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If $G$ is a finitely generated abelian group then $G$ is residually finite.

I don't know if the result holds or not. I tried to follow the definition but could not go far. Any hint will be highly appreciated.

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Hint: Let $G$ be a finitely generated abelian group. Then $G \simeq \mathbb{Z}^n \times \mathbb{Z}_{n_1} \times \cdots \times \mathbb{Z}_{n_r}$ for some $n,n_1,\dots,n_r$. Now consider the epimorphism $$G \twoheadrightarrow \mathbb{Z}_k^n \times \mathbb{Z}_{n_1} \times \cdots \times \mathbb{Z}_{n_r}$$ with $k$ arbitrarily large.

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  • $\begingroup$ I see, this epimorphism is in fact an isomorphism. How to proceed further? $\endgroup$ – schzan Apr 1 '14 at 7:24
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    $\begingroup$ It cannot be an isomorphism when $n>0$ because $G$ is infinite but the image is finite. You need to prove that for any $g \in G$ there exists a $k$ such that $g$ is not in the kernel of this epimorphism. $\endgroup$ – Derek Holt Apr 1 '14 at 7:48
  • $\begingroup$ Since G is hopfian, it must be an isomorphism $\endgroup$ – schzan Apr 2 '14 at 3:12
  • $\begingroup$ and also how did you get G is infinite? $\endgroup$ – schzan Apr 2 '14 at 3:36
  • $\begingroup$ You should read again the definition of a hopfian group: the epimorphism must be from a group to itself, otherwise you could conclude that any hopfian group is trivial since there is always an epimorphism $G \twoheadrightarrow \{1\}$. $\endgroup$ – Seirios Apr 2 '14 at 7:46
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Here's a proof without classification.

Let $G$ be f.g. abelian and $g\in G$ nonzero. Let $N$ be a maximal subgroup not containing $g$. Let us prove that $G'=G/N$ is finite: hence $g$ survives in the finite quotient and we are done.

Let $g$ be the image of $g$ in $G'$. Hence $g$ is contained in every nonzero subgroup of $G'$. If some element of $G'$ has infinite order, then it generates a cyclic subgroup, in which we can find a deep enough subgroup, not containing $g$, contradiction. Hence every element in $G'$ has finite order. If $g_1,\dots,g_k$ are finitely many generators for $G'$, say $n_ig_i=0$ for some $n_i>0$, hence if $n$ is the lcm of the $n_i$ we have $nG'=0$. Hence $G'$ is a quotient of $(\mathbf{Z}/n\mathbf{Z})^k$ thus is finite.

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