Timeline for 〈x,y : x^p = y^p = (xy)^p = 1〉
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 25, 2010 at 8:26 | vote | accept | Martin Brandenburg | ||
| Apr 25, 2010 at 8:26 | comment | added | Martin Brandenburg | finally I found my error. okay $ba$ has a similar decomposition into $p$-cycles as $a$, $b$, but always $2$ instead of $1$ is added. | |
| Apr 25, 2010 at 1:51 | comment | added | Victor Miller | I gave a few example to kbmag -- a program for dealing with automatic groups. Use what's called recursive ordering, it found that the given presentation (with some very minor tweaking) is what's called "confluent". The upshot is that there's a very simple automaton which solves the word problem for this group. Looking at that might help give a simple combinatorial proof. | |
| Apr 25, 2010 at 0:33 | comment | added | Steve D | OK, but where do you see $ba$ not having order $p$? | |
| Apr 25, 2010 at 0:19 | comment | added | KConrad | If (ab)^N = 1 then(ba)^N = 1 too since (ba)^N = b(ab)^{N-1}a = b(ab)^(-1)a = b(b^(-1)a^(-1))a = 1, so if ab in some example has order p then ba must as well. | |
| Apr 25, 2010 at 0:10 | comment | added | Steve D | I don't know which example you're referring to. If you mean the infinite permutations, then $ab$ does have order $p$, if you let $b$ act first, then $a$. | |
| Apr 25, 2010 at 0:02 | comment | added | Martin Brandenburg | thank you. but $ba$ does not have order $p$. I've checked that many times. is there are typing error? | |
| Apr 24, 2010 at 23:15 | history | edited | Steve D | CC BY-SA 2.5 |
added 860 characters in body
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| Apr 24, 2010 at 23:06 | comment | added | Steve D | Hmm, it works for me. I will edit to add the ideas of the answers there. | |
| Apr 24, 2010 at 23:04 | comment | added | Martin Brandenburg | currently the site seems to be offline. | |
| Apr 24, 2010 at 23:01 | history | answered | Steve D | CC BY-SA 2.5 |