Is there a construction that will give a non-abelian group of order $p^mr$ where $p$ is a prime, $r$ and $p$ are relatively prime and $m$ is an arbitrary non-negative integer? I suspect in this generality there is no simple construction so feel free to restrict $m$ and $r$. 


I'm reading some notes on group theory and so far I've only seen the group $G=SL(2,p)$ which has order $p(p+1)(p-1)$. This is great because it gives me a class of groups to play with and test out the various theorems. It is a little annoying to have all these theorems and no concrete non-trival examples to test them out on to see all the subtleties since for the abelian groups all these theorems reduce to saying something trivial.


Edit: Mariano makes a good point and I'm not sure how to rule out silly examples like $G\times\mathbb{Z}\_p^{m-3}\times\mathbb{Z}\_r$. These aren't bad per se but what matters for me is an explicit description of $G$, the non-commutative part, so I have some hope of carrying out some calculations. In essence what I would really like is a construction that parametrizes the non-commutative part depending on all the parameters. In Mariano's example the non-commutative part has no dependence on $m$ and this simplifies the structure of the resulting group.




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Thanks for the examples and references. This gives me a lot more concrete stuff to work with. Now hopefully I can work out some of the reasons for the various assumptions used in the proofs.



non-abelian groups of prescribed order
I am also unsure of what "nontriviality" conditions you want to impose. Without any further conditions, the following answers your question:


Call a positive integer $n$ **nilpotent** if every group of order $n$ is nilpotent.


Call a positive integer $n$ **abelian** if every group of order $n$ is abelian.


Suppose that the prime factorization of $n$ is $p\_1^{a\_1} \cdots p\_r^{a\_r}$. Then:


1. $n$ is nilpotent iff for all $i,j,k$ with $1 \leq k \leq a\_i$, $p\_i^k \not \equiv 1 \pmod{p\_j}$.
2. $n$ is abelian iff it is nilpotent and $a\_i \leq 2$ for all $i$.


These results are proved in




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Pakianathan, Jonathan(1-WI); Shankar, Krishnan(1-MI)
Nilpotent numbers.
Amer. Math. Monthly 107 (2000), no. 7, 631--634.




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The proofs are constructive: for any $n$ which is not nilpotent (resp. abelian), they give an explicit group of that order which is not nilpotent (resp. abelian).


The paper is available at


[http://alpha.math.uga.edu/~pete/nilpotentnumbers.pdf](http://alpha.math.uga.edu/%7Epete/nilpotentnumbers.pdf)


