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Jack is trying to prove:

Let $G$ be an abelian group, and $n\in\Bbb Z$. Denote $nG = \{ng \mid g\in G\}$.

(1) Show that $nG$ is a subgroup in $G$.

(2) Show that if $G$ is a finitely generated abelian group, and $p$ is prime, then $G/pG$ is a $p$-group (a group whose order is a power of $p$).

I think $G/pG$ is a $p$-group because it is a direct sum of cyclic groups of order $p$. But I cannot give a detailed proof.

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    How is the operation of $n \in Z$ on $g \in G$ defined?2012-11-08
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    $$\forall\,g\in G\;\;,\;pg\in pG\Longrightarrow p(g+pG)=pG\Longrightarrow$$ the element $\,p(g+pG)\, $ is zero in the quotient $\,G/pG\,$ and from here that all the elements in this quotient have order a power of p, which is precisely the definition of p-group, no matter if it is finitely generated or not.2012-11-08
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    @HerpDerpington: I suspect $G$ is taken to be an additive group, so that $ng$ is simply adding up $n$ terms $g$ for $n>0$ and adding up $n$ terms $-g$ for $n<0$.2012-11-08
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    @DonAntonio: Why not make that an answer?2012-11-09
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    @CameronBuie, I will. It's just that there were already several answers...2012-11-09

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