Let $G$ be a finite by cyclic group.Prove that its automorphism group $\operatorname{Aut}(G)$ is finite.

A finite by cyclic group is a group G that has a normal subgroup $N$ such that $N$ is finite and $G/N$ is cyclic group.

Any ideas?

  • $\begingroup$ So, if $G$ is finite, then the result is obvious. If $G$ is infinite, then $G/N$ is also infinite, which means that it's $\Bbb Z$. $\endgroup$
    – Arthur
    Commented Aug 25, 2015 at 10:26

1 Answer 1


Since $N$ is normal, $N=G_{tor}$, the torsion subgroup of $G$. It is finite by hypothesis.

Every automorphism of $G$ sends torsion elements to torsion elements, thus restricts to an automorphism of $N$ and defines an automorphism of $G/N\simeq\Bbb Z$. But ${\rm Aut}({\Bbb Z})=\{\pm1\}$ is finite.

Thus, if $t\in G$ denotes a representative of the generator of $G/N$ and $\phi$ is an automorphism, when computing $\phi$ on the generic element $g=nt^k$ there are only finitely many possibilities for $\phi(n)$ and for $\phi(t)$ making for but a finitely many of $\phi$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .