Subgroups of a finite abelian group - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T14:05:22Z http://mathoverflow.net/feeds/question/46115 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46115/subgroups-of-a-finite-abelian-group Subgroups of a finite abelian group uuu 2010-11-15T14:47:48Z 2010-11-19T12:24:51Z <p>Let $$G=\mathbb{Z}/p_1^{e_1}\times\cdots\times\mathbb{Z}/p_n^{e_n}$$ be any finite abelian group.</p> <p>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).</p> <p>Which are the subgroups $H$ in $G$ for which $G/H$ is primary cyclic?</p> http://mathoverflow.net/questions/46115/subgroups-of-a-finite-abelian-group/46300#46300 Answer by Mark Sapir for Subgroups of a finite abelian group Mark Sapir 2010-11-16T23:54:24Z 2010-11-19T12:24:51Z <p>These three answers were originally comments. I am answering the part of the question which was deleted:</p> <p><b>Is there anything else (interesting) to say about the collection of subgroups of an [finite] abelian group. </b></p> <ol> <li>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. </li> </ol> <p><b> Update.</b> The answer by Amritanshu Prasad, comments by Derek Holt and Robin Chapman here provide much more (very interesting) information about enumerating subgroups. </p> <ol> <li><p>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 $A$).</p></li> <li><p>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. </p></li> </ol> <p><b> Update </b> 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 $(A_i,H_i)$.</p> http://mathoverflow.net/questions/46115/subgroups-of-a-finite-abelian-group/46323#46323 Answer by Amritanshu Prasad for Subgroups of a finite abelian group Amritanshu Prasad 2010-11-17T03:58:16Z 2010-11-17T03:58:16Z <p>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.</p> http://mathoverflow.net/questions/46115/subgroups-of-a-finite-abelian-group/46604#46604 Answer by Neil Strickland for Subgroups of a finite abelian group Neil Strickland 2010-11-19T08:41:04Z 2010-11-19T09:42:57Z <p>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> <p>$(p^r-1)(p^r-p)\dotsb(p^r-p^{n-1})$ </p> <p>(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> <p>$(p^n-1)(p^n-p)\dotsb(p^n-p^{n-1})$ </p> <p>by the same argument. It follows that the number $N_n$ of subspaces of dimension $n$ is</p> <p>$N_n = \frac{(p^r-1)\dotsb(p^r-p^{n-1})}{(p^n-1)\dotsb(p^n-p^{n-1})} = \frac{(p^r-1)(p^{r-1}-1)\dotsb(p-1)}{(p^n-1)(p^{n-1}-1)\dotsb(p-1)}$ </p>