# 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?

• Hmm. Couldn't you take any cyclic group with odd order $q$, then semidirect product it with the two element group (where one element acts as the identity, the other acts as inversion) to get a non-abelian group of order $2q$? I would think that there are a lot of different cases. – Cameron Williams Dec 3 '15 at 18:28
• It seems that any useful classification would assume some additional information about $q$. – Taylor Dec 3 '15 at 19:02
• I don't think you can say any more than that they are solvable. In general the classification of solvable groups is hopeless; even, say, groups of order $3^k$ are hopeless to classify for sufficiently large $k$. Can you ask a more specific question? For example, you might hope for a classification relative to the classification of groups of order $q$. – Qiaochu Yuan Dec 3 '15 at 19:19
• @ all, thanks for your attempts. – Stan Dec 3 '15 at 19:38
• @ Qiaochu, the same way we know that there is only one nonabelian group of order 2p (for p odd prime), I also want to know how many nonabelian groups of order 2q (where q is an odd positive integer) there are? – Stan Dec 3 '15 at 19:40

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:

1. $N$ is cyclic , so $N\cong C_q$ and $\text{Aut}(N)\cong(\ZZ/q\ZZ)^\times$, and
2. $\rho$ 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 $N\rtimes C_2$.