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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?2012-02-05
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    Can you give a link to the notes?2012-02-05
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    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 $n$.2012-02-05
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    @SteveD: unfortunately they are of paper-form and not in english!2012-02-05
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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.2012-02-05

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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 $3 \cdot 5$.2012-02-05
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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 $3\cdot5$.2012-02-05
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    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).2012-02-05