Timeline for Minimal generation for finite abelian groups
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 27, 2016 at 23:03 | comment | added | Geoff Robinson | Yes, that is included in my argument above, but here we are looking at the maximum possible size of n irredundant generating set. | |
| Jun 27, 2016 at 21:53 | comment | added | yakov | It is well known that if an abelian group $G=P_1\times\dots\times P_k$, where $P_i\in\text{Syl}(G)$, and $\text{d}(P_1)\ge\dots\ge\text{d}(P_k)$, then the minimal number of generators of $G$ is equal to $\text{d}(P_1)$. This is also true for nilpotent groups $G$. | |
| Dec 7, 2011 at 11:28 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
stated second sentence more precisely
|
| Dec 5, 2011 at 11:18 | vote | accept | Calc | ||
| Dec 5, 2011 at 11:15 | comment | added | Geoff Robinson |
The maximal cardinality of a minimal generating set is a fixed positive integer. There may be many minimal ghenerating sets of that cardinality. Every minimal generating set has cardinality less than or equal to $|G|,$ so there are only finitely many minimal generating sets, and we can decide in principle the maximum size of any of them which is a fixed integer. Among all minimal gnerating sets of that fixed maximum size, we can choose one $S$ so that $ \sum_{s \in } o(s)$ is as small as possible.
|
|
| Dec 5, 2011 at 11:11 | comment | added | Calc | Is this not a restriction on the set of all minimal generating sets of maximal cardinality? Is the cardinality of all minimal generating sets of maximal cardinality all the same? | |
| Dec 5, 2011 at 11:00 | comment | added | Geoff Robinson |
It means that among all minimal generating sets of maximal cardinality, we choose one, $S$ say, so that $ \sum_{ s \in S} o(s)$ is minimal, where $o(s)$ is the order of $s$.
|
|
| Dec 5, 2011 at 10:55 | comment | added | Calc | Sorry, one more thing. What does "minimize the sum of the orders of elements of $S$ subject to that" mean? | |
| Dec 5, 2011 at 10:52 | vote | accept | Calc | ||
| Dec 5, 2011 at 10:52 | |||||
| Dec 5, 2011 at 10:33 | comment | added | Geoff Robinson | Well, for example, let the order of $s$ be $p^c.q$, where $p$ is a prime, $c$ is a positive integer, and $q$ is a positive integer which is not divisible by $p.$ We may write $1= xp^c + yq$ for integers $x$ and $y.$ Take $t = xp^c .s,$ which has order $q$ and $u = yq.s ,$ which has order $p^c.$ This is a standard decomposition of the element $s$ into the sum of its $p$-part and $p'$-part, and may be found in most group theory texts in a more general form. | |
| Dec 5, 2011 at 10:05 | comment | added | Calc | Thanks for your answer! How can you write $s=t+u$ with orders of $t$ and $u$ respectively $a>1, b>1$ and such that $\gcd(a,b)=1$ and $a \dot b$ is the order of $s$? | |
| Dec 4, 2011 at 19:52 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
minor typo
|
| Dec 4, 2011 at 19:46 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
clarified that $t$ and $u$ are nonidentity elements
|
| Dec 4, 2011 at 17:45 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
Minor typo
|
| Dec 4, 2011 at 13:08 | history | answered | Geoff Robinson | CC BY-SA 3.0 |