# Finite nonabelian groups of odd order

For every even $n$ there exists nonabelian group. As example of such group we can take dihedral group.

The question is about odd $n$. For some of them there are no nonabelian groups of order $n$ (for example, if $n$ is prime then the group of order $n$ is cyclic and hence abelian).

For what odd $n$ are there known examples of nonabelian finite groups of order $n$?

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n=4 no nonabelian group. Also n=2 of course. –  Gerald Edgar Jul 12 '10 at 14:14
The answer is easy with sylows theorems. Use that there are nonabelian groups of order $p^3$ for every prime and look at all prime factorizations with exponents $\leq 2$. For such $n$, the conditions of sylow for normality of all the sylow-groups are necessary and sufficient for all groups of order $n$ being abelian. –  Johannes Hahn Jul 12 '10 at 14:34
This appears to be an exact duplicate of mathoverflow.net/questions/11001/… –  Pete L. Clark Jul 12 '10 at 16:51
@Gerald Edgar Groups of order $p^2$ for $p$ prime are always abelian and hence your comment can be easily generalized. –  Amitesh Datta Jul 14 '10 at 4:01

It's well-known that for a natural number $n$ with prime factorization $n=\prod_i p_i^{r_i}$, all groups of order $n$ are abelian if and only if all $r_i\le 2$ and $\gcd(n,\Phi(n))=1$ where $\Phi(n)=\prod_i (p_i^{r_i}-1)$. (See http://groups.google.co.uk/group/sci.math/msg/215efc43ebb659c5?hl=en)

For other $n$ there are non-abelian groups. If some $r_i\ge3$ then we can take a direct product of a non-abelian group of order $p_i^3$ and a cyclic group. There are always non-abelian groups of order $p^3$; when $p=2$ take the quaternion group, and when $p$ is odd the group of upper triangular matrices with unit diagonal over $\mathbb{F}_p$.

Otherwise $G$ will have a factor $pq$ with $p\mid(q-1)$ or $pq^2$ with $p\mid(q^2-1)$. In the first case the group of all maps $x\mapsto ax+b$ for $a$, $b$, $x\in\mathbb{F}_q$ and $a\ne 0$ has a non-abelian subgroup of order $pq$. In the second case replace $\mathbb{F}_q$ by $\mathbb{F}_{q^2}$ and then get a non-abelian group of order $pq^2$. In both cases multiply by a cyclic group to get an order $n$ non-abelian group.

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Thank you for your answer! –  falagar Jul 12 '10 at 16:58