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Dec 5, 2011 at 11:18 vote accept Calc
Dec 5, 2011 at 10:52 vote accept Calc
Dec 5, 2011 at 10:52
Dec 4, 2011 at 21:41 comment added Arturo Magidin As noted in math.SE, $k\leq c$ holds: pick a prime $p$ that divides $d_1$. Then $G/pG$ is a vector space over $\mathbf{F}_p$ of dimension $k$, hence $G$ needs at least $k$ generators.
Dec 4, 2011 at 21:38 comment added Arturo Magidin Also posted in math.SE: math.stackexchange.com/questions/88106/…
Dec 4, 2011 at 13:13 comment added Geoff Robinson @Boris: Not sure that I agree with the above description of what $n$ is. The number of summands when $G$ is expressed as a direct sum of cyclic groups is not uniquely determined.For example, in general the number $k$ and $n$ above are usually different.
Dec 4, 2011 at 13:08 answer added Geoff Robinson timeline score: 6
Dec 4, 2011 at 12:18 comment added Boris Novikov I don't know. :-(
Dec 4, 2011 at 11:59 comment added Calc Ok. Sorry. How can one erase a question? ;)
Dec 4, 2011 at 11:46 comment added Boris Novikov OK, so $n$ is the number of summands in the direct sum of cyclic groups, isn't it? Then the inequality $c\le n$ is evident.
Dec 4, 2011 at 11:31 comment added Calc It is part of the fundamental theorem for finitely generated abelian groups that the $d_i, p_i, \alpha_i$ are uniquely determined by $G$ itself (up to reordering in the second case). If instead you are asking about a name, I heard of people calling $k$ and $n$ respectively the minimal and maximal rank.
Dec 4, 2011 at 11:15 comment added Boris Novikov How do you define $n$?
Dec 4, 2011 at 10:17 history edited Calc CC BY-SA 3.0
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Dec 4, 2011 at 8:58 history edited Calc CC BY-SA 3.0
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Dec 4, 2011 at 8:35 history asked Calc CC BY-SA 3.0