Extensions of finite cyclic groups

Question

Consider the exact sequence

$1\to C_n\to G\to C_m\to 1$,

where $C_n$ and $C_m$ are finite cyclic groups. Then, what $G$ can be? Is it true that it is either dihedral, or a (semi)direct product of my cyclic groups?

Answer 1

The general term for such groups $G$ is "metacyclic".

If $m = 2$ and $G$ is nonabelian, then $G$ must be dihedral. If there exists a homomorphism $C_m\hookrightarrow G$ such that the composite $C_m\hookrightarrow G\rightarrow C_m$ is the identity, then the sequence is said to be split, which is equivalent to saying that $G\cong C_n\rtimes C_m$.

In general, metacyclic groups need not be a semidirect product. For example, the quaternion group of order 8 is metacyclic, with sequence: $$1\rightarrow\langle i\rangle\rightarrow Q_8\rightarrow Q_8/\langle i\rangle\rightarrow 1$$ It's easy to verify that this sequence is not split.