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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 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 .

Consider . We will show that this group is non-Hopfian (i.e., that there exists a homomorphism that is a surjection with nontrivial kernel), hence is not RF. Consider defined by and . First, we show this is a homomorphism, then a surjection, then that it has nontrivial kernel.

Homomorphism: Hence, the map is a homomorphism.

Epimorphism: To see that is a surjection, we check that . This is obvious for . Further and so . But then .

Nontrivial kernel: Consider . Then . So we just need to show that is nontrivial. We use Britton’s Lemma (Normal Form Theorem). What Britton’s Lemma says is that there will be that can be removed from our word using the relation if it is trivial. But this can’t be done with , so we must have as required.

**Remark:** can never embed in a word-hyperbolic group by a previous Lemma.

**Exercise 20:** Show that , where is the dyadic rationals and we have acting via multiplication by 2. One can deduce that is linear, hence RF by Selberg’s Lemma.

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

**Proof: **Let . Consider . This has some nice free subgroups, unlike : Let . Let . By the Normal Form Theorem, (basically since there will always be some g’s between the s’s whenever you multiply any two elements together). Let . Again, . Since $latex\Sigma_{1}$ and are both countable, .

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 , . But is generated by , , and : , so or . So by induction . for all n. But by construction is generated by .

**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 be a tree and let . The translation length of is defined to be . Let .

**Definition: **If , is called *semisimple*. If is semisimple, and , is called elliptic. If , it is called loxodromic.

**Theorem 16: **Every is semisimple. If is loxodromic, is isometric to , and acts on as translation by .

Notation: If is elliptic, we write . If is loxodromic$, we write . Note that is connected since if we have two points fixed by any path between them must be fixed, for otherwise we would not be in a tree.

**Proof of Theorem 16:** Consider any and the triangle with vertices (i.e., ). Let , and let be the midpoint of .

Case 1:

In this case we have . So , and is elliptic.

Case 2:

Let . Now . Therefore is isometric to and acts as translation by . Furthermore, . Therefore unless is on the line just constructed, .

**Lemma 4: **Let be surjective. Then is a separable subgroup of if and only if is a separable subgroup of .

**Proof.** If , then . So there is of finite index such that , but . So is as required.

Suppose is separable. Let . There is so that . Therefore, there exists of finite index in , with . Now . Then , and you can check that . **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 with nontrivial kernel.

**Example.** Consider . Then the map defined by is an epimorphism with nontrivial kernel. Hence is not Hopf.

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

Let be finitely generated and let be a finite index subgroup. The characteristic core of 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 is an epimorphism and is the characteristic core of , then .

**Proof of Lemma 5. **Let be a surjection with . So there exists a finite index subgroup such that . Let be the characteristic core of . Then . Now descends to a homomorphism . is a surjection, but kills . **QED**

**Theorem 4:** Finitely generated free groups are Hopf.

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

**Corollary:** If , then .

**Proof.** There is a surjection with nontrivial kernel. If , then we have contradicted the Hopf property. **QED**

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

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