Timeline for How large can the smallest generating set of a group $G$ of order $n$ be?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 4, 2015 at 14:27 | comment | added | Peter | I looked at this site : groupprops.subwiki.org/wiki/Groups_of_order_2048 and I did not find the upper bound anywhere. And I think I made it quite clear that I am interested in the worst case. I have chosen the example $2048$ because I did not know if the maximum number of generators is known. And the "lack of understanding" is the reason for this question. If I would have known all this, I would not have asked the question. | |
| Dec 4, 2015 at 11:32 | comment | added | Frieder Ladisch | @FedorPetrov and Geoff Robinson: Agreed! | |
| Dec 4, 2015 at 10:39 | comment | added | Fedor Petrov | I would formulate it in the following way: for groups with at most $n$ elements, maximum is $[\log_2 n]$. While for groups with exactly $n$ elements it is a delicate question. What may be said for sure is that the answer is either $m$ or $m+1$, where $m$ is maximal exponent of primes in factorization of $n$. | |
| Dec 4, 2015 at 10:22 | comment | added | Derek Holt | @FriederLadisch Yes of course you are right! I don't really understand why such a simple and well-known argument merits 18 points. | |
| Dec 4, 2015 at 10:21 | history | edited | Derek Holt | CC BY-SA 3.0 |
added 74 characters in body
|
| Dec 3, 2015 at 23:01 | comment | added | Geoff Robinson | @FriederLadisch: It's fair to say that in general, $\log_{2}(n)$ is an upper bound which can't (in general) be sharpened, and is attained infinitely often. Though it is true that we can usually do much better if we know the prime factorization of $n$. | |
| Dec 3, 2015 at 22:35 | comment | added | Lucia | @FriederLadisch: You're right of course! | |
| Dec 3, 2015 at 22:27 | comment | added | Frieder Ladisch | @Lucia: When $|G| = p_1^{a_1} \dotsm p_k^{a_k}$, then the argument of this answer yields the bound $a_1 + \dotsb + a_k$ (in the worst case, as you say), while Geoff Robinson's answer yields a bound $ 1 + \max\{ a_i\} $. | |
| Dec 3, 2015 at 19:12 | comment | added | Frieder Ladisch | With all due respect, I do not agree that $\log_2 n$ is "the general answer", since in the question it is assumed that the factorization of $n$ is known, and for example if $n$ is a big prime, then $\log_2 n$ is pretty far of the right answer. $\log_2 n$ is only the answer for $n$ a power of $2$, and also $2$-powers yield the biggest values. | |
| Dec 3, 2015 at 16:12 | comment | added | Richard Stanley | $\log_2\,n$ is an upper bound, but sometimes one can do better. For instance, if $n$ is a product of distinct primes $p_i$, and no $p_i|(p_j-1)$, then every group of order $n$ is cyclic. | |
| Dec 3, 2015 at 14:52 | comment | added | Igor Rivin | @Peter Classification of finite simple groups. | |
| Dec 3, 2015 at 14:51 | comment | added | Peter | What does $CFSG$ mean ? | |
| Dec 3, 2015 at 14:48 | comment | added | Stefan Kohl♦ | .You could add: "... and unlike Geoff's answer, this doesn't require CFSG!" | |
| Dec 3, 2015 at 14:19 | history | answered | Derek Holt | CC BY-SA 3.0 |