4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 Aug 25 '15 at 10:26
2
$\begingroup$

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 $\phi$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.