MathOverflow is a question and answer site for professional mathematicians. 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

If $p < q$ are primes then there is a nonabelian group of order $pq$ iff $q = 1 \pmod p$, in which case the group is unique. If $p = 2$ we obtain the dihedral group of order $2q$, which generalizes first to the dihedral group of order $2n$ and then even further to the "generalized dihedral group" where the cyclic group of order $n$ is replaced with any abelian group.

What if $p > 2$? Is there a natural generalization of the groups of order $pq$ to a family of groups of order $pn$? Maybe more than one possible generalization? Is it maybe even meaningful to talk about the "generalized $p$-hedral group"?

share|cite|improve this question
To define the generalized dihedral group , don't you need to start with an abelian group? – Dan Ramras Jun 27 '10 at 1:54
Sorry, yes, you're right. I've corrected the mistake above. – Robin Saunders Jun 27 '10 at 2:01
The group of order $pq$ is the semidirect product of $Z/p$ with $Z/q$ via the homomorphism $Z/p \to Aut(Z/q)$. The latter homomorphism also makes sense if $p,q$ are integers such that $Aut(Z/q)$ has an element of order $p$. – Martin Brandenburg Jun 28 '10 at 10:44
Thanks, Martin. So if I have a general group G such that Aut(G) has an element a of order n and define the semidirect product of Z/nZ with G via the homomorphism Z/nZ \to <a>, I suppose this is in some sense a "generalized n-hedral group". I wonder, under what conditions is such a semidirect product unique? – Robin Saunders Jun 28 '10 at 19:31
up vote 7 down vote accepted

The nonabelian group of order $pq$ is given by generators $a$, $b$, with relations $a^p=1$, $b^q=1$, $a^{-1}ba=b^r$, where $r$ is chosen so $r^p$ is 1 mod $q$. If there is an element $r$ of order $p$ mod $n$, then there is a nonabelian group of order $pn$ with generators $a$, $b$, and relations $a^p=1$, $b^n=1$, $a^{-1}ba=b^r$.

share|cite|improve this answer
You can even let p be composite, and choose r mod n to have order dividing p. The groups with G' cyclic of order coprime to G/G' are given a nice description in Hall's Theory of Groups, Th 9.4.3, one of which is exactly your description. In general, he might be interested in metacyclic groups, which in the finite case have a fairly similar description (King and Liebeck have some nice parameterizations of the metacyclic p-groups). – Jack Schmidt Jun 27 '10 at 3:11

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.