Timeline for How large can the smallest generating set of a group $G$ of order $n$ be?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 30, 2020 at 7:50 | answer | added | Jeppe Stig Nielsen | timeline score: 1 | |
| Dec 4, 2015 at 14:30 | comment | added | Peter | Why should I have asked the question, if I would have known all this ? And the link below I mentioned did not help me either. | |
| Dec 4, 2015 at 14:28 | comment | added | Benjamin Steinberg | Anyway I did not intend to criticize the closers although I see my comment could be read that way. I was really just curious since I can see the argument in both directions | |
| Dec 4, 2015 at 14:24 | comment | added | Benjamin Steinberg | The choice of a p-group shows a lack of knowledge of the Frattini subgroup, but on the other hand the prime factorization is in the problem and then CFSG is needed. | |
| Dec 4, 2015 at 14:22 | comment | added | Peter | @Benjamin me, too. I would like to understand the criteria for so-called good and bad-questions. But, at least, $13$ voters had another oppinion. | |
| Dec 4, 2015 at 11:44 | comment | added | Frieder Ladisch | @BenjaminSteinberg: It's maybe also because, while determining the maximum value of $d(G)$ from the factorization of the order $n$ is a delicate and interesting question, the choice of the particularly bad example $n=2^{11}$ shows a certain lack of understanding. (I did not vote to close.) | |
| Dec 4, 2015 at 10:26 | comment | added | Derek Holt | @BenjaminSteinberg I have to confess that I voted to close, because I just thought it was a well-known problem with the well-known and easy answer $\log_2 n$, described in my answer. I think Geoff's answer has made it into a more interesting question. | |
| Dec 4, 2015 at 1:24 | comment | added | Benjamin Steinberg | I am curious why there are 3 votes to close this? | |
| Dec 3, 2015 at 14:19 | answer | added | Derek Holt | timeline score: 25 | |
| Dec 3, 2015 at 14:17 | review | Close votes | |||
| Dec 3, 2015 at 16:45 | |||||
| Dec 3, 2015 at 14:11 | vote | accept | Peter | ||
| Dec 3, 2015 at 14:06 | answer | added | Geoff Robinson | timeline score: 44 | |
| Dec 3, 2015 at 13:44 | comment | added | Fedor Petrov | For group of order $p^n$ simply choose elements $a_1,a_2,\dots$ such that $a_k$ does not lie in a subgroup $G_{k-1}$ generated by $a_1,\dots,a_{k-1}$. Then $|G_0|=1$, $|G_k|\geq p|G_{k-1}|$, hence this process stops on at most $n$ steps. | |
| Dec 3, 2015 at 13:39 | comment | added | Noah Snyder | For p-groups, the Burnside Basis Theorem tells you exactly how many generators you need (and the elementary abelian case is indeed the worst case). | |
| Dec 3, 2015 at 13:31 | comment | added | Gordon Royle | I expect $Z_2^{11}$ will be the extremal group for order 2048. | |
| Dec 3, 2015 at 13:30 | comment | added | Richard Stanley | For $n=2048$ the maximum value of $d(G)$ is 11, obtained by $(\mathbb{Z}/2\mathbb{Z})^{11}$. | |
| Dec 3, 2015 at 13:22 | history | asked | Peter | CC BY-SA 3.0 |