It is well known that every finite abelian group is a direct product of cyclic groups. So for every $n$ every finite abelian group of exponent $n$ is a direct product of cyclic groups of order at most $n$.

Consider pairs $(A,H)$ where $A$ is a finite abelian group, $H$ is its subgroup. One can define a direct product $(A_1,H_1)\times (A_2,H_2)=(A_1\times A_2, H_1\times H_2)$. Let $n$ be a natural number.

Is there a finite number of pairs $(A_i,H_i)$ such that every pair $(A,H)$ where $A$ is of exponent $n$ is a subpair of a direct product of several copies of $(A_i,H_i)$ (each pair can be used several times)?

Here we say that $(A, H)$ is a subpair of $(A', H')$ if $A<A'$ and $H=H'\cap A$.

  • $\begingroup$ Yes, just take the family of pairs $(A,H_i)$ with $H_i$ ranging over subgroups of $A$. $\endgroup$
    – YCor
    Jun 10, 2020 at 22:17
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    $\begingroup$ The finite set of pairs should be independent of $A$. It shuld be clear from the question: every pair $(A,H)$ where $A$ of exponent $n$ should be equal to a direct product of copies of pairs from a finite set. $\endgroup$
    – user158834
    Jun 10, 2020 at 22:24
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    $\begingroup$ Not every subgroup of $A\times B$ is of the form $H\times K$ where $H\leq A$ and $K\leq B$. If you want equality, you’re probably out of luck. Shouldn’t you want some kind of equivalence, e.g., $(A,H)\sim (B,K)$ iff there exists an isomorphism $A\to B$ whose restriction maps $H$ isomorphically onto $K$? $\endgroup$ Jun 10, 2020 at 22:44
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    $\begingroup$ I think you still want some kind of “isomorphism” somewhere. Otherwise, for abstract groups, asking for inclusion is way too restrictive. $\endgroup$ Jun 10, 2020 at 23:22
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    $\begingroup$ Another way to formulate the question: view a pair $(A, H)$ as an algebraic system with a binary operation defining an Abelian group and a unary predicaebdegining $H$. If we assume the exponent equal $n$ we get a variety of algebraic systems. The quedtion asks for which $n$ thia variety has only finitely many subdirectly irreducible finite systems. $\endgroup$
    – user158834
    Jun 10, 2020 at 23:56

1 Answer 1


Yes, this is true for any $n$, and one only needs twice as many building blocks as in the ordinary case.

I’m going to state it for not necessarily finite abelian groups. The basic observation here is that the diagonal map embeds any pair $(A,H)$ into $(A,A)\times(A/H,0)$. In view of $\prod_i(A_i,A_i)=\bigl(\prod_iA_i,\prod_iA_i\bigr)$ and $\prod_i(A_i,0)=\bigl(\prod_iA_i,0\bigr)$, this reduces the problem to the known classification of abelian groups:

Proposition: Any structure $(A,H)$ where $A$ is an abelian group and $H\le A$ is a subdirect product of such structures where

  • $A$ is subdirectly irreducible (i.e., a cyclic group of prime power order, or a Prüfer $p$-group), and
  • $H=A$ or $H=0$.

There is nothing special about abelian groups: the same argument applies to noncommutative groups and normal subgroups, and more generally, to arbitrary algebras and congruences. Moreover, it applies even if we have an indexed sequence of congruences, where we may additionally impose inclusion constraints between them. Let $\Delta_A=\mathrm{id}_A$ and $\nabla_A=A\times A$ denote the smallest and largest equivalence relations on $A$, respectively.

Proposition: Let $K$ be a quasivariety and $(I,\le)$ a fixed partial order. Consider structures $(A,\{\theta_i:i\in I\})$ such that

  • $A\in K$,
  • each $\theta_i$ is a $K$-congruence on $A$ (i.e., $A/\theta_i\in K$), and
  • $i\le j\implies\theta_i\subseteq\theta_j$.

Then any such structure is a subdirect product of structures of the form $(A,\{\theta_i:i\in I\})$ where

  • $A$ is relatively subdirectly irreducible in $K$, and
  • either $\theta_i=\nabla_A$ for all $i\in I$, or there exists $j\in I$ such that $$\theta_i=\begin{cases}\Delta_A,&i\le j,\\\nabla_A,&\text{otherwise.}\end{cases}$$

Proof: For any $A\in K$, let us denote $A^\varnothing=(A,\{\nabla_A:i\in I\})$, and for $j\in J$, $A^j=(A,\{\theta_i^j:i\in I\})$, where $$\theta_i^j=\begin{cases}\Delta_A,&i\le j,\\\nabla_A,&\text{otherwise.}\end{cases}$$ Then given any structure $(A,\{\theta_i:i\in I\})$, the diagonal map embeds it (as a subdirect product) into $$A^\varnothing\times\prod_{i\in I}(A/\theta_i)^i,$$ and since the $B^\varnothing$ and $B^i$ operators commute with products and substructures, we can write each factor as a subdirect product of $B^\varnothing$ or $B^i$ where the $B$’s are relatively subdirectly irreducible in $K$.

  • $\begingroup$ One could further generalize it from partial orders to finitary closure operators $C\colon\mathcal P(I)\to\mathcal P(I)$, where we require $\bigcap_{j\in J}\theta_j\subseteq\theta_i$ whenever $i\in C(J)$. The subdirect decomposition is then to $(A,\{\theta_i:i\in I\})$ where $A$ is sdi in $K$, each $\theta_i$ is $\Delta_A$ or $\nabla_A$, and $\{i\in I:\theta_i=\nabla_A\}$ is either $I$ or a completely meet-irreducible closed set. However, this all appearently works only if $K$ is congruence-distributive, hence it is not a straightforward generalization. $\endgroup$ Jun 12, 2020 at 13:51

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