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Is it known an explicit formula for the number of subgroups of a given exponent of a finite abelian $p$-group?

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up vote 12 down vote accepted

I had to look this up as well at some point in my research. The answer is yes, and a Google search for "number of subgroups of an abelian group" leads to several downloadable papers, not all of them easy to read. The paper "On computing the number of subgroups of a finite abelian group" by T. Stehling, in Combinatorica 12 (1992), contains the following formula and (I think) references to where it has appeared earlier in the literature.

Let $\alpha = (\alpha_1,\dots,\alpha_\ell)$ be a partition, so that $\alpha_1\ge\cdots\ge\alpha_\ell$. (In this formula it is convenient to allow some of the parts of the partition at the end to equal 0.) Define the notation $$ {\mathbb Z}_\alpha = {\mathbb Z}/p^{\alpha_1}{\mathbb Z} \times \cdots \times {\mathbb Z}/p^{\alpha_\ell}{\mathbb Z} $$ for a general $p$-group of type $\alpha$. Define similarly a partition $\beta$, and suppose that $\beta\preceq\alpha$, meaning that $\beta_j\le\alpha_j$ for each $j$. We want to count the number of subgroups of ${\mathbb Z}_\alpha$ that are isomorphic to ${\mathbb Z}_\beta$.

Let $a=(a_1,\dots,a_{\alpha_1})$ be the conjugate partition to $\alpha$, so that $a_1=\ell$ for example; similarly, let $b$ be the conjugate partition to $\beta$. Then the number of subgroups of ${\mathbb Z}_\alpha$ that are isomorphic to ${\mathbb Z}_\beta$ is $$ \prod_{i=1}^{\alpha_1} \genfrac{[}{]}{0pt}{}{a_i-b_{i+1}}{b_i-b_{i+1}}p^{(a_i-b_i)b_{i+1}}, $$ where $$ \genfrac{[}{]}{0pt}{}nm = \prod_{j=1}^m \frac{p^{n-m+j}-1}{p^j-1} $$ is the Gaussian binomial coefficient.

To answer your specific question, you'd want to sum over subpartitions $\beta\preceq\alpha$ such that $\beta_1$ equals the exponent in question.

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I don't understand what is meant by the conjugate partition. –  Martin Brandenburg Oct 24 '11 at 9:21
    
Transpose. $\,\,$ –  Allen Knutson Oct 24 '11 at 15:49
    
Googling "conjugate partition" works too.... –  Greg Martin Oct 24 '11 at 22:49
    
It's a good answer. Thanks. –  Marius Tarnauceanu Oct 26 '11 at 6:53
    
@GregMartin Please be informed that a copy of your answer is now on Mathematics.SE, namely here. –  Lord_Farin Jul 1 '13 at 18:14
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Greg's answer is a particular case of one of the many beautiful identities that arise in the study of Hall algebras http://en.wikipedia.org/wiki/Hall_algebra with full details in Ian Macdonald's Symmetric Functions and Hall Polynomials ISBN 978-0-19-853489-1 http://ukcatalogue.oup.com/product/9780198504504.do

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Thanks for the mention of Hall algebras. I suspect I'm not the only one wondering: "Any relation to Greg Martin?" Gerhard "It's None Of My Business" Paseman, 2011.10.24 –  Gerhard Paseman Oct 24 '11 at 22:42
    
Not to the best of my knowledge :) –  Greg Martin Oct 24 '11 at 22:47
    
No relation AFIK (but I'm always on the lookout for Martins who have any family connections with the village of Cuddesdon in Oxfordshire). –  Ursula Martin Oct 25 '11 at 4:46
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