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?
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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.
Anne Bauval
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1«If $m=2$…» and the extension is not abelian. – Mariano Suárez-Álvarez Sep 06 '16 at 18:43
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If $m=2$ and $G$ is nonabelian, then $G$ can be the dicyclic group, which is not dihedral. https://en.wikipedia.org/wiki/Dicyclic_group Or am I missing something? – Andrius Kulikauskas Oct 11 '24 at 09:45