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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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    @CameronBuie, I will. It's just that there were already several answers...2012-11-09

4 Answers 4

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Following my comment:

$∀g∈G,pg∈pG⟹p(g+pG)=pG⟹ $ 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.

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$G/pG$ can be regarded as a finite dimensional vector space over $\mathbb{Z}/p\mathbb{Z}$. Suppose its dimension is $n$. Then $|G/pG| = p^n$.

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$G/pG$ is a direct sum of a finite number of cyclic groups by the fundamental theorem of finitely generated abelian groups. Since every non-zero element of $G/pG$ is of order $p$. It is a direct sum of a finite number of cyclic groups of order $p$.

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    @All **There is now an associated meta question** [here.](http://meta.math.stackexchange.com/q/6525/242) Please take any further meta discussion there.2012-11-08
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Since $G/pG$ is a finitely generated torsion group, it is finite. Let $q$ be a prime number which divides $|G/pG|$. Then it has an element of order $q$ by the theorem of Cauchy. Hence $q = p$. Hence $G/pG$ is a $p$-group.

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    @Jack If every element of a group has finite order, it is called a torsion group.2012-11-08