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 $p_1^{\alpha_1}, \ldots, p_n^{\alpha_n}$.
Let $S$ be the collection of all possible minimal generating sets. Minimal here means that removing an element of $s \in S$ makes it into a non generating set for $G$.
The isomorphisms of 1) and 2) provide two elements $s_1, s_2 \in S$.
a) Is it true that the cardinality of $s_1$ is the minimum among the cardinalities of all $s \in S$?
b) The second description provides us with lots of elements of $S$. Indeed, by using the chinese reminder theorem, summing two generators for $\mathbb{Z}_{p_i^{\alpha_i}}$ and $\mathbb{Z}_{p_j^{\alpha_j}}$ if $p_i \neq p_j$, we get a generator for $\mathbb{Z}_{p_i^{\alpha_i} p_j^{\alpha_j}}$. Is it true that all elements in $S$ can be obtained in this way?
(I hope it is clear what I mean with b)
Best Answer
a. Yes, this is true. Choose a prime $p|d_1$. Then $G/pG$ will be a vector space over $\mathbb{F}_p$ of dimension $k$. This requires at least $k$ generators, so the same is true of $G$.
b. If I understand the question correctly, no. Consider $G=\mathbb{Z}/p^2 \oplus \mathbb{Z}/p$. The generating set $\{(1,0),(1,1)\}$ is minimal, but cannot be obtained in the way you have described.