Timeline for How do I find hyperbolic generating triples for a group using GAP?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 11, 2021 at 15:29 | comment | added | ahulpke | @Kris. Yes. Do you want them pairwise coprime, or just coprime over all? | |
| Sep 11, 2021 at 8:10 | comment | added | Kris | is it possible to carry out a further filter on t:=Filtered(UnorderedTuples(o,3),x->1/x[1]+1/x[2]+1/x[3]<1); so that we can remove any triples whose orders are not coprime? | |
| Sep 1, 2021 at 14:38 | comment | added | ahulpke | @Kris GQuotients will find epimorphisms, that is each homomorphism is onto, and thus the elements generate. If only non-generating triples are found, it returns an empty list. | |
| Sep 1, 2021 at 8:56 | comment | added | Kris | Thank you @ahulpke. In your example above, I can assign generators x, y and z to show that their product is the identity. But not too sure how I would show that these elements generate the group G. Would each mytup from 1 to 12 do this and how can I be sure? I'm obviously missing some theory here. | |
| Aug 31, 2021 at 13:57 | history | answered | ahulpke | CC BY-SA 4.0 |