TITLE: Finite nonabelian groups of odd order
QUESTION [14 upvotes]: 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 $n$?

REPLY [7 votes]: 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.