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19

$\langle x,y \; | \; x^2=y^3=1 \rangle \cong \operatorname{PSL}_2(\mathbb Z)$ and this isomorphism identifies G with $\operatorname{PSL}_2/T^7=1$ (where $T:z\mapsto z+1$). Result is the symmetry group of the tiling of the hyperbolic plane. From this description one can see that G is infinite (e.g. because there are infinitely many triangles in the tiling and ...

14

$F_2$ acts on a certain tiling of the hyperbolic plane. It looks sort of like this: The above tiling is acted on by the modular group $\Gamma \cong \text{PSL}_2(\mathbb{Z})$, which naturally sits as a subgroup inside of the full group $\text{PSL}_2(\mathbb{R})$ of isometries of the hyperbolic plane. Abstractly, this group is the free product $C_2 \ast ... 13 Grigory has already answered your particular question. However, I wanted to point out that your question "How do you prove that a group specified by a presentation is infinite?" has no good answer in general. Indeed, in general the question of whether a group presentation defines the trivial group is undecidable. 11 First consider an unpainted$2\times 2\times 2$Rubik's cube: What is the "symmetry group"? Before we can discuss the symmetry group of this cube (or any Rubik's cube), we must be clear about what operations are allowed. There are three pertinent questions: Do rigid rotations of the cube count as "symmetries"?(If no, we must somehow exclude them.) Are ... 10 As I said in the comments above, this has a positive solution due to a recent preprint of Lars Louder, which can be found here. The paper proves that surface groups have a "single Nielsen equivalence class of generating$2g$-tuples". I'll explain what this means, and then I will explain why this solves the problem. (In the comments below, Lars Louder has ... 10 The answer to your question is yes: See Bekka–de la Harpe–Valette, Kazhdan's property$(T)$, page 69 (the standard reference on property (T), freely available on Bekka's homepage): The proof is not very difficult, and it is given in a clear fashion in the book, so it doesn't make much sense to reproduce it here. Note also ... 8 Here is a different way of doing the second problem. How many homomorphisms are there from$G = \langle a,b,c \mid a^3b^3 \rangle$to$C_6$(cyclic group of order$6$)? We can map$a$and$c$to each of the$6$elements independently, and then we must map$b$to an element whose cube is the same as the cube of the image of$a^{-1}$. In$C_6 = \langle x ...

7

Solving the relation for $c$, we conclude that there is a homomorphism $\langle\, a,b,c\mid a^2cb^3\,\rangle\to \langle a,b\rangle$ given by $a\mapsto a$, $b\mapsto b$, $c\mapsto a^{-2}b^{-3}$, which is an isomorphism. There exists a homomorphism $\langle \,a,b,c\mid a^3b^3\rangle \to \mathbb Z/3\mathbb Z\times \mathbb Z/3\mathbb Z$ given by $a\mapsto ... 7 I'm an outsider, so maybe what I'm saying is silly. I would just think about the Tits alternative application since that really gives all the intuition. To apply it---and we'll just look at the complex case---you should produce two matrices$A$and$B$that have the following properties.$A$has a dominant eigenvalue$\lambda$(i.e., the eigenspace of ... 7 This counterexample seems to work for$\frac{1}{2}$, and beyond: $$x_g:=Cr^{\ell(g)}\qquad\forall g\in F_2$$ for the right choices of constants $$C>0\qquad\mbox{and}\qquad 0\leq r<\frac{1}{\sqrt{3}}.$$ Here$\ell(g)$denotes the word length of the element$g$, i.e. the minimal number of letters needed in the alphabet ... 7 This might not be the best argument, but it seems to work. First, no nontrivial finite, connected, vertex-transitive graph has a cutpoint (nontrivial here means something like cardinality at least$3$to avoid having to define cutpoint carefully). If it did, then every vertex would be a cutpoint, so you could inductively build an arbitrarily long finite ... 7 You can do this using metric currents in the sense of Ambrosio-Kirchheim. This is a rather new development of geometric measure theory, triggered by Gromov and really worked out only in the last decade. I should warn you that this is rather technical stuff and nothing for the faint-hearted. Urs Lang has a set of nice lecture notes, where you can find most ... 7 More generally, any group$G$defined by a finite presentation with more generators than relations is infinite - in fact$G/[G,G]$is infinite. That follows from the proof of the fundamental theorem of abelian groups. You can prove it directly, by showing that there is a nontrivial epimorphism$\phi$onto${\mathbb Z}$. Let$\phi:G \to {\mathbb Z}$be any ... 6 Let$z = x^2 = y^3$. This element clearly commutes with$x$and$y$. Therefore$z$lies in the center and$\langle z \rangle$is a normal subgroup. We have: $$G / \langle z \rangle = \langle x, y \mid x^2 = y^3 = 1\rangle.$$ Now the abelianization of this quotient group is$\Bbb Z_2 \times \Bbb Z_3$, which is clearly non-trivial. 6 You need to modify the definition of$\Psi$: put $$\Psi(t^na^xt^{-n})=xk^n$$ (instead of$\Psi(t^na^xt^{-n})=xk^{-n}$). The homomorphic property should follow easily along the lines of the computations given in the original question. Alternatively, you could replace$t$by$t^{-1}$in the definition of the Baumslag-Solitar group, in which case the$\Psi$... 6 If you think of$F$as acting by left multiplication on the vertices of$\Delta$, then the natural interpretation of the quotient graph is as a graph whose vertices are the distinct cosets$Gg$of$G$in$F$, with an edge (labelled$x$) from$Gg \to Ggx$for each$x \in \{a,a^{-1},b,b^{-1}\}$. So, if you ignore the labels, it is still a an infinite ... 5 The amalgamation only identifies two isomorphic subgroups, it doesn't perform any further quotienting. So the group you are amalgamating over has to be isomorphic to a subgroup of both$A$and$B$. Thus your first statement is correct. You cannot form an amalgamated free product of$F_2$and a finite group with non-trivial amalgamation, because$F_2$... 5 Remark: For those unacquainted with Property$(T)$, the standard reference is the freely available book by Bekka-de la Harpe-Valette. Since groups with property$(T)$are finitely generated, we can assume that the rank of the free group$F$is finite. If a group$G$has property$(T)$then so does every quotient(1), recall also my answer to your ... 5 Gromov in his original 1987 book (Section 3.1) wrote a classification for arbitrary isometric group actions on hyperbolic spaces (with no further assumption) into 5 main classes. It goes at follows (the terminology is borrowed from here) 1: bounded: orbits are bounded 2: horocyclic: orbits are unbounded,$G$acts with no hyperbolic isometry (hence there's ... 5 (Of course the trivial group acts freely on all spheres! Let's assume$G$is finite and nontrivial.) Since$G$is finite, the action is certainly properly discontinuous, and since the action is given to be free, it follows that the quotient$q: X \rightarrow X/G$is a finite covering map. Recalling that in any$n$-sheeted covering map$q: X \rightarrow Y$... 5 Yes, your interpretation is correct. There are no further relations since there are no loops in the Cayley graph except those deducible from the relations$a^2 = b^2 = e$. The group$\langle a, b \mid a^2, b^2 \rangle$is known as the infinite dihedral group. 5 Take$G = S_{3}$,$H = U = \langle (12) \rangle = \{ 1, (12) \}$. As a system of right coset representatives, take$R = \{ 1, (123), (132) \}$. Now choose$u = (12) \in U$, and$s = (123) \ne (132) = s'$. We have $$u s H = (12) (123) H = (13) H = \{ (13), (132) \} = (132) H = s' H.$$ Barring mistakes. 5 This is probably the explanation you don't like, since it involves using the Lie structure, but it seems incredibly simple to me and I don't see what you could ask for that is simpler. Let$G$be a Lie group.$G$acts on itself by conjugation, fixing the identity, and thus$G$acts on$T_e(G)$, the tangent space at the identity. Write$Ad(g)$for the action ... 5 Given a group$G$and a topological space$X$, an action of$G$on$X$is, formally, a homomorphism from$G$to the group$\text{Homeo}(X)$of all homeomorphisms from$X$to itself. This can also be expressed with more notation as a function which associates to each$g \in G$and each$x \in X$an element$g \cdot x \in X$subject to various properties: (1) ... 5 Inspiration: in the free product$H=C_2*C_2=\langle a,b|a^2,b^2\rangle$the element$ab$has infinite order. And furthermore the only nontrivial coset of$\langle ab\rangle$is$b\langle ab\rangle$(every word in$H$is either a product of$ab$s or it is$b$followed by a product of$ab$s), so$\langle ab\rangle$has index two. The given information in this ... 5 If$G,H$are groups then$H$is a double cover of$G$iff$H$has a normal subgroup$K$of order 2 such that$K \leq [H,H] \cap Z(H)$and$H/K \cong G$. In other words,$H$has an element$z$that (1) commutes with every element of$H$, (2) can be written as a product of commutators$z=\left(h_1^{-1} h_2^{-1} h_1 h_2\right)\cdots\left(h_{2n-1}^{-1} ...

5

There is an expression for this, if $r$ is an integer. See projecteuclid. Also, OEIS A016725. I am not a number theorist. Just discovered this while searching for something else.

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It's open too. Indeed, the question whether $Aut(F_n)$ has the Haagerup Property is open as well (actually for all $n\ge 2$). So this one (for $n\ge 4$) is just intermediate between the two. [For an arbitrary infinite discrete group $G$ we have implications: $G$ has Kazhdan's T $\Rightarrow$ $G$ has an infinite subgroup with Kazhdan's T $\Rightarrow$ $G$ ...

4

Yes if you assume a group has a normal subgroup $Z$ of finite index which is infinite cyclic then it's much easier. If $Z$ is central, then a classical theorem shows that the derived subgroup is finite. So modding out by the derived subgroup, you get an abelian group, and eventually get that the group has a homomorphism onto $\mathbf{Z}$ (with finite kernel, ...

4

Here is an expansion of Derek Holt's comment, with added details regarding how to deal with various finite index issues that arise along the way. I will use your terminology "undistorted", although the more standard terminology is "quasi-isometry" and "quasi-isometric embedding". The group $G$ has a finite index free abelian subgroup which I will denote ...

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