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For every even $n>4$ 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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    $\begingroup$ n=4 no nonabelian group. Also n=2 of course. $\endgroup$
    – Gerald Edgar
    Commented Jul 12, 2010 at 14:14
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    $\begingroup$ 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. $\endgroup$
    – Johannes Hahn
    Commented Jul 12, 2010 at 14:34
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    $\begingroup$ This appears to be an exact duplicate of mathoverflow.net/questions/11001/… $\endgroup$
    – Pete L. Clark
    Commented Jul 12, 2010 at 16:51
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    $\begingroup$ @Gerald Edgar Groups of order $p^2$ for $p$ prime are always abelian and hence your comment can be easily generalized. $\endgroup$
    – Amitesh Datta
    Commented Jul 14, 2010 at 4:01

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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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You might be interested in the result that if n is odd, |G| = n for a finite group G, and if every subgroup of G is normal, then G is abelian. (This does not hold if the hypothesis that n is odd is ommitted as the quaternion group of order 8 demonstrates.)

A group whose every subgroup is normal is called a Dedekind group. A non-abelian Dedekind group is called a Hamiltonian group. With this terminology the result simply states that a Dedekind group of odd order is abelian.

The proof is not immediately obvious. It relies on a classification result that states that every Hamiltonian group is a direct product of the quaternion group of order 8, an elemetary abelian 2-group, and a periodic abelian group of odd order. Once this classification result is established, however, the result can be seen easily.

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    $\begingroup$ Actually, the proof that all Dedekindian groups of odd order are abelian is somewhat simpler than the classification and can be used to prove the classification. What you do is, you prove it for p-groups using that if a p-group has just one subgroup of order p and p is odd, then the group is cyclic. Since the group is clearly the direct product of its Sylow-subgroups it then follows. $\endgroup$
    – Tobias Kildetoft
    Commented Jul 14, 2010 at 10:03
  • $\begingroup$ That is indeed true. Thanks for noting this! $\endgroup$
    – Amitesh Datta
    Commented Jul 14, 2010 at 12:08

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