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Is the following statement true? I think it's a characterization of finite groups.

A group $G$ is finite if and only if every subgroup generated by some $g \in G$ is finite.

The "only if" is immediate, but I can't convince myself of the "if" and am therefore unsure about the claim. The book I'm reading is of the math for physics variety and so doesn't spell this out, though I suspect it might be true.

As I think about it more, perhaps I can argue by the contrapositive? If $G$ is infinite then there must exist an infinite subgroup generated by some $g$ for, if not, then $G \subset \bigcup C_g$ where $C_g$ is the subgroup generated by $g$. But this is absurd since $\bigcup C_g$ is a finite group. Does this reasoning even make sense, and is there a clearer way to see this?

Edit: The proof doesn't make sense since the indexing set $G$ for the union is assumed infinite. Can someone supply a proof or a counterexample?

Shaun
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EE18
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  • I'm afraid I'm unfamiliar with "torsion" here but if you are saying that a counterexample to my proposed theorem exists then I'm happy to accept. @AnneBauval – EE18 Mar 15 '23 at 22:37
  • A torsion group is a group where every element is of finite order, i.e. generates a finite subgroup. So, your conjecture "every torsion group is finite" is false. Counterexamples given in the duplicate are the additive group $\Bbb Q/\Bbb Z,$ and the multiplicative group of all complex roots of $1.$ (Actually, they are isomorphic.) – Anne Bauval Mar 15 '23 at 22:38
  • Awesome, thank you for letting me know! @AnneBauval – EE18 Mar 15 '23 at 22:44
  • Another example that you might find more elementary is $\Bbb Z/(2)\oplus \Bbb Z/(3)\oplus \Bbb Z/(5)\oplus\Bbb Z/(7)\oplus\cdots$ – pancini Mar 15 '23 at 23:48
  • Or just $\Bbb Z/(2)\oplus\Bbb Z/(2)\oplus \Bbb Z/(2)\oplus\cdots$. Note that a direct sum only contains elements with finitely many nonzero parts. (So you don't have to worry about the order of $(1,1,1,\ldots)$ in my first example.) – pancini Mar 15 '23 at 23:48

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