(\gamma) = \{ x \in T \vert d(x, \gamma x) = |\gamma| \}.

Definition: If \mathrm{Min}(\gamma) \neq \emptyset, \gamma is called semisimple.  If \gamma is semisimple, and |\gamma| = 0, \gamma is called elliptic.  If |\gamma| > 0, it is called loxodromic.

Theorem 16: Every \gamma \in \mathrm{Isom}(T) is semisimple.  If \gamma is loxodromic, \mathrm{Min}(\gamma) is isometric to \mathbb{R}, and \gamma acts on \mathrm{Min}(\gamma) as translation by |\gamma|.

Notation: If \gamma is elliptic, we write \mathrm{Min}(\gamma) = \mathrm{Fix}(\gamma).  If \gamma is loxodromic$, we write \mathrm{Min}(\gamma) = \mathrm{Axis}(\gamma).  Note that \mathrm{Fix}(\gamma) is connected since if we have two points fixed by \gamma any path between them must be fixed, for otherwise we would not be in a tree.

Case 1 of Theorem 16

Case 1 of Theorem 16

Proof of Theorem 16: Consider any x \in T and the triangle with vertices x, \gamma x, \gamma^{2} x (i.e., [x, \gamma x] \cup \gamma [x, \gamma x]).  Let [ x, \gamma x] \cap [ \gamma x, \gamma^{2} x ] \cap [ \gamma^{2} x, x ] = \{ O \}, and let M be the midpoint of [ x, \gamma x ].

Case 1: d(M, \gamma x) \leq d(O,\gamma x)

In this case we have d(M, \gamma x) = d(M, x) = d(\gamma M, \gamma x).  So \gamma M = M, and \gamma is elliptic.

Case 2: d(M, \gamma x) > d(O, \gamma x)

Case 2 of Theorem 16

Case 2 of Theorem 16

Let I = [\gamma^{-1} O, O] \subseteq [x, \gamma x].  Now I \cap \gamma I = \{ O \}.  Therefore \bigcup\limits_{n \in \mathbb{Z}} \gamma^{n} I is isometric to \mathbb{R} and \gamma acts as translation by [\gamma^{-1}O, O]. Furthermore, d(x, \gamma x) = d(\gamma^{-1}O, O) + 2 d(O, \gamma x).  Therefore unless x is on the line just constructed, d(x, \gamma x) > d(\gamma^{-1} O, O) = |\gamma|. \square

Lemma 4: Let \varphi: G_{1} \longrightarrow G_{2} be surjective.  Then H is a separable subgroup of G_{2} if and only if \varphi^{-1}(H) is a separable subgroup of G_{1}.

Proof. (\Longrightarrow) If g_{1} \in G_{1} \smallsetminus \varphi^{-1}(H), then \varphi(g_{1}) \notin H. So there is K_{2} < G of finite index such that H_{2} \subset K_{2}, but \varphi(g_{1}) \notin K_{2}.  So K_{1} = \varphi^{-1}(K_{2}) is as required.

(\Longleftarrow) Suppose \varphi^{-1}(H) < G_{1} is separable.  Let g_{2} \in G_{2} \smallsetminus H.  There is g_{1} \in G_{1} \smallsetminus \varphi^{-1}(H) so that \varphi(g_{1}) = g_{2}.  Therefore, there exists K_{1} \supset \varphi^{-1}(H) of finite index in G_{1}, with g_{1} \notin K_{1}.  Now \varphi(K_{1}) = K_{2}.  Then H \subset K_{2}, and you can check that  g_{2} \notin K_{2}QED

This lemma gives us examples of some ERF groups.  For example, any finitely generated abelian group is ERF.

The Hopf Property

Definition. A group is not Hopf if and only if there is an epimorphism G \longrightarrow G with nontrivial kernel.

Example. Consider G = \mathbb{Z}^{\infty}.  Then the map G \longrightarrow G defined by (n_{1}, n_{2}, \ldots) \longmapsto (n_{2}, n_{3}, \ldots) is an epimorphism with nontrivial kernel.  Hence G is not Hopf.

Lemma 5 (Malcev): If a finitely generated group G is RF, then it is Hopf.

Let G be finitely generated and let K be a finite index subgroup.  The characteristic core of K is defined to be


The characteristic core is of finite index by Exercise 2.  It is also a normal subgroup, and one can check that if \alpha is an epimorphism and H is the characteristic core of K, then \alpha(H) = H.

Proof of Lemma 5. Let f: G \longrightarrow G be a surjection with g \in \ker(f) \smallsetminus \left\{ 1 \right\}. So there exists a finite index subgroup K < G such that g \notin K. Let H be the characteristic core of K. Then g \notin H.  Now f descends to a homomorphism \widehat{f}: G/H \longrightarrow G/H\widehat{f} is a surjection, but kills gH \neq H. QED

Theorem 4: Finitely generated free groups are Hopf.

Proof. Follows from the fact that free groups are RF and Lemma 5.

Corollary: If k > 1, then F_{k} \not\cong F_{l}.

Proof. There is a surjection F_{k} \longrightarrow F_{l} with nontrivial kernel.  If F_{k} \cong F_{l}, then we have contradicted the Hopf property. QED

Exercise 5. Give a different proof of the corollary above using homology and the Hurewicz Theorem.