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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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  • $\begingroup$ How is the operation of $n \in Z$ on $g \in G$ defined? $\endgroup$
    – dpington
    Nov 8, 2012 at 0:00
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    $\begingroup$ $$\forall\,g\in G\;\;,\;pg\in pG\Longrightarrow p(g+pG)=pG\Longrightarrow$$ 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. $\endgroup$
    – DonAntonio
    Nov 8, 2012 at 2:32
  • $\begingroup$ @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 $n<0$. $\endgroup$ Nov 8, 2012 at 8:37
  • $\begingroup$ @DonAntonio: Why not make that an answer? $\endgroup$ Nov 9, 2012 at 6:53
  • $\begingroup$ @CameronBuie, I will. It's just that there were already several answers... $\endgroup$
    – DonAntonio
    Nov 9, 2012 at 9:07

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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    $\begingroup$ Three answers, Makoto? Why don't you just combine them into one, and just give them as three ways to see it? $\endgroup$ Nov 8, 2012 at 0:22
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    $\begingroup$ Why would anyone downvote extra effort? By contrast, when one posts multiple answers, it looks like reputation-farming. $\endgroup$ Nov 8, 2012 at 3:47
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    $\begingroup$ It looks like you might be trying to get more reputation by posting several answers. It's off-putting. I personally liked both your first and last answer, but I don't want to reinforce behavior that I know isn't acceptable here, so I didn't upvote either of them. If you merge your answers, I'll gladly upvote, and I suspect that others will do the same. I recommend merging this answer and the vector space answer into the last one, so that you don't lose the comments there. $\endgroup$ Nov 8, 2012 at 8:14
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    $\begingroup$ @CameronBuie Since I'm not rep hungry, I decline your proposal. Please answer what's wrong with posting several answers. $\endgroup$ Nov 8, 2012 at 8:26
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    $\begingroup$ @Makoto: I strongly disagree in this case. You gave three two-liner answers. You can easily say: "I will give three different ways to look at the problem." And divide your answer using the convenient formatting afforded to you by MarkDown. $\endgroup$ Nov 8, 2012 at 9:01
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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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  • $\begingroup$ Because $G$ is finitely generated and $G/pG$ is a torsion group(i.e. every element has finite order). $\endgroup$ Nov 8, 2012 at 0:21
  • $\begingroup$ Consider generators $x_1, \dots, x_n$ of $G/pG$. $\endgroup$ Nov 8, 2012 at 0:27
  • $\begingroup$ A finitely generated torsion abelian group is finite, @Jack. It's an easy proof, try it. $\endgroup$
    – DonAntonio
    Nov 8, 2012 at 3:57
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    $\begingroup$ @Jack If every element of a group has finite order, it is called a torsion group. $\endgroup$ Nov 8, 2012 at 6:51

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