Any factor group of a finite abelian group is isomorphic to some subgroup - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T11:25:47Z http://mathoverflow.net/feeds/question/34874 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/34874/any-factor-group-of-a-finite-abelian-group-is-isomorphic-to-some-subgroup Any factor group of a finite abelian group is isomorphic to some subgroup Daniel Donnelly 2010-08-08T00:41:24Z 2010-08-11T14:54:13Z <p>If you visit this <a href="http://www.springerlink.com/content/ug8h1563j3484211/" rel="nofollow">link</a>, you'll see at the top of the PDF view. Basic properties of finite abelian groups:</p> <p>Every quotient group of a finite abelian group is isomorphic to a subgroup.</p> <p>If the above statement true, it would make some proofs in Serge Lang's Algebra easier, particularly in the p-Sylow groups section.</p> <p>I know that there is a correspondence between subgroups of G/N and subgroups of G containing N, but the corresponding groups are not necessarily isomorphic or are they?</p> http://mathoverflow.net/questions/34874/any-factor-group-of-a-finite-abelian-group-is-isomorphic-to-some-subgroup/34876#34876 Answer by Keivan Karai for Any factor group of a finite abelian group is isomorphic to some subgroup Keivan Karai 2010-08-08T01:15:06Z 2010-08-08T01:15:06Z <p>The quotients of an abelian group are in bijection with the subgroups of its Pontryagin dual. Now, every finite abelian group is isomorphic to its dual.</p> http://mathoverflow.net/questions/34874/any-factor-group-of-a-finite-abelian-group-is-isomorphic-to-some-subgroup/34877#34877 Answer by Pete L. Clark for Any factor group of a finite abelian group is isomorphic to some subgroup Pete L. Clark 2010-08-08T01:16:21Z 2010-08-11T14:54:13Z <p>The result you are interested in is Theorem 19 on page 8 of </p> <p><a href="http://www.math.uga.edu/~pete/4400algebra2point5.pdf" rel="nofollow">http://www.math.uga.edu/~pete/4400algebra2point5.pdf</a></p> <p>As I explain there, this fact is a kind of duality statement, but it lies deeper than the fact that passage to the dual group takes injections to surjections and conversely (Proposition 16). To deduce Theorem 19 from Proposition 16, one needs the fact that a finite abelian group is [oy vey -- <em>at least</em>] non-canonically isomorphic to its own dual group (Theorem 20), which I go on to prove in Section 5 of these notes in the most elementary way I know how. </p> <p>Note that the first step in the proof of Theorem 20 develops the Sylow theory of finite abelian groups from scratch -- this is much easier than the nonabelian case.</p>