4
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ How do you define $n$? $\endgroup$ Dec 4, 2011 at 11:15
  • $\begingroup$ 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. $\endgroup$
    – Calc
    Dec 4, 2011 at 11:31
  • $\begingroup$ 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. $\endgroup$ Dec 4, 2011 at 11:46
  • $\begingroup$ Ok. Sorry. How can one erase a question? ;) $\endgroup$
    – Calc
    Dec 4, 2011 at 11:59
  • 1
    $\begingroup$ Also posted in math.SE: math.stackexchange.com/questions/88106/… $\endgroup$ Dec 4, 2011 at 21:38

1 Answer 1

6
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ 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 $s$? $\endgroup$
    – Calc
    Dec 5, 2011 at 10:05
  • $\begingroup$ 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. $\endgroup$ Dec 5, 2011 at 10:33
  • $\begingroup$ Sorry, one more thing. What does "minimize the sum of the orders of elements of $S$ subject to that" mean? $\endgroup$
    – Calc
    Dec 5, 2011 at 10:55
  • $\begingroup$ 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 $s$. $\endgroup$ Dec 5, 2011 at 11:00
  • $\begingroup$ 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? $\endgroup$
    – Calc
    Dec 5, 2011 at 11:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.