# Orders of Elements in Minimal Generating sets of Abelian p-Groups

I'm looking for as much information about the orders of elements in minimal generating sets of finite abelian $p$-groups as possible. What I really need is complete knowledge about the possible orders of elements in such groups and how many of each order there can be.

I know that every group of the form $(\mathbb Z_{p^{n_1}})^{m_z} \times \cdots \times (\mathbb Z_{p^{n_z}})^{m_z}$ has a minimal generating set of exactly $\sum_{k=1}^{z}m_k$ elements, and cannot be generated by less than this many elements. The obvious minimal generating set has exactly $m_k$ elements of order $p^{n_k}$, and every generating set must clearly contain an element of order $p^{n_k}$ since otherwise that set would at most generate a subgroup.

In specific, I am wondering whether every minimal generating set has exactly $\sum_{k=1}^{z}m_k$ elements as is the case for elementary abelian groups, and exactly what combinations of element orders are possible. My initial (uncertain) conjecture is that yes, every minimal generating set has exactly $\sum_{k=1}^{z}m_k$ elements and the only restriction on the orders are:

There are at least $m_k$ elements of order $p^{n_k}$ or higher.

Example: The obvious minimal generating set for $Z_3 \times Z_9 \times Z_{27}$ can be modified as follows: $\{(1,0,0), (0,1,1), (0,0,1)\}$ to get a set with one element of order 3, and two of order 27.

The ultimate goal is try to find, if possible, a paper or textbook I might be able to cite so that I don't have to re-prove theorems which are already known.

The most useful article I have found so far is this which is unnervingly recent.

• In any finite $p$-group, abelian or not, it is true that all minimal generating sets have the same number of elements. If $|G/\Phi(G)! =p^n$ (where $\Phi(G)$ is the Frattini subgroup of $G$) then all minimal generating sets have size $n$. In an abelian group $\Phi(G)$ is equal to the subgroup $\{g^p : g \in G \}$. – Derek Holt Nov 18 '13 at 21:55
• Thank you! This helps some. – Rachael Alvir Nov 23 '13 at 17:30