# Minimal generation for finite abelian groups

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)

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In general, generator of type 2) do not provide a minimal set (take for example $G = \mathbb{Z} / 6 \mathbb{Z} \simeq \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 3 \mathbb{Z}$). Also I don't think type 1) is the only way to have a minimal set of generators (take for example $G = \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 30 \mathbb{Z} \simeq \mathbb{Z} / 6 \mathbb{Z} \times \mathbb{Z} / 10 \mathbb{Z}$) –  Joel Cohen Dec 3 '11 at 22:05
Minimal $\neq$ minimum. If in your example of $\mathbb{Z}_6$ you remove one of the two generators, you no longer generate the whole group. So the set of generators $\{\overline{2}, \overline{3}\}$ is minimal. –  calc Dec 4 '11 at 8:17
Also posted in MO. In the future, please indicate on both sites that the question has been asked elsewhere (I do appreciate that you waited before posting in MathOverflow after posting here). By letting both sites know that the question has been posted in the other site, you prevent duplication of effort. You should note in your MO post that your question has been answered. –  Arturo Magidin Dec 4 '11 at 21:40
Thank you Arturo, and sorry for that. The two questions are almost, but not exactly the same. –  calc Dec 5 '11 at 10:06

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.