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 $s_1, s_2 \in S$.

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?

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.

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)

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 $s_1, s_2 \in S$.

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?