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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.

user390026
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    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 '19 at 06:40
  • 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 '19 at 10:54
  • @DerekHolt thanks. You reminded me of Fermat’s claim on his famous last theorem. – user390026 Apr 23 '19 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 $\mathbb{Z}_p$. – user390026 Apr 23 '19 at 14:59

1 Answers1

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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.

Andreas Caranti
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