Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
@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

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

share|cite|improve this answer
Even worse: there is no non-abelian group of order $3 \cdot 5$. – Dylan Moreland Feb 5 '12 at 18:36
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.