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?
Best Answer
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