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Consider the category $\mathsf{FinAb}$ of finite abelian groups. The structure theorem tells us that we can write down a skeleton for this category (a set of representatives for the isomorphism classes), which consists of the groups $\mathbb{Z}/n_1 \oplus \cdots \oplus \mathbb{Z}/n_s$ with positive integers satisfying $n_1 | \cdots | n_s$. But this does not tell us anything about the morphisms between finite abelian groups.

Question. Can we improve the structure theorem in such a way that we find an explicit and easy to describe (this also includes the morphisms), countable category $\mathcal{C}$ together with an equivalence of categories $\mathcal{C} \cong \mathsf{FinAb}$?

Of course, "the full subcategory consisting of all $\mathbb{Z}/n_1 \oplus \cdots \oplus \mathbb{Z}/n_s$" is no answer.

Variants:

a) There is an equivalence of categories $\mathsf{FinAb} \cong \mathsf{FinAb}^{op}$, given by $\hom(-,\mathbb{Q}/\mathbb{Z})$. We could require that $\mathcal{C}$ also comes equipped with an explicit anti-equivalence and that $\mathcal{C} \cong \mathsf{FinAb}$ is compatible.

b) Actually $\mathsf{FinAb}$ is a symmetric monoidal (abelian) category with the usual tensor product $\otimes_{\mathbb{Z}}$ (without unit). It would be great if we can set up an equivalence of symmetric monoidal categories $\mathcal{C} \cong \mathsf{FinAb}$; in particular $\mathcal{C}$ should be symmetric monoidal.

c) Probably everything may be reduced to $\mathsf{FinAb}_p$, the category of finite abelian $p$-groups, where $p$ is a prime number.

d) I am also happy with the category of finitely generated abelian groups. This is the initial finitely cocomplete symmetric monoidal abelian category.

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Isn't a homomorphism going to be specified by a matrix? I guess the hard part is knowing when two homomorphisms are the same... –  Zhen Lin Apr 22 '12 at 8:39
    
At least for endomorphisms of $\mathbb{Z}/n_1 \oplus \dotsc \oplus \mathbb{Z}/n_s$, a matrix representation is known. See msri.org/people/members/chillar/files/autabeliangrps.pdf –  Martin Brandenburg Apr 22 '12 at 9:12
    
On further reflection I think it's actually easy to tell when two matrices represent the same homomorphism; the hard part is knowing whether a matrix represents a homomorphism in the first place. But isn't the only obstruction to the construction of homomorphisms the orders of the generators? –  Zhen Lin Apr 22 '12 at 9:22
    
@Zhen: Sure this is only application of universal properties. But this is not quite what I'm looking for. –  Martin Brandenburg Apr 22 '12 at 11:23

1 Answer 1

up vote 5 down vote accepted

As you notice in a comment, $FinAb$ is the direct sum of the $FinAb_p$, so we need only deal with the latter.

It should not be difficult to describe the full subcategory $S_p$ of $FinAb_p$ spanned by the set $\{\mathbb Z/p^r\mathbb Z:r\geq0\}$, which is a skeleton, by generators and relations as a category enriched in abelian groups. Once you do that, $FinAb_P$ is (equivalent to) what I would write $S^\oplus$, the category whose objects are finite sequences of objects of $S$ and whose morphisms are matrices.

The tensor product "comes" from the functor on $S$ which we can write $\min$ for hopefully obvious reasons. In the generating set for $S$ there will be one map $\alpha_k:\mathbb Z/p^r\mathbb Z\to \mathbb Z/p^{r+1}\mathbb Z$ and one map $\beta_r:\mathbb Z/p^r\mathbb Z\to\mathbb Z/p^{r-1}\mathbb Z$ (which can be taken to be the maps such that $1\mapsto p$ and $1\to 1$, respectively): the Pontryagin duality is induced by a functor on $S$ which is the identity on objects and on endomorphism rings, and which swaps the $\alpha$s and the $\beta$s.

Later. I think $S_p$ can be presented as the quotient of the $\mathbb Z$-span of the category generated by the graph $$ 0\underset{\alpha_0}{\overset{\beta_1}\leftrightarrows} 1\underset{\alpha_1}{\overset{\beta_2}\leftrightarrows} 2\underset{\alpha_2}{\overset{\beta_3}\leftrightarrows} 3\underset{\alpha_3}{\overset{\beta_4}\leftrightarrows} \cdots$$ modulo the relations $$p^rI_r=0, \qquad \alpha_{r-1}\beta_r=pI_r=\beta_{r+1}\alpha_r,$$ with $I_r$ the identity of the object $r$. I did not really check, but the tensor product seems to be the unique $\mathbb Z$-linear functor which acts like $\min$ on objects.

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I like this decomposition, thanks! Is there some simplical interpretation of $S_p$? –  Martin Brandenburg Apr 22 '12 at 13:39
    
The edit is very interesting, thanks! I will check the details later. –  Martin Brandenburg Apr 24 '12 at 8:22

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