Subgroups of a finite abelian group

Question

Let $$G=\mathbb{Z}/p_1^{e_1}\times\cdots\times\mathbb{Z}/p_n^{e_n}$$ be any finite abelian group.

What are $G$'s subgroups? I can get many subgroups by grouping the factors and multiplying them by constants, for example: If $$G=\mathbb{Z}/3\times \mathbb{Z}/9\times \mathbb{Z}/4\times \mathbb{Z}/8,$$ then I can take $$H=3(\mathbb{Z}/3\times \mathbb{Z}/9)\times 2(\mathbb{Z}/4) \times \mathbb{Z}/8.$$ Do I get all subgroups that way? (I'm interested in all subgroups, not just up-to-isomorphism).

Which are the subgroups $H$ in $G$ for which $G/H$ is primary cyclic?

Answer 1

These three answers were originally comments. I am answering the part of the question which was deleted:

Is there anything else (interesting) to say about the collection of subgroups of an [finite] abelian group.

1. This paper: Ganjuškin, A. G. Enumeration of subgroups of a finite abelian group (theory). Computations in algebra and combinatorial analysis, pp. 148–164, Akad. Nauk Ukrain. SSR, Inst. Kibernet., Kiev, 1978 gives an algorithm for enumerating all subgroups of a finite Abelian group.

Update. The answer by Amritanshu Prasad, comments by Derek Holt and Robin Chapman here provide much more (very interesting) information about enumerating subgroups.

1. On the other hand, the elementary theory of pairs $(A,H)$ where $A$ is a finite Abelian group and $H$ is its subgroup is undecidable (see Taĭclin, M. A. Elementary theories of lattices of subgroups, Algebra i Logika 9 1970 473–483 and references there). Hence there cannot be a nice description of subgroups of finite Abelian groups (say, one cannot represent a pair $(A,H)$ as a direct product of pairs of sizes bounded in terms of the period of $A$).

2. This is a better reference than Taiclin. Slobodskoĭ, A. M.; Fridman, È. I.: Theories of abelian groups with predicates that distinguish subgroups. Algebra i Logika 14 (1975), no. 5, 572–575.

Update In fact, in the paper, Sapir, M. V. Varieties with a finite number of subquasivarieties. Sibirsk. Mat. Zh. 22 (1981), no. 6, 168–187, I proved that for every prime $p$ one cannot find finitely many pairs $(A_i,H_i)$ such that every pair $(A,H)$ where $A$ is Abelian group of period $p^6$ (it is not true for $p^5$) is a direct product of copies of $(A_i,H_i)$.

Answer 2

The problem of enumerating subgroups of a finite abelian group is both non-trivial and interesting. It is also worthwhile to study this set as a lattice. As far as I know, the reference "Subgroup lattices and symmetric functions" by Lynne M. Butler (Memoirs of the AMS, no. 539; MR1223236) reflects the state of art. It is beautifully written and has a good survey of the history of the problem.

Answer 3

It's worth noting this special case: if $G$ is isomorphic to $(\mathbb{Z}/p)^r$ then it can be regarded as a vector space of dimension $r$ over the field $\mathbb{Z}/p$, and the subgroups are just the subspaces. The number of linearly independent lists of length $n$ is

$(p^r-1)(p^r-p)\dotsb(p^r-p^{n-1})$

(choose a nonzero vector $v_1$, then a vector $v_2$ not in the one-dimensional space spanned by $v_1$, then $v_2$ not in the 2-dimensional space spanned by $v_1$ and $v_2$, and so on). For any subspace $V$ of dimension $n$, the number of bases is

$(p^n-1)(p^n-p)\dotsb(p^n-p^{n-1})$

by the same argument. It follows that the number $N_n$ of subspaces of dimension $n$ is

$ N_n = \frac{(p^r-1)\dotsb(p^r-p^{n-1})}{(p^n-1)\dotsb(p^n-p^{n-1})} = \frac{(p^r-1)(p^{r-1}-1)\dotsb(p^{r-n+1}-1)}{(p^n-1)(p^{n-1}-1)\dotsb(p-1)} $