Timeline for Freiheitssatz implies a finitely generated one relator group embeds in a two-generator one relator group?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Apr 19, 2011 at 21:48 | vote | accept | JeremyKun | ||
| Apr 19, 2011 at 19:46 | comment | added | Ian Agol | The point of the second paragraph is that one can apply the construction after doing a change of basis (I've added in a bit more explanation). There is no canonical choice of generating sets for a 1-relator group, but I think it's known that any two generating sets are Nielsen equivalent, so it's natural to allow Nielsen moves. | |
| Apr 19, 2011 at 19:44 | history | edited | Ian Agol | CC BY-SA 3.0 |
added 167 characters in body
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| Apr 19, 2011 at 18:49 | comment | added | JeremyKun | I think I need a slight clarification, because I was promised this works for every one relator group, regardless of the relator. I know the Freiheitssatz extends to the case when the relator does not necessarily involve all generators, but the chosen subset still excludes a generator involved in $r$. However I'm not so sure this proof extends in a similar way. Take the case where $r = g_1^k$ involves a single generator, then it seems to me that any HNN extension (at least, the construction used here) would necessarily have at least three generators. Is there something I'm missing here? | |
| Apr 19, 2011 at 12:15 | comment | added | user6976 | @Ian: Yes, I missed it. Unfortunately this simple way cannot be used in our paper with Spakulova because your construction increases the Magnus level of the group (how many Magnus rewriting is needed to get a free group). | |
| Apr 19, 2011 at 4:37 | history | answered | Ian Agol | CC BY-SA 3.0 |