# Any factor group of a finite abelian group is isomorphic to some subgroup

If you visit this link, you'll see at the top of the PDF view. Basic properties of finite abelian groups:

Every quotient group of a finite abelian group is isomorphic to a subgroup.

If the above statement true, it would make some proofs in Serge Lang's Algebra easier, particularly in the p-Sylow groups section.

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?

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Since your motivation is to "make some proofs easier", you should explain what can be assumed. The statement you've asked about follows from any of the following: (a) classification of finite abelian groups; (b) self-injectivity of the ring $\mathbb{Z}/M\mathbb{Z}$ for any integer $M$ (applied with $M=|G|$); (c) the fact that a finite abelian group is isomorphic to its Pontryagin dual (the isomorphism is not canonical).The last sentence isn't very meaningful: if a subgroup of $G/N$ has order $k$ then its pre-image in $G$ has order $k|N|,$ so they can only be isomorphic if $N$ is trivial. –  Victor Protsak Aug 8 '10 at 1:05
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## 2 Answers

The result you are interested in is Theorem 19 on page 8 of

http://www.math.uga.edu/~pete/4400algebra2point5.pdf

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 -- at least] 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.

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.

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@Pete: I don't think the trivial group is non-canonically isomorphic to its own dual group ;-) –  Kevin Buzzard Aug 8 '10 at 20:57
@Kevin: Nor Z_2! –  Allen Knutson Aug 11 '10 at 8:31
@Pete I've never seen the proof you give in the handout of the fact that finite subgroups of the multiplicative of a field are cyclic. That was really slick. I'd only ever seen ones that invoked the classification theorem. Cool! –  Keenan Kidwell Aug 11 '10 at 19:14
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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.

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