$\DeclareMathOperator\Aut{Aut}\newcommand\card[1]{\lvert#1\rvert}$So, after going over the classification of finite abelian groups in a class I was teaching this winter, I got curious about whether it could be used to obtain a 'nice' value for the groupoid cardinality of the class of abelian groups of fixed size, that is if one could compute :

$$ \sum_{\card G=m} \frac{1}{\card{\Aut(G)}} $$

where the sum is over the isomorphism class of abelian groups with $m$ elements. That is just pure curiosity, I had absolutely no real motivation for this. One relatively quickly see that it is enough to compute it for $m=p^n$, so I started doing it for $p$, $p^2$ and $p^3$. Unsurprisingly, the results look messy at first, but then a quite surprising number of simplifications occured and I arrived at the formula :

$$ \sum_{\card G=p^n} \frac{1}{\card{\Aut(G)}} = \prod_{k=1}^n \frac{p^{k-1}}{p^k-1}. $$

I checked it by hand up to $n=5$.

In any case the simplicity of the formula suggests that there is a better way to compute this than going over all isomorphism class of abelian groups and computing $\Aut(G)$ and then taking the sum.

So basically I'm wondering if this formula is known, and if not if someone has an idea of how to prove it — ideally with an argument that doesn't involve summing over partitions of $n$.

Note : there are known formulæ for $\card{\Aut(G)}$ see for example the very end of Hillar and Rhea - Automorphisms of finite Abelian groups.

  • 4
    $\begingroup$ This other question seems highly relevant (including remarks in the question itself, its comments, and the paper by Philip Hall referred to in Richard Stanley's answer). $\endgroup$
    – Gro-Tsen
    Apr 10 at 13:47
  • $\begingroup$ @Gro-Tsen thanks. I don't why I missed it when searching on the topic ! $\endgroup$ Apr 10 at 18:41

2 Answers 2


This is Corollary 3.8 in Cohen, H.; Lenstra, H. W., Jr. Heuristics on class groups. Number theory (New York, 1982), 26–36, Lecture Notes in Math., 1052, Springer, Berlin, 1984 (MR0750661).

In this paper you will find many formulas for averages of various functions over abelian groups weighted by inverse size of the automorphism group.


Hall's original paper gives a nice combinatorial proof of this, which seems not to be mentioned in any of the follow up papers.

Given a partition $\lambda = (\lambda_1, \lambda_2, \ldots, \lambda_n)$, define the Durfee square of $\lambda$ to be the largest $\mu_1$ such that the diagram of $\lambda$ contains a $\mu_1 \times \mu_1$ square. Below is a $4 \times 4$ square, fitting inside the partition $8+7+5+4+4+3+1+1$. I took this diagram from Andrews' paper Partitions and Durfee dissection

enter image description here

If one removes $\mu_1 \times \mu_1$ from $\lambda$, one is left with two partitions, a partition $\lambda_{\downarrow}$ below the Durfee square and a partition $\lambda_{\rightarrow}$ to its right. Let $\mu_2$ be the Durfee square of $\lambda_{\downarrow}$ and continue in this way to define $\mu_3$, $\mu_4$, .... The partition $(\mu_1, \mu_2, \ldots)$ is called the Durfee dissection of $\lambda$ and was introduced by Andrews in the paper above. Here (also from that paper) is an image showing that the Durfee dissection of $8+7+5+4+4+3+1+1$ is $4+2+1+1$. I'll write $D(\lambda)$ for the Durfee dissection of $\lambda$.

enter image description here

Then Hall proves:

Theorem For any partition $\mu$, we have $$\frac{1}{\left|\text{Aut} \prod_j (\mathbb{Z}/p^{\mu^T_j} \mathbb{Z})\right|} = \sum_{D(\lambda) = \mu} \frac{1}{p^{|\lambda|}}.$$ Here $\mu^T$ is the transpose partition to $\mu$.

For example, take $\mu = (2)$. The partitions with $D(\lambda) = (2)$ are of the form $a+b$ with $a \geq b \geq 2$, and $$\sum_{a \geq b \geq 2} \frac{1}{p^{a+b}} = \frac{1}{p^4(1-p^{-1})(1-p^{-2})} = \frac{1}{(p^2-p)(p^2-1)} = \frac{1}{\left| \text{Aut}((\mathbb{Z}/p \mathbb{Z})^2 )\right|}.$$

To give another example, take $\lambda = 1+1$. The partitions with $D(\lambda) = 1+1$ are of the form $a+1$ with $a \geq 1$, and $$\sum_{a \geq 1} \frac{1}{p^{a+1}} = \frac{1}{p^2(1-p^{-1})} = \frac{1}{p^2-p} = \frac{1}{\left| \text{Aut}(\mathbb{Z}/p^2 \mathbb{Z}) \right|}.$$

Let's see how this implies the product formula in the original post. As you can see, the squares in the Durfee dissection stretch from the top of the partition to the bottom, so $|D(\lambda)|$ is the number of parts of $\lambda$. Thus, $\sum_{|G| = p^n} \frac{1}{\left| \text{Aut}(G) \right|}$ is the sum of $p^{-|\lambda|}$ over partitions $\lambda$ with $n$ parts, which is well known to be $$\frac{1}{p^{-n} \prod_{j=1}^n (1-p^{-j})} = \prod_{j=1}^n \frac{p^{j-1}}{p^j-1}$$ as desired.

I am writing this up here because I rediscovered it about five years ago, and did a lot of google and MathSciNet searches on words like "Durfee", "square", "automorphism", "Cohen-Lenstra" before finally reading Hall's paper and finding it was there. It seems to be a fact that has passed out of the range of search engines, so I thought I would put it back in.

  • $\begingroup$ This is very nice. But it seems like there is one more step to get the product formula in question, no? $\endgroup$ Apr 11 at 14:27
  • 2
    $\begingroup$ Good point, I should add this. $\endgroup$ Apr 11 at 14:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.