It is true that every group that has a finite number of subgroups is finite?
I think not, but I can not find counterexamples.
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See http://math.stackexchange.com/questions/322713/if-a-group-g-has-only-finitely-many-subgroups-then-show-that-g-is-a-finite – Jérémy Blanc Jun 21 '14 at 13:21
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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!
Nicky Hekster
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This is not a proof. It only shows that every element is of finite order. – Jérémy Blanc Jun 21 '14 at 13:01
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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
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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
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