# Is the number of finite subgroups of a finite-by-cyclic group finite?

Let $$G$$ be a finite-by-cyclic group, i.e. it contains a finite normal subgroup $$N$$ such that $$G/N$$ is cyclic.

My question is that: Does $$G$$ contain only finitely many finite subgroups?

What I've tried: Assume that $$G/N$$ is finite cyclic. Then $$G$$ is a finite group and then it has only a finite number of subgroups.

• "Assume that $G/N$ is finite cyclic. Then $G$ is a finite group" : why ? Commented Aug 22, 2022 at 11:07
• @Arthur It was a question for the OP... Thanks for the answer though. Commented Aug 22, 2022 at 11:13
• @TheSilverDoe Fair enough. I usually address such questions more explicitly to the OP precisely to avoid these misunderstandings. It is impossible for me to know from your original comment whether you want my help or not. Commented Aug 22, 2022 at 11:15
• (Although if you care to look, you will see that just my previous comment on this site is exactly on the opposite side of this issue, speaking of the merits of letting the OP answer the questions in the comments. So I'm not exactly consistent about it.) Commented Aug 22, 2022 at 12:18

Yes, $$G$$ contains only finitely many finite subgroups. As you pointed out, this is trivial if $$G/N$$ is finite, so we will consider the case $$G/N \cong \mathbb{Z}$$.
Let $$K$$ be a finite subgroup of $$G$$, then $$KN$$ is also a finite subgroup of $$G$$ (since $$N$$ is finite and normal) and therefore $$KN/N$$ is a finite subgroup of $$G/N \cong \mathbb{Z}$$. We conclude that $$KN/N = \left\{0\right\}$$, hence $$KN = N$$ and so $$K < N$$. This means that we have only a finite number of choices for $$K$$.