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Let $G$ be a finite abelian group. I know of two ways of writing it as a direct sum of cyclic groups:

1) With orders $d_1, d_2, \ldots, d_k$ in such a way that $d_i|d_{i+1}$,

2) With orders that are powers of not necessarily distinct primes $p_1^{\alpha_1}, \ldots, p_n^{\alpha_n}$.

Is it true, and how can one prove that the cardinality $c$ of any minimal generating set for $G$ satisfies $k \leq c \leq n$ (I am most concerned about the second inequality)? Here minimal means irredundant.

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How do you define $n$? –  Boris Novikov Dec 4 '11 at 11:15
    
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. –  Calc Dec 4 '11 at 11:31
    
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. –  Boris Novikov Dec 4 '11 at 11:46
    
Ok. Sorry. How can one erase a question? ;) –  Calc Dec 4 '11 at 11:59
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Also posted in math.SE: math.stackexchange.com/questions/88106/… –  Arturo Magidin Dec 4 '11 at 21:38
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1 Answer 1

up vote 6 down vote accepted

Note that $n$ is the sum over prime divisors $p$ of $|G|$ of the minimal number of generators of the distinct Sylow $p$-subgroups of $G.$ The sizes of all minimal generating sets of a finite $p$-group are the same by properties of the Frattini subgroup. Use of the Frattini subgroup helps to prove the leftmost inequality: take a prime $p$ which divides $d_1 .$ Then a Sylow $p$-subgroup of $G$ can't be generated by fewer than $k$ elements, so $G$ itself certainly can't be generated by fewer than $k$ elements, as each Sylow $p$-subgroup of $G$ is a homomorphic image of $G.$ On the other hand, take a minimal generating set $S$ for $G$ of maximal cardinality, and minimize the sum of the orders of elements of $S$ subject to that. Then each element of $S$ must have prime power order, for if $s \in S$ has order divisible by more than one prime, then we may write $s = t + u $ where $t$ and $u$ have coprime orders (each greater than one) whose product is the order of $s$. Then $(S \backslash \{ s \}) \cup \{t,u\}$ is still a minimal generating set for $G,$ contradicting the maximality of the cardinality of $S.$ The fact that $S$ is a minimal generating set means that if we now collect the elements of $S$ whose orders are powers of a fixed prime $p$, we must obtain a generating set for a Sylow $p$-subgroup of $G,$ and this must be minimal by the choice of $S$. Hence the cardinality of $S$ is at most $n,$ as defined above.

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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 $s$? –  Calc Dec 5 '11 at 10:05
    
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. –  Geoff Robinson Dec 5 '11 at 10:33
    
Sorry, one more thing. What does "minimize the sum of the orders of elements of $S$ subject to that" mean? –  Calc Dec 5 '11 at 10:55
    
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 $s$. –  Geoff Robinson Dec 5 '11 at 11:00
    
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? –  Calc Dec 5 '11 at 11:11
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