I am interested in finding all of the subgroups (up to isomorphism) of a finite Abelian group $A$.

I know the following:

-- A finite Abelian group $A$ can be represented as a direct product of cyclic groups, $A_1, A_2, A_3, ...$.

-- The direct product of subgroups of a set of groups is always a subgroup of the direct product of the groups.

-- Each of the cyclic groups $A_1, A_2, A_3, ...$ will have a unique subgroup for each divisor of its order.

To start finding subgroups of $A$ I can therefore:

-- Write down all the subgroups of each of $A_1, A_2, A_3, ...$

-- Form all possible direct products of these subgroups

I realise that this method will not necessarily find all subgroups of $A$. (For example, if $A = \mathbb{Z}_2 \times \mathbb{Z}_2$, then it has a subgroup $ \{(0, 0), (1, 1) \}$ which is not a direct product of subgroups of $\mathbb{Z}_2$.)

However my question is this: will the above method find representatives from all isomorphism classes of subgroups of $A$? If no, can you provide a counterexample? If yes, can you provide a proof?

  • 1
    $\begingroup$ Some clarity is needed about your question. Initially you say you want to count upto isomorphism; that means two different subsets if they are subgroups and isomorphic you want to count them as one, and not as 2. But later you object you are not able to find a subgroup which is isomorphic to one already found. $\endgroup$ – P Vanchinathan Apr 5 '14 at 9:22
  • $\begingroup$ Thanks for your comment. I am only interested in establishing a general method for finding all subgroups of a finite abelian group up to isomorphism. $\endgroup$ – user123473 Apr 5 '14 at 9:32

The first thing I would do is to reduce to abelian $p$-groups: If $A = G_1 \times \cdots \times G_r$ where each $G_i$ is a $p_i$-Sylow subgroup of $A$, then every subgroup $B$ of $A$ is a product of subgroups $B_i$ of $G_i$'s. (in fact $B_i$ will be the $p_i$-Sylow subgroup of $B$.)

So assume $A$ is an abelian $p$-group, say $A=\prod_{i=1}^k (\mathbb{Z}/p^{e_i} \mathbb{Z})^{r_i}$. In this case, clearly every $i$ can contributes up to $r_i$ direct factors of order dividing $p^{e_i}$. I think it is straightforward, although tedious, to show that no other subgroups can occur. (I haven't wrote the details, so one have to check...)

So I think the answer to your question is: YES

  • $\begingroup$ In the first paragraph you say that "every subgroup $B$ of $A$ is a product of subgroups $B_i$ of $G_i$'s." How do you know that this is always true? $\endgroup$ – user123473 Apr 6 '14 at 1:42
  • $\begingroup$ I don't really understand your second paragraph. Would it be possible to explain a little more? Thanks! $\endgroup$ – user123473 Apr 6 '14 at 1:43
  • $\begingroup$ For your first question: Take B_i to be the p_i sylow subgroup of B, then it is contained in the unique p_i subgroup of A as it is a p_i-group. $\endgroup$ – Lior B-S Apr 16 '14 at 9:20

Birkhoff, Garrett; Subgroups of Abelian Groups. Proc. London Math. Soc. S2-38 no. 1, 385.


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