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Here is my problem:

Let $G$ is an infinite abelian group. Prove that if every proper quotient is finite, then $G\cong\mathbb Z$.

And here is my incompleted approach:

I know that the quotient subgroup $\frac{G}{tG}$ wherein $tG$ is torsion subgroup of $G$ is always torsion-free. So, if $tG\neq\{0\}$ then here we have $\frac{G}{tG}$ torsion-free and finite simultonously which is a contradiction. Then $G$ is itself a torsion-free group.

Moreover, I assume $G$ be a divisible group, so: $$G\cong\sum\mathbb Q\oplus\sum_{p\in P}\mathbb Z(p^{\infty})$$ As any proper quotient of $G$ is infinite, so I concluded it is not divisible. I confess that I am missing the final part. If my way to this problem untill my last conclusion is valid logically, please help me about the last part of the proof. Thanks

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up vote 6 down vote accepted

Let $a_0\in G$ be a nonzero element. Then $\langle a\rangle\cong \mathbb Z$ as $G$ is torsion-free (which you have shown). Now $Q_0=G/\langle a_0\rangle$ is a finite abelian group. If $Q_0\cong 1$, we are done. Otherwise, select $a_1\in G\setminus\langle a_0\rangle$. Then let $Q_1=G/\langle a_0, a_1\rangle$, etc.

The orders of the finite groups $Q_0, Q_1, \ldots$ are strictly decreasing as long as they are $>1$, hence we ultimately find an $a_n$ with $G=\langle a_1, \ldots,a_n\rangle$. Thus $G$ is a finitely generated abelian group. A quick check with the classification theorem shows that $G\cong \mathbb Z$.

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+1 Liked this simple, straightforward one. – DonAntonio Oct 26 '12 at 18:05
Thanks Hagen!. And this quick check is related to quotient of $G$? – Babak S. Oct 26 '12 at 18:09

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