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Recently I've been looking through Lang's Algebra, and I encountered a problem in the proof of Proposition 4.3.6 in Chapter I Groups.

Let G be a finite Abelian group. If G is not cyclic, then there exists a prime p and a subgroup of G isomorphic to C x C, where C is cyclic of order p.

I understand that one way to prove this is using the structure theorem, however in Lang it is said that a direct proof, without proving the structure theorem is possible. I understand that, the subgroup of G should contain p^2 elements, and should not be cyclic but I have no idea on how to proceed. Any help would be appreciated.

Abhijith Rao
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    You can use the proof that every group of order $p^2$ is abelian and either cyclic or isomorphic to $C_p\times C_p$, see here at this site. – Dietrich Burde May 11 '21 at 16:10
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    @DietrichBurde did you post the link to the site? – Abhijith Rao May 11 '21 at 16:16
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    Yes, I did. See above. You can also search this site. Most of the standard results are explained here. – Dietrich Burde May 11 '21 at 16:29
  • I really do not "see" the fact that a group with $$2^3\cdot 227^4\cdot 2027^{17}$$ elements, known to be not cyclic, has a subgroup isomorphic to either $(\Bbb Z/2)^{\times 2}$, or $(\Bbb Z/227)^{\times 2}$, or $(\Bbb Z/2027)^{\times 2}$, when i am allowed to only use the linked closing reason, which mainly considers only groups of order $p^2$ for some prime $p$... (And moreover not use structure theorems...) @DietrichBurde – dan_fulea May 11 '21 at 16:45
  • @dan_fulea Yes, but the OP said, that Lang's proof already has done it, up to the point where we have an abelian group of order $p^2$. Then the OP doesn't know how to continue, and wanted a link. He said "I understand that, the subgroup of G should contain p^2 elements, and should not be cyclic but I have no idea on how to proceed". – Dietrich Burde May 11 '21 at 16:47
  • @DietrichBurde My understanding is, that the OP wants a proof without the structural "final state of the art" for the proposition: Every non-cyclic, finite, abelian group has - for some suitable prime $p$ - a subgroup isomorphic to $(\Bbb Z/p)\times (\Bbb Z/p)$. Then, the OP only "understands" that such a candidate should have $p^2$ elements and should be non-cyclic. (But not how to get here with an ad-hoc proof.) Well... if there is no further comment, closing the issue is OK. – dan_fulea May 11 '21 at 16:55

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Consider two factors a,b of G that their period are not coprime.

Notice that if such a,b does not exist, then G is a cyclic group.(using induction and the fact that if a,b has coprime period, then they generates a cyclic subgroup of G)

(I come to answer this because I have encountered the same question :D)

Sinnaruil
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