Timeline for Zero-sum partition of an abelian group
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Apr 21, 2022 at 10:34 | answer | added | alpmu | timeline score: 2 | |
| Mar 3, 2018 at 23:26 | answer | added | Sylvia | timeline score: 2 | |
| Feb 20, 2012 at 21:46 | comment | added | Geoff Robinson | @Zack: Yes, I came at it from the point of view of the case $n =3$ and $G$ being an Abelian group of order congruent to 1 (mod $3$) which doe not admit an automorphism of order $3$, as in my modified answer below. Such a $G$ has pretty restricted structure. I thought the group of order $16$ would be OK, but I hadn't checked. I'm not sure how to be systematic in general. | |
| Feb 20, 2012 at 21:17 | comment | added | Zack Wolske | @Geoff: For that we have $(1,3,4), (2,9,13), (5,12,15), (6,8,10), (7,11,14)$. This comes from considering how many of each triple (either $(0,0,0)$ or $(0,1,1)$) we must use for $\mathbb{Z}/2 \mathbb{Z}$. There's only one option, and the rest of it falls into place. You're right, I wasn't considering even values, and I also don't know $\mathbb Z / \left(10\right) \times \mathbb Z / \left(4\right)$ for $n=3$ or $\mathbb Z / \left(18\right) \times \mathbb Z / \left(2\right)$ for $n=5$. They seem to boil down to finding positive solutions to underdetermined linear systems. | |
| Feb 20, 2012 at 18:17 | comment | added | Geoff Robinson | @Zack Wolske : A similar question arises for the group $\left( \mathbb{Z}/8\mathbb{Z} \right) \times \left( \mathbb{Z}/2 \mathbb{Z} \right) .$ | |
| Feb 20, 2012 at 10:40 | comment | added | user9072 |
@Denis Serre: yes, as I said, the two-rank must not be one. But you are right the explict form is helpful. Perhaps even more explcitly: $d_{r-1}$ is odd and $d_r$ even.
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| Feb 20, 2012 at 9:19 | comment | added | Denis Serre | About the zero-sum assumption: We know that $G\sim({\mathbb Z}/(d_1))\times\cdots({\mathbb Z}/(d_r))$ with $d_1|d_2$, ..., $d_{r-1}|d_r$. Then the sum of all elements in $G$ equals zero unless one $d_\ell$ is even and all the other ones are odd. | |
| Feb 20, 2012 at 1:00 | comment | added | Zack Wolske | Can you solve $n=3$ when $G = \mathbb Z / \left(5\right) \times \mathbb Z / \left(11\right)$? Cases $\mathbb Z / \left(p\right) \times \mathbb Z / \left(q\right)$ where $n$ divides both of $p-1, q-1$ are solved by patching together the partitions given by each prime. $\left|G\right| = 55$ is the smallest case for $n=3$ where this doesn't apply. I haven't made it work, but I also can't say that it doesn't. | |
| Feb 19, 2012 at 23:57 | answer | added | Geoff Robinson | timeline score: 4 | |
| Feb 19, 2012 at 22:42 | comment | added | Gerhard Paseman | I find investigating the problem for small cyclic groups to be instructive (meaning it torpedoes my naive attempts for solving). For example, for the cyclic group of order 13, starting with sets (1,2,-3), (-1,-2,-3) leads to an unhappy conclusion. Gerhard "Ask Me About Exploratory Failures" Paseman, 2012.02.19 | |
| Feb 19, 2012 at 20:48 | history | edited | darij grinberg | CC BY-SA 3.0 |
added 70 characters in body
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| Feb 19, 2012 at 20:47 | comment | added | darij grinberg | Seva: good question, but none I can answer... | |
| Feb 19, 2012 at 20:43 | answer | added | user9072 | timeline score: 8 | |
| Feb 19, 2012 at 20:35 | comment | added | Seva | Is the case of $G$ cyclic easy? | |
| Feb 19, 2012 at 20:28 | comment | added | Gerhard Paseman | Also, this problem has a feel of projective geometry. You might consider a solution involving a subset of a finite geometry. Gerhard "Ask Me About System Design" Paseman, 2012.02.19 | |
| Feb 19, 2012 at 20:10 | comment | added | darij grinberg | If there are no proper nontrivial divisors of |G|-1, the better... in this case the problem holds vacuously. | |
| Feb 19, 2012 at 20:09 | comment | added | Gerhard Paseman | I should add the words "Mersenne prime" to my previous comment. Gerhard "Ask Me About System Design" Paseman, 2012.02.19 | |
| Feb 19, 2012 at 20:06 | comment | added | Gerhard Paseman | You need to take care that p is not 2, or otherwise modify the conditions. Otherwise there are no choices for n. Gerhard "Ask Me About System Design" Paseman, 2012.02.19 | |
| Feb 19, 2012 at 20:01 | history | asked | darij grinberg | CC BY-SA 3.0 |