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The following is a question that was asked by my teacher as a ponder-upon question:to which unfortunately I have not been able to put a single forward step.

If an abelian group has subgroups of order $m\ \&\ n$, then is it true that it must have a subgroup of order $= LCM(m,n)$?

I am a beginner in group theory, so please consider explaining in detail your answer. Any reference to the proof is okay as well, but please don't down vote if its too simple.

Qwerty
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  • Yes, it's true, and it's either easy to prove, or virtually impossible to prove, depending on what you've seen so far. – Matt Samuel May 12 '16 at 18:51
  • @MattSamuel : Can you give me a proof? Or at least tell me what I need to learn before attempting its proof? – Qwerty May 12 '16 at 18:55
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    This is a double starred problem in my favorite textbook and it has a disclaimer saying the author has no idea how to prove it using only the material given in the book up to that point. Once you see the classification theorem for finite/finitely generated abelian groups, though, it's straightforward. – Matt Samuel May 12 '16 at 18:57

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For abelian groups converse of langranges theorm holds.Hence corrresponding to each divisor of of order of the group there exsists a subgroup.

Rayees Ahmad
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