Seirios
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 1d awarded Necromancer Apr16 comment A small cancellation group does not contain $\mathbb{Z}^3$ Finitely presented small cancellation groups are hyperbolic, but what about the general case? Apr16 asked A small cancellation group does not contain $\mathbb{Z}^3$ Apr15 awarded Revival Apr15 comment Examples for Burnside problem. An example, based on symmetries of a tree, may be found in John Meier's book, Groups, Graphs and Trees. However, the torsion is not uniformely bounded. Apr1 comment Decomposition of a group whose Cayley graph is a tree @user1729: I verified in the mentionned book, the question is correctly formulated. Mar27 comment If $(M,d)$ is a compact metric space and $f:M\rightarrow M$ is an isometry ($d(x,y)=d(f(x),f(y))$ for any $x,y\in M$), then $f$ is a homeomorphism @BasantSharma: The property $d(x,A)=0 \Rightarrow x \in A$ is true whenever $A$ is closed. In my answer, because $X$ is compact, $f(X)$ is indeed closed. Mar23 answered Prove that $PSL(2,\mathbb{Z})$ is free product of $C_2$ and $C_3$ Mar22 comment Let $G$ be a group. Prove or disprove that $H=\{g^2~ |~g\in G\}$ is a subgroup of $G$ Yes, it works. You can notice that $(12)(34)$ is not a square from the Cayley table you wrote or solving the equation $(12)(34)= \sigma^2$: decomposing $\sigma$ into disjoint cycles, notice that $\sigma$ must be the product of two permutations, which implies $\sigma^2=1$, a contradiction. Mar22 answered A problem on free products Mar22 answered Let $G$ be a group. Prove or disprove that $H=\{g^2~ |~g\in G\}$ is a subgroup of $G$ Mar13 comment Gromov's boundary at infinity, drop the hypothesis on hyperbolicity Which definition of boundary do you use? I think they are not all equivalent for non-hyperbolic spaces. Mar9 comment Are all $\delta$-hyperbolic groups CAT(0)? Button and Kropholler just showed that $G$ is linear: see theorem 4.5 of arxiv.org/pdf/1503.01989v1.pdf. Mar7 comment Determinant for Area and Volume? May be relevant: chiasme.wordpress.com/2013/07/16/… Feb18 answered Are countably infinite, compact, Hausdorff spaces necessarily second countable? Feb17 revised The free group given by $\langle a,b:a^2=b^3=e\rangle$ is not abelian. added 341 characters in body Feb17 answered The free group given by $\langle a,b:a^2=b^3=e\rangle$ is not abelian. Feb13 awarded Nice Question Feb13 awarded Enlightened Feb13 awarded Nice Answer