# non-abelian groups of prescribed order

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.

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.

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You may want to be more stringent about what you want, lest you get silly examples like $G\times\mathbb Z_p^{m-3}\times\mathbb Z_r$ with $G$ a non abelian group of order $p^3$ (which you can get as $(\mathbb Z_p\times\mathbb Z_p)\rtimes\mathbb Z_p)$. –  Mariano Suárez-Alvarez Jan 7 '10 at 2:57
Very good point. –  davidk01 Jan 7 '10 at 3:18
If Mariano's G is the G that I'm thinking of, it has the following explicit description: it's the group of p x p matrices generated by the identity matrix, the diagonal matrix with all entries zeta_p, and the diagonal matrix (1, zeta_p, zeta_p^2, ...). –  Qiaochu Yuan Jan 7 '10 at 3:39
Whoops. Replace "the identity matrix" with "a permutation matrix of order p." –  Qiaochu Yuan Jan 7 '10 at 3:41
Also, maybe you should be more explicit about which theorems you want to see the subtleties of. –  Qiaochu Yuan Jan 7 '10 at 3:48

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

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

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://www.math.uga.edu/~pete/nilpotentnumbers.pdf

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Also, by a theorem of Burnside (1890-ish), n is cyclic iff n and \phi(n) are relatively prime. –  S. Carnahan Jan 7 '10 at 4:04
Right. That is also mentioned in the paper linked to above. –  Pete L. Clark Jan 7 '10 at 14:08

This does not answer your question but may be quite useful anyways: there are much worse things you can do than learn how to use GAP, which, among many marvels, has a library of all (!) groups of small order, and lets you construct the 'usual' groups, do operationx with them, &c, &c.

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This is nice. I want a concrete feel for these kinds of things that the theoretical development doesn't provide so I'll definitely look at GAP. –  davidk01 Jan 7 '10 at 3:47
Of course how impressive this is depends on the meaning of "small". To see that it is impressive, note that the list contains all groups "of order at most 2000 except 1024 (423 164 062 groups)" as well as many others. –  Pete L. Clark Jan 7 '10 at 3:52
... and 95% of them have order 1536. –  S. Carnahan Jan 7 '10 at 4:08

I think you can generate what you're looking for using the Wreath Product. The linked Wikipedia article is fairly well written and includes some examples, so I won't duplicate everything.

If G is any group and H is any subgroup of the permutation group on n elements, then the wreath product of G with H has order $|G|^n |H|$. Multiplication works by using elements of H to permute n copies of G. The result is non-commutative as long as H is non-commutative. You can take, e.g., |G|=p, H any non-commutative order r subgroup of the symmetric group on m elements. This imposes some relations between r and m, but there's plenty of non-trivial examples to play with. In fact, a lot of interesting examples of groups can be expressed as wreath products of smaller groups.

Also note that by Cayley's Theorem, every group of order r is a subgroup of the symmetric group on r elements (and the group multiplication tells you how it acts to permute its own elements). So if you take m=r you can use any non-commutative group H you like.

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Yes, this is essentially the first construction I gave. –  Qiaochu Yuan Jan 7 '10 at 5:35
@Qiaochu: You're right, I didn't read your reply carefully enough. Sorry! –  Ian Weiner Jan 7 '10 at 12:54
Duplicating part of an answer and providing a different perspective is fine and welcomed. –  davidk01 Jan 7 '10 at 22:22

Given p and r: Pick your favorite group $G$ of order $r$. It has a faithful transitive action on a set of size $m$ for some $m$, so you can take the semidirect product $\mathbb{F}_p^m \rtimes G$. Alternately, if $G$ has an interesting automorphism group, pick one of the Sylow $p$-subgroups $H$ of $\text{Aut}(G)$ and you can take the semidirect product $G \rtimes H$.

Given p and m: The group $GL_m(\mathbb{F}_p)$ contains a (say, Sylow) subgroup $G$ of order relatively prime to $p$, so again you can take the semidirect product $\mathbb{F}_p^m \rtimes G$.

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Also, if all you want to do is test out theorems, it's a fun exercise to construct the Sylow p-subgroups of the symmetric groups explicitly. –  Qiaochu Yuan Jan 7 '10 at 3:42
If $m$ is big, you get a large family of nonabelian unipotent algebraic groups over $\mathbb{F}\_p$, and these yield the nonabelian p-groups. The standard examples include the group of $k \times k$ upper triangular matrices with ones along the diagonal and entries in $\mathbb{F}\_p$ for $k \geq 3$, and subgroups like Heisenberg (whose elements are nonzero in a hollow upper triangle). By a theorem of Sims (1965), as $m$ grows large, most isomorphism types will end up being 3-step nilpotent, with exponent $p^3$ (i.e., not of the specific matrix form I specified). You can ask for a group of order $r$ to act on such a group by automorphisms, and you get a semidirect product.
You also might want to consider central extensions of $p$-groups by abelian groups of order $r$. There can be interesting homological calculations there.