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I have an abelian group $G$ of order $m$. And I want to know if there is any subgroup $H$ with order $n$. The condition of the Lagrange's theorem ($m = 0\ (mod\ n)$) seems to be necessary but insufficient.

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Do you know that finite abelian groups can be written as a direct product? – mixedmath♦ Sep 12 '11 at 13:24
What examples have you seen of insufficiency? – lhf Sep 12 '11 at 13:36
@lhf, I didn't find any. So as a proof of insufficient. – Ximik Sep 12 '11 at 13:38

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The converse of Lagrange's theorem holds for finite abelian groups. Here is an outline of a proof:

  • Every finite abelian $p$-group is a product of cyclic subgroups and so the converse of Lagrange's theorem holds for abelian $p$-groups.
  • Every finite abelian group is the product of its Sylow subgroups. By the previous result, the converse of Lagrange's theorem holds for all abelian groups.
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Can you please add a "finite" in your first line... – user1729 Sep 12 '11 at 14:24
@Swlabr, done, thanks for the nudge. – lhf Sep 12 '11 at 14:27
You're welcome. – user1729 Sep 12 '11 at 14:59

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