Let $p$ be a prime number and let $n\in \mathbb{N}$. I know that every abelian group of order $p^n$ is uniquely a direct sum of cyclic groups of order $p^{\alpha_i}$ where $\sum \alpha_i = n$. Now the question:
Among all abelian groups of order $p^n$ which one has the most number of subgroups? Actually, I am looking for the Max number of subgroups so a close upper bound for the maximum number of subgroups would also be appreciated.
ADDED LATER: So far two persons submitted a solution, suggesting that the maximum number of subgroups is $2^n$ (Which is not true, consider $\mathbb{Z}_2\times\mathbb{Z}_2$, an abelian group with $2^2$ elements and $5$ subgroups). They deleted their solution because there were some gaps.
A slightly more general result is contained in the paper
I understand from the MathSciNet review that in this paper it is proved that among all finite groups of order $p^{n}$, where $p$ is a prime, the elementary abelian one has the maximum number of subgroups.
The same result should also appear in a more general form in the following paper.
I understand from the review that the following result is proved there. If $G$ is a group of order $p^{n}$, and for some $k$, with $1 < k < n$, the number of subgroups of $G$ of order $p^{k}$ is at least the corresponding number for the elementary abelian group of order $p^{n}$, then $G$ is elementary abelian itself.
The paper
appears also to be relevant.