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It's easy to know that the sequence of number of subgroups is unimodal for elementary abelian $p$-groups. I want to know if the result is true for any $p$-group.

More, precisely, let $G$ be a finite $p$-group with $|G|=p^n$. Write $$s_{i}(G)=\big|\{H\leq G:|H|=p^i\}\big|.$$ Is sequence $(s_{i}(G))_{1\leq i\leq n-1}$ is unimodal?

This means there exist $1\leq t\leq n-2$ such that $s_{i}(G)\leq s_{i+1}(G)$ for $i\leq t-1$ and $s_{i}(G)\geq s_{i+1}(G)$ for $i\geq t$. It's easy to show that the answer is positive when $G$ is elementary abelian.

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    $\begingroup$ I think the property you're describing is unimodality, not convexity. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jun 24, 2023 at 14:42
  • $\begingroup$ I've fixed the terminology. $\endgroup$
    – YCor
    Commented Jun 27, 2023 at 11:29

2 Answers 2

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It's true for any finite abelian $p$-group by a result of Lynne Butler, Proc. Amer. Math. Soc. 101 (1987), 771-775. See here.

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  • $\begingroup$ Thanks your answer.Is this the latest achievement in this question? $\endgroup$
    – gdre
    Commented Jun 24, 2023 at 15:01
  • $\begingroup$ @gdre: it is to my knowledge, but I haven't tried to keep up in this area. $\endgroup$
    – Richard Stanley
    Commented Jun 24, 2023 at 18:30
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Your function is not always unimodal.

The groups SmallGroup(64,222) with 7, 35, 27, 35, 15 and SmallGroup(64,245) of order $64$ with 3, 31, 15, 35, 15 are not unimodal.

There are many $2$-groups for which your function is unimodal but not convex. A random example of order $32$ is SmallGroup(32,15) with 3, 3, 7, 3.

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  • $\begingroup$ The finite 2-groups I've found for which the number of subgroups is not unimodal are as follows: SmallGroup(64,222), (64,245), (128,72), (128,98), (128,116), (128,890), (128,943), (128,961), (128,969), (128,974), (128,981), (128,985), (256,216), (256,411), (256,435), (256,442), (256,463), but then my laptop got bored. $\endgroup$
    – Dave Benson
    Commented Jun 25, 2023 at 22:34
  • $\begingroup$ But I should tell you that for $p$ odd, the problem is harder. My laptop got bored before it found any non-unimodal examples. $\endgroup$
    – Dave Benson
    Commented Jun 25, 2023 at 22:51
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    $\begingroup$ the odd $p$ case is quite intriguing (for my taste, particularly when the nilpotency class is $<p$, so that things can be rephrased in terms of subrings of Lie rings). It would be worth a separate question. [For instance if one finds a finite-dimensional nilpotent complex Lie algebra for which the sequence of dimensions of spaces of subalgebras is not unimodal then I think there also exist finite groups of exponent $p$ with the failure of unimodality, for large $p$.] $\endgroup$
    – YCor
    Commented Jun 27, 2023 at 11:26
  • $\begingroup$ @YCor I agree, the odd $p$ case seems quite delicate, and I'm not sure which way to think it'll go. $\endgroup$
    – Dave Benson
    Commented Jun 28, 2023 at 8:55

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