I have a finite abelian group $G$ of order $n$.
What is the maximum number of minimal ways to construct a non zero element $g \in G$ from elements in $G$?
By "way" and "minimal", I mean the following. A way to construct $g$ can be described by a subset $H \subset G$ that is summed to form $g$. $H$ is considered minimal if there is no subset of $H$ such that the sum of this subset is equal to the zero element. Perhaps there is a better way to name this, but minimal is the best I could come up with.
Example:
Let $G = \mathbb{Z}^2 \mod 2$, and $g = (0,1)$. Then there are two minimal ways to construct $g$. $H_1 = \{(0,1)\}$ and $H_2 = \{(1,0),(1,1)\}$. For any $g$ there are maximally 2 ways, so 2 is the value I'm looking for.
For $G = \mathbb{Z}^3 \mod 2$, I got 8 (identical for all nonzero $g$).
Thoughts so far:
At first I was thinking it should be related to the minimal generating set of $G$, which is bounded in size by $\log_2 n$ (link). But when I partially worked out the case of $\mathbb{Z}^2_4$ I was getting numbers greater than $n$. I also found this stack answer but I honestly can't tell if it's related to my question or not.
Is there an easy way to bound this value?
