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ADL
  • Member for 14 years, 6 months
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Why are free groups residually finite?
For the final step, which feels quite different from Malcev: $G=\mathbb{Z}^2$ is a finitely generated, residually finite group, therefore an (elementary) theorem of Baumslag says that $\operatorname{Aut}(G)$ is residually finite. The result now follows as $\operatorname{Aut}(G)\cong\operatorname{GL}(2, \mathbb{Z})$. (This is the proof in Magnus, Karrass and Solitar, Section 6.5.)
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All group structures on a set with cardinality $\aleph_0$
@tobias Every infinite countable group is in bijection with $\mathbb{Z}$, and this bijection can be arranged so that the identity of the group is mapped to $0$ (there is no group theory here). This gives a group structure on $\mathbb{Z}$ with identity $0$, one structure for each countable group. As Peter says, there are continuously many countable groups up to isomorphism, and hence this construction gives continuously many group structures on $\mathbb{Z}$ with your required property.
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A curious group presentation
@HJRW They state that $H$ is normal in $G$ with $G_2\leq G$, so normality in $G_2$ is surely immediate? (Possibly they actually constructed a normal chain $H\lhd G_2\lhd G_1\lhd G$, as Reidemeister-Schreier is easier to handle when we have normality.)
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Nontriviality of one-relator products
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Nontriviality of one-relator products
@HJRW I can't see this immediately, so I'd have to think properly about it. Theorem 6.1 of Fenn and Rourke's paper proves the result for words $w$ of a certain form, but really, both in this theorem and in their whole paper, they're focusing on embedding $A$ (i.e. proving a Freiheitssatz) rather than proving non-triviality. So their methods are stronger than needed here, which makes me less confident that they can be applied.
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Nontriviality of one-relator products
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Nontriviality of one-relator products
@HJRW It isn't a complete solution, so wasn't sure! I've posted it now.
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Nontriviality of one-relator products
There are no counter-examples for $A$ torsion-free and $B\cong\mathbb{Z}$. This corresponds to the Kervaire-Laudenbach Conjecture for torsion-free groups, which was proven by Klyachko; see Fenn and Rourke, Klyachko's methods and the solution of equations over torsion-free groups, Enseign. Math. (2) 42 (1996), no. 1-2, 49–74.
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When is a generalised Baumslag-Solitar group linear?
Thanks for this! This all makes sense, and the three cases in fact correspond to a classification of Whyte (arxiv.org/abs/math/0405272).
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When is a generalised Baumslag-Solitar group linear?
@HJRW I thought both those questions seemed extremely hard, so wondered if the question I asked might have been easier. I'll look up the work of Woodhouse+others, thanks for pointing it out!
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When is a generalised Baumslag-Solitar group linear?
@YCor You're right, I've changed the word "classification" to "characterisation".
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