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I learned from MO Subgroups of a finite abelian group that the problem of enumerating subgroups (not up to isomorphism) of finite abelian groups is a difficult one.

Are there simple formulas if one restricts to low rank for the subgroups? For example, are there formulas for enumerating cyclic subgroups, or subgroups whose minimal number of generators is $2$?

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Some results in this direction were obtained by P.E. Djubjuk (IAN SSSR, Ser. Mat. (\bf12} (1948, 352-378 DAN SSSR {\bf137} (1961), 506-508. See also Sec. 198 in `Reviews on Finite Groeps' (editor D. Gorenstein). – yakov Jun 27 at 21:40
up vote 7 down vote accepted

I think that you can find the formulas that you are looking for in the paper "An arithmetic method of counting the subgroups of a finite abelian group" by Marius Tarnauceanu, Bull. Math. Soc. Sci. Math. Roumanie (N.S.) 53(101) (2010), no. 4, 373–386.

In particular, Theorem 4.3 seems relevant, but there are other results that might be of interest to your question.

The paper can be downloaded from the journal's website.

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Thank you very much for your answer! I am very surprised that the answer to this question seems to be the main topic of a contemporary research paper. I would have expected that at least the cyclic case would be reasonably doable. – user22518 Apr 4 '12 at 12:57

If $G$ is an abelian $p$-group and $p^k\le\exp(G)$, then the number of cyclic subgropups of order $p^k$ in $G$ is $$ {\rm c}_k(G)=\frac{|\Omega_k(G)-\Omega_{k-1}(G)|}{(p-1)p^{k-1}}. $$ If the type of $G$ is given, it is easy to compute $|\Omega_k(G)|$. The displayed formula is also suit for the regular $p$-groups. Computation of subgroups of given order is known for not very complicated $p$-groups, for example, for metacyclic $p$-groups (see \S 124 in the book of Berkovich-Janko; in that series a great number of counting theorems is proved, in particular, celebrated Kulakoff's theorem for all $p$).

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