Timeline for Deriving a relation in a group based on a presentation
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
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| Feb 14, 2010 at 14:58 | vote | accept | Steve D | ||
| Feb 13, 2010 at 18:07 | history | edited | Steve Huntsman | CC BY-SA 2.5 |
Removed reference to my deleted incorrect answer
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| Feb 13, 2010 at 17:00 | history | edited | Victor Miller | CC BY-SA 2.5 |
inserted derivation trace from kbmag
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| Feb 13, 2010 at 16:56 | comment | added | Victor Miller | @Steve: I just noticed when looking at the above system of equations that it had derived $x^3 = x^{-3}$. I did rerun it with trace on and it gives the series of derivations (which can probably be edited down to only point to the relevant ones). | |
| Feb 13, 2010 at 16:49 | comment | added | Victor Miller | @Steve, I think (but don't know for sure) that kbmag has a program that will give you the trace of its calculations from the word problem. If not, it should :-). | |
| Feb 13, 2010 at 16:48 | comment | added | Victor Miller | @Steve, I had the wrong presentation for the first group. BTW I was pretty sure that kbmag would work in this case because it's known that $SL(n,p)$ is automatic, and being automatic is a property independent of the presentation. | |
| Feb 13, 2010 at 16:46 | comment | added | Steve D | Yes, GAP is capable of telling me the group is $SL(2,5)$. I wanted to derive the relation $x^6=1$ from the presentation itself. That would entail either following GAP's calculations (probably very difficult) or doing something like you mentioned with the automaton (which sounds very interesting). | |
| Feb 13, 2010 at 16:45 | history | edited | Victor Miller | CC BY-SA 2.5 |
output from the correct presentation
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| Feb 13, 2010 at 16:43 | comment | added | Victor Miller | I followed your link, and found that the original question involved the group $\langle x,y | x^3 y^{-5}, x^3 (y x)^2 \rangle$. I gave that presentation to kbmag and it came back with the answer that it was a finite group of cardinality 120, so it is isomorphic to $\text{SL}(2,5)$. | |
| Feb 13, 2010 at 16:36 | comment | added | Steve D | Hi Victor, can you explain a little bit more how this can be used to derive the identity? Also, does [x^2,xi] mean x^2 = xi? Because that's not true. | |
| Feb 13, 2010 at 16:35 | comment | added | Victor Miller | I should add that kbmag also produces two finite state automata the first of which solves the word problem (i.e. given two words are they the same element) in time at most quadratic in the length of the words. By tracing how that automaton functions you will get a proof of your desired assertion. The second automaton tells you what happens when you add a generator onto the end of a word. | |
| Feb 13, 2010 at 16:23 | history | answered | Victor Miller | CC BY-SA 2.5 |