Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question already has an answer here:

It is true that every group that has a finite number of subgroups is finite?

I think not, but I can not find counterexamples.

share|cite|improve this question

marked as duplicate by Martin Brandenburg abstract-algebra Jun 21 '14 at 13:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


1 Answer 1

up vote 6 down vote accepted

Yes, true - look at $\langle g\rangle$ for every $g \in G$ and note that an infinite cyclic group has an infinite number of subgroups.

$\langle g \rangle$ must be finite, and since $G$ has only a finite number of subgroups, we get that $G=\bigcup_{g \in G}\langle g \rangle$ is a finite union. hence $G$ must be finite after all!

share|cite|improve this answer
This is not a proof. It only shows that every element is of finite order. – Jérémy Blanc Jun 21 '14 at 13:01
This is a hint for a proof, I do not want to give away the proof. Will add spoiler though. – Nicky Hekster Jun 21 '14 at 13:04
Ah OK. I get the idea. – Jérémy Blanc Jun 21 '14 at 13:04
This part of that an infinite cyclic group has an infinite number of subgroups is because each $g^n$ would create a new subgroup $<g^n>$? – Croos Jun 21 '14 at 13:53
Yes correct Croos! – Nicky Hekster Jun 21 '14 at 14:09

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