# Generalizations of the nonabelian group of order $pq$

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"?

• 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

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 $a^p=1$, $b^n=1$, $a^{-1}ba=b^r$.