Show that an Infinite abelian group all of whose proper quoteints are finite $\cong \mathbb{Z}$

So I'm a little confused on how to get started here. Perhaps something like:

Let $x \in G$, then $|G/<x>|$ is finite.... Now what?

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    $\begingroup$ That's not true. Take any infinite simple group as an example. $\endgroup$ – freakish Apr 22 at 21:33
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    $\begingroup$ @freakish You missed “abelian”. $\endgroup$ – egreg Apr 22 at 21:36
  • $\begingroup$ @egreg Oh, right. $\endgroup$ – freakish Apr 22 at 22:05
  • $\begingroup$ @Mark There are no infinite abelian simple groups, $\mathbb{Z}$ is not simple. $\endgroup$ – freakish Apr 22 at 22:06
  • $\begingroup$ @freakish Yes, you are right, sorry. If there was such an infinite simple abelian group then it would be isomorphic to $\mathbb{Z}$. But it is indeed nothing but a contradiction. $\endgroup$ – Mark Apr 22 at 22:08

I'll use additive notation. Let $x\in G$, $x\ne0$; then $G/\langle x\rangle$ is finite; suppose $$ G/\langle x\rangle=\{x_1+\langle x\rangle,x_2+\langle x\rangle,\dots,x_n+\langle x\rangle\} $$ Then $G=\langle x,x_1,x_2,\dots,x_n\rangle$ is finitely generated.

What do you know about finitely generated abelian groups?

  • $\begingroup$ That you can write them as direct products of $\mathbb{Z}$ and $\mathbb{Z_{p^i}}$?? So if they are finite then they must be direct products of $\mathbb{Z_{p^i}}$? Still don't see why $G \cong \mathbb{Z}$ $\endgroup$ – Mathematical Mushroom Apr 22 at 21:39
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    $\begingroup$ Because you could quotient by some of the torsion groups and this quotient will not be finite (as it contains a copy of the integers). $\endgroup$ – Severin Schraven Apr 22 at 21:41
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    $\begingroup$ @MathematicalMushroom $G\cong\mathbb{Z}^n\oplus T$, where $T$ is the torsion part; also $n\ge1$ because $G$ is infinite. What can you say about $G/T$? $\endgroup$ – egreg Apr 22 at 21:48
  • $\begingroup$ ah of course thank you guys. $\endgroup$ – Mathematical Mushroom Apr 22 at 21:52

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