# Number of subgroups of an abelian p-group

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.

• The subgroups of $\Bbb{Z}_p^n$ are the elements of $M_n(\Bbb{Z}_p)/GL_n(\Bbb{Z}_p)$. From $G = \prod_j \Bbb{Z}_{p^{a_j}}$ and $\phi_j$ the morphism reducing $\bmod p$ the $j$-th one and $\Phi_j$ the corresponding map on the set of subgroups, how to evaluate $|\Phi_j^{-1}(H)|$ ? If we show it is $\le \frac{p^{a_j}-1}{p-1}$ the number of ways to embed $\Bbb{Z}_p$ in $(\Bbb{Z}_p)^{a_j}$ then we are done. – reuns Apr 23 at 6:40
• The number of subgroups is certainly largest when the group is elementary abelian. But I am afraid I am not going to write down a proof. – Derek Holt Apr 23 at 10:54
• @DerekHolt thanks. You reminded me of Fermat’s claim on his famous last theorem. – Sara.T Apr 23 at 14:48
• @DerekHolt If you are right, then the maximum number of subgroups is equal to the number of subspaces of a vector space of dimension $n$ over $\mathbb{Z}_p$. – Sara.T Apr 23 at 14:59

A slightly more general result is contained in the paper

Yun Fan, A characterization of elementary abelian $$p$$-groups by counting subgroups. (Chinese) Math. Practice Theory 1988, no. 1, 63–65.

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.

Yakov Berkovich and Zvonimir Janko, Structure of finite p-groups with given subgroups. Ischia group theory 2004, 13–93, Contemp. Math., 402, Israel Math. Conf. Proc., Amer. Math. Soc., Providence, RI, 2006.

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

Haipeng Qu, Finite non-elementary abelian $$p$$-groups whose number of subgroups is maximal. Israel J. Math. 195 (2013), no. 2, 773–781

appears also to be relevant.