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18

<x,y|x^2=y^3=1>≅PSL_2(Z) and this isomorphism identifies G with $PSL_2/T^7=1$ (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 G acts on them transitively).


13

$F_2$ 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 ...


12

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.


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 ...


9

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 ...


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

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

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 ...


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 ...


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 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 ...


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 ...


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 $\Psi$ ...


5

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 ...


5

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 ...


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

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

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

(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.


4

From linear algebra we know that every invertible matrix can be obtained from the identity matrix by a sequence of elementary row operations. The corresponding elementary matrices therefore generate the general linear group. Now you can find pre-images in ${\mathrm Aut}(F_{n})$ by considering Nielsen transformations, which match up fairly directly with the ...


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

1) $G = \bigoplus_{i=1}^{\infty} \mathbb{Z}/2\mathbb{Z}$ is residually finite but not Hopfian. It is clearly not Hopfian, and any non-identity element $g$ must have $g_i \neq 0$ for at least one $i$. Then $\pi_i(G) \cong \mathbb{Z}/2\mathbb{Z}$ is a finite quotient in which the image of $g$ is nontrivial. 2) An infinitely generated free group is ...


4

Here is one way to prove it: 1) If the triangle group is uniform, then Milnor-Schwarz shows that it is word-hyperbolic (since it acts properly discontinuously and cocompacty by isometries on the hyperbolic plane). 2) If the triangle group is nonuniform, then we can argue as follows: a) There is a single quasi-isometry class of nonuniform lattices in ...


4

There are a lot of connections in geometric group theory, which studies things like free groups. Every hyperbolic group (an infinite group with a special condition) has a space at infinity that is either a sphere or a fractal. For instance, the free group has a cantor set at infinity. Other groups have Sierpinski curves and Menger sponges. Space-filling ...


4

I think the answer is yes unless one of the groups is trivial or both have order 2. Suppose $1\neq g\in G$ and $1\neq h_1,h_2\in H$ with $h_1\neq h_2$. Then for any $n$, the $2^n$ (exponential in $n$) elements $gh_{i_1}gh_{i_2}\dots gh_{i_n}$ of $G\ast H$ are all different.



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