Timeline for What are the possible finite non-solvable quotients of one relator groups?
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| Jul 8, 2014 at 20:12 | comment | added | moshe newman | Specifically, I'm trying the case aab=bba, which has some non-solvable quotients, but it does not seem to be known if it has all Alt(n) as quotients. | |
| Jul 8, 2014 at 20:11 | comment | added | moshe newman | I'm corresponding with Button already. Thanks all the same. | |
| Jul 8, 2014 at 17:13 | comment | added | HJRW | My impression is that these kind of questions are wide open. Indeed, I think that every known 1-relator group either: (i) has every finite quotient metabelian; or (ii) is large (in the sense of Pride). You might like to look at the following paper of Jack Button: 'Largeness of LERF and 1-relator groups', Groups Geom. Dyn. 4 (2010), 709–738. | |
| Jul 7, 2014 at 17:21 | comment | added | moshe newman | Correct. If the powers are both divisible by p, say p>2, then you can take b to be of order p and the relation is trivially satisfied, so you can get pretty much anything as a quotient. | |
| Jul 7, 2014 at 16:54 | review | First posts | |||
| Jul 7, 2014 at 17:16 | |||||
| Jul 7, 2014 at 16:49 | comment | added | YCor | Moshe: I think it's only true for Baumslag-Solitar $BS(m,n)$ with $m,n$ coprime. | |
| Jul 7, 2014 at 16:43 | comment | added | moshe newman | G = < a,b | aa=bbb> and G = <a,b,c,d | abcd = dcba> have the alternating groups Alt(n) as quotients for all sufficiently large n. Is there something 'in between'? | |
| Jul 7, 2014 at 16:40 | comment | added | moshe newman | The Baumslag- Solitar groups are examples of one-relator groups with only solvable finite quotients. The groups | |
| Jul 7, 2014 at 16:36 | history | asked | moshe newman | CC BY-SA 3.0 |