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 Yearling
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12h
answered Open math problems which high school students can understand
Jul
23
answered Subgroups of Prüfer Group
Jul
20
revised Isometric group of hyperbolic 3-dim manifolds
added 8 characters in body; edited tags
Jul
20
answered Isometric group of hyperbolic 3-dim manifolds
Jul
18
comment $G=\langle x,y\ |\ x^{-1}y^2x=y^{-2}, y^{-1}x^2y=x^{-2} \rangle$ is torsion-free.
For me, it seems very natural, because a motivation behind small cancellation is to determine whether or not an element belongs to the normal closure of some subgroup. Afther that, I agree that my answer is probably not the simplest solution to the problem. But I found interesting to give a "standard" approach to this kind of question.
Jul
18
answered $G=\langle x,y\ |\ x^{-1}y^2x=y^{-2}, y^{-1}x^2y=x^{-2} \rangle$ is torsion-free.
Jul
1
awarded  Yearling
Jun
20
accepted When is a right-angled Coxeter group one-ended?
Jun
19
answered When is a right-angled Coxeter group one-ended?
Jun
19
comment When is a right-angled Coxeter group one-ended?
Six hundred pages on the geometry of Coxeter groups, what a huge work! Thank you for the reference, I didn't know it. I will add an answer applying these results to right-angled Coxeter groups. +1
Jun
19
asked When is a right-angled Coxeter group one-ended?
Jun
7
revised Monomorphisms into direct products
added 43 characters in body
Jun
7
asked Monomorphisms into direct products
Jun
4
revised Presentation of the additive group of the rational numbers
added 14 characters in body
Jun
4
answered Presentation of the additive group of the rational numbers
Jun
4
revised $\mathbb{R}$ is not a direct sum of its subgroups
added 1 character in body
Jun
2
answered Prove that $G = \langle x,y\ |\ x^2=y^2 \rangle $ is torsion free.
May
31
comment Around braid groups
Right, the description is rather ambiguous. What I had in mind is to consider braids where (up to isotopy) some fixed pairs of strands never cross in diagrams. But it is only an example, my question is general: it deals with groups which are defined by families of objects which are more or less braids.
May
31
comment Around braid groups
I am not sure. If we want the product to be well-defined, the subgroup we consider must be included into the pure braid subgroup, ie., in the kernel of the usual morphisme $B_n \to S_n$.
May
31
awarded  Necromancer