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.
Best Answer
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.