Timeline for Exponent of a group
Current License: CC BY-SA 2.5
28 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 17, 2010 at 9:42 | comment | added | Amitesh Datta | @Pete L. Clark People generally like to comment a lot on the "easier questions" simply because it takes less time and effort not to mention that it also might get you free reputation points! | |
| Jul 17, 2010 at 6:47 | answer | added | Primoz | timeline score: 5 | |
| Jul 17, 2010 at 0:33 | comment | added | Pete L. Clark | @marwalix: For every finite group of exponent $3$, $3$ divides the order of $G$. So what you say is true but vacuous. @everybody: I am frankly surprised at the number of comments on this question: there's no research-level math issue here. | |
| Jul 16, 2010 at 22:19 | vote | accept | marwalix | ||
| Jul 16, 2010 at 22:19 | answer | added | marwalix | timeline score: 1 | |
| Jul 16, 2010 at 21:44 | comment | added | marwalix | The result I have seen is that every finite group $G$ of exponent $3$ such as $3$ does not divide $o(G)$ is abelian. | |
| Jul 16, 2010 at 17:11 | comment | added | Yemon Choi | Wadim: I did not vote to close because this is a bad question. I voted to close because it seems to already have been answered by the comments and answers to mathoverflow.net/questions/31797/… | |
| Jul 16, 2010 at 15:12 | vote | accept | marwalix | ||
| Jul 16, 2010 at 22:19 | |||||
| Jul 16, 2010 at 15:12 | vote | accept | marwalix | ||
| Jul 16, 2010 at 15:12 | |||||
| S Jul 16, 2010 at 15:12 | vote | accept | marwalix | ||
| Jul 16, 2010 at 15:12 | |||||
| Jul 16, 2010 at 15:12 | vote | accept | marwalix | ||
| S Jul 16, 2010 at 15:12 | |||||
| Jul 16, 2010 at 12:52 | comment | added | user717 | @Wadim...or check Alan's answer. | |
| Jul 16, 2010 at 12:37 | comment | added | Wadim Zudilin | @Armi: There is only one group up to isomorphism. :-) If the author means the exponent, then he can construct a nonabelian example by hand... | |
| Jul 16, 2010 at 12:26 | comment | added | Doug Chatham | @marwalix: The result you "remember having seen" for n=3 might have been Jacobson's result for rings, not groups. See mathoverflow.net/questions/29590/… | |
| Jul 16, 2010 at 11:58 | comment | added | user717 | @Wadim: I don't think the author meant "order" 3. There is only one group of order 3, namely Z/(3) which is pretty abelian... | |
| Jul 16, 2010 at 11:36 | answer | added | ADL | timeline score: 16 | |
| Jul 16, 2010 at 11:12 | answer | added | Ashot Minasyan | timeline score: 6 | |
| Jul 16, 2010 at 10:20 | comment | added | Wadim Zudilin | @Robin: thank you very much for refreshing my memory (I don't think I met the notion of "exponent" in the last 20 years!). Of course, with exponent 3 there is much room for noncommutativity! (But probably the author meant "order"?!) @Yemon: you are extremely active in closing but not in clarifying to poor me what was wrong with my understanding. I was saved thanks to Robin. Since this isn't a homework, but an ordinary question, no reason (for me) to close. I recalled "exponent" from my childhood! :-) | |
| Jul 16, 2010 at 9:06 | comment | added | Yemon Choi | Martin: he asks how "this" generalizes to higher exponent. The question I linked to asks: what can we say about finite groups of fixed exponent (save for the identity). I therefore see no reason for this question to remain open, especially when the Heisenberg example you give is given in the comments to that other question. | |
| Jul 16, 2010 at 8:55 | comment | added | Martin Brandenburg | This is not an exact dublicate. Here, only the property "abelian" is asked. | |
| Jul 16, 2010 at 8:54 | comment | added | Yemon Choi | In case my first comment wasn't clear: I think this is in effect a duplicate question, which is why I voted to close. Do people disagree? | |
| Jul 16, 2010 at 8:53 | comment | added | Martin Brandenburg | The group $G$ of upper triangular matrices over $\mathbb{F}_3$ with diagonal $(1,1,1)$ is of exponent $3$. | |
| Jul 16, 2010 at 8:05 | comment | added | Benoît Kloeckner | The key word is "Burnside groups", you should easily find your answer in the literature by searching this. | |
| Jul 16, 2010 at 7:20 | comment | added | Robin Chapman | Wadim, "exponent" and "order" are not synonyms in finite group theory. The order of a finite group is its cardinality, while the exponent is the least $n$ with $x^n$ the identity for all $x$ in the group. | |
| Jul 16, 2010 at 6:04 | comment | added | Yemon Choi | Wadim, I don't follow. See the link in my comment and also en.wikipedia.org/wiki/Exponent_%28group_theory%29 | |
| Jul 16, 2010 at 5:56 | comment | added | Wadim Zudilin | Your "exponent" is known as "order". There is not too much room for a group of 3 elements, say $e$, $a$ and $b$: you automatically get $ab=ba$... | |
| Jul 16, 2010 at 5:43 | comment | added | Yemon Choi | I'm not sure I agree with your claim about exponent 3. See mathoverflow.net/questions/31797/… in particular the comments of Pete Clark and the answer of Francesco Polizzi | |
| Jul 16, 2010 at 5:39 | history | asked | marwalix | CC BY-SA 2.5 |