Timeline for Is Higman's group a free product?
Current License: CC BY-SA 4.0
9 events
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| Jan 15, 2021 at 16:35 | comment | added | YCor | Btw this promotes to a proof that Higman's group $G$ has no quotient $Q$ that is a nontrivial free product. Indeed, let $R$ be the image of the Baumslag-Solitar group $H$ in $Q$. It is part of the original argument that $G$ has no nontrivial finite quotient, that the only quotient of $G$ in which some of the four generators is torsion, is trivial. Since any proper quotient of the Baumslag-Solitar group $H$ makes the second generator torsion, it follows that $R$ is a copy of $H$, hence my previous comment applies. | |
| Jan 15, 2021 at 16:27 | comment | added | Guest | Thanks a lot! This is exactly what I wanted to know. | |
| Jan 15, 2021 at 16:26 | vote | accept | Guest | ||
| Jan 13, 2021 at 8:15 | comment | added | YCor | This applies more generally as follows: if $G$ is a group, $H$ a subgroup with $H$ freely indecomposable and not infinite cyclic and $H$ generates $G$ normally (i.e. killing $H$ kills $G$): then $G$ is freely indecomposable. | |
| Jan 13, 2021 at 3:01 | history | edited | Ian Agol | CC BY-SA 4.0 |
added 226 characters in body
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| Jan 13, 2021 at 2:56 | comment | added | Ian Agol | @BenjaminSteinberg Of course you're right. In fact, that was the argument that I started writing up, then thought that this gave a slicker proof (forgetting about the statement of Kurosh). I'll fix, that, thanks. | |
| Jan 13, 2021 at 2:50 | comment | added | Benjamin Steinberg | But the subgroup generated by a,b is a solvable Baumslag Solitar group and so one-ended and you can apply your argument to that. | |
| Jan 13, 2021 at 2:42 | comment | added | Benjamin Steinberg | Why must the subgroup generated by a be conjugate to a subgroup of a factor? Its an infinite cyclic group if I am not mistaken and so the Kurosh theorem says nothing about it as far as being conjugate into a factor. If you take a product is a nontrivial element of A with a nontrivial element of B you get an element of infinite order which is hyperbolic in its action on the free product Bass Serre tree | |
| Jan 13, 2021 at 2:15 | history | answered | Ian Agol | CC BY-SA 4.0 |