Nonabelian groups of order $2q$, where $q$ is an odd number. We know that every non-abelian group of order $2p$ (where $p$ is an odd prime) is dihedral. Is there any classification of non-abelian groups of order $2q$, where $q$ is an odd number, not necessarily prime?
[The well-known Feit-Thompson theorem tells us that groups of odd orders are soluble; so groups of order $2q$ (where $q$ is odd) are soluble too.]
But what I am interested in is: 
What are the non-abelian groups of order $2q$, where $q$ is an odd positive integer?
 A: No, not every group $G$ of order $2q$ for odd $q$ is dihedral. 
Firstly, $G$ must be solvable. This is because a nonabelian simple group must have order divisible by 4, see $\S$4 of https://www.maa.org/sites/default/files/pdf/upload_library/22/Allendoerfer/0025570x.di021096.02p01365.pdf
A solvable group $G$ of order $2q$ for odd $q$ must have a normal subgroup $N$ of order $q$. This is the prime-to-2 Hall subgroup (https://en.wikipedia.org/wiki/Hall_subgroup), which is normal because it has index 2.
$\newcommand{\ZZ}{\mathbb{Z}}$
Thus, we have an exact sequence
$$1\rightarrow N\rightarrow G\rightarrow C_2\rightarrow 1$$
which by the Schur-Zassenhaus theorem must be split by a section $C_2\rightarrow G$, and thus $G$ is a semidirect product $N\rtimes C_2$, which is completely described by the action homomorphism $\rho : C_2\rightarrow\text{Aut}(N)$ (which is induced by the section $C_2\rightarrow G$)
Then $G$ is dihedral if and only if we have the following two additional properties:


*

*$N$ is cyclic , so $N\cong C_q$ and $\text{Aut}(N)\cong(\ZZ/q\ZZ)^\times$, and

*$\rho$ sends the nontrivial element of $C_2$ to "inversion", ie $-1\in(\ZZ/q\ZZ)^\times$.


Assuming (1) holds, then when $q$ is odd, $(\ZZ/q\ZZ)^\times$ is cyclic if and only if $q$ is a prime power. Thus, when $q$ is a prime power, "-1" is the unique element of order 2 in $(\ZZ/q\ZZ)^\times$, and hence $G$ must be dihedral. When $q$ is not a prime power, there are many choices for $\rho$. For example, take $q = 3\cdot 5 = 15$, then $(\ZZ/q\ZZ)^\times \cong C_2\times C_4$. There are then three nontrivial choices for $\rho$, only one of which results in $G$ being dihedral.
This also gives you all the ways of constructing groups of order $2q$. First take a group $N$ of order $q$, a homomorphism $\rho : C_2\rightarrow\text{Aut}(N)$, and form the semi-direct product $N\rtimes C_2$.
