# Show that an infinite abelian group all of whose proper quotients are finite $\cong \mathbb{Z}$

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 $$|G/|$$ is finite.... Now what?

• That's not true. Take any infinite simple group as an example. – freakish Apr 22 at 21:33
• @freakish You missed “abelian”. – egreg Apr 22 at 21:36
• @egreg Oh, right. – freakish Apr 22 at 22:05
• @Mark There are no infinite abelian simple groups, $\mathbb{Z}$ is not simple. – freakish Apr 22 at 22:06
• @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. – 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.
• 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 $G \cong \mathbb{Z}$ – Mathematical Mushroom Apr 22 at 21:39
• @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 $G/T$? – egreg Apr 22 at 21:48