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Original (Flawed) Question

Why is it that if a group is of the order $pqn$ where $p, q$ are distinct primes and $n$ is some integer coprime to $p$ and $q$, then there is a non-abelian subgroup of order $pq$? (I am reading some notes and the author says this without proving it, so I assume it is very elementary.)

Revised Question

Sorry about this confusing question, I think I have misunderstood it. (As @QiaoChuYuan kindly suggested.) It should be saying for a cyclic group of order $pqn$ where $p,q,n$ are as described above, AND $p$ divides $q-1$ then there is a non-abelian subgroup of order $pq$. Does this make sense now? If so could someone please tell me why it is true? Thank you.

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This is clearly false in general (take the cyclic group of order $pqn$). What source is this from? Are you sure you aren't misreading it? –  Qiaochu Yuan Feb 5 '12 at 18:25
    
Can you give a link to the notes? –  user641 Feb 5 '12 at 18:50
    
Even if the original group is non-abelian, this is not true. Take the direct product product of a cyclic group of order $pq$ and an nonabelian group of order $n$. –  N. S. Feb 5 '12 at 18:54
    
@SteveD: unfortunately they are of paper-form and not in english! –  User1835639 Feb 5 '12 at 20:08
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@User1835639: your edited statement is still clearly false, and the cyclic group is still a counterexample. Can you quote from the relevant section of the notes? I think you are misunderstanding something still. –  Qiaochu Yuan Feb 5 '12 at 20:13
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1 Answer 1

The original claim is just not true. There is an abelian group of every order.

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Even worse: there is no non-abelian group of order $3 \cdot 5$. –  Dylan Moreland Feb 5 '12 at 18:36
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To piggyback on Dylan's comment, the dihedral group of order 30 is nonabelian of order $3\cdot5\cdot2$, and has no nonabelian subgroup of order $3\cdot5$. –  user641 Feb 5 '12 at 18:50
    
Yup. It sounds like he is trying to say that for a fixed order there exists a group with such properties (which probably follows from some semidirect product construction -- given the "dividing $q-1$" condition). –  Matthew Pancia Feb 5 '12 at 21:27
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