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$\begingroup$

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

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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    Let $x \in G$. Since $px \in pG$, $p[x] = 0$, where $[x]$ is the image of $x$ in $G/pG$. Hence the order of $[x]$ is $p$ or $1$.2012-11-08
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    Three answers, Makoto? Why don't you just combine them into one, and just give them as three ways to see it?2012-11-08
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    @CameronBuie They are different and independent answers. I think combining them into one answer may cause some confusions. Is there any merit in doing so?2012-11-08
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    Just label them in **boldface** and put lines between them, with `***`.2012-11-08
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    @CameronBuie I'm not talking about such a thing. For example, what about reps? One of them may get up votes while others may get down votes.2012-11-08
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    Why would anyone downvote extra effort? By contrast, when one posts multiple answers, it looks like reputation-farming.2012-11-08
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    @MakotoKato , the very first day I participated in this site I was said that sneding ,ore than one answer to the same question is extremely unpolite and confusing, and I couldn't but a gree with that. I think that what Cameron is telling you is the correct and used thing to do here.2012-11-08
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    @CameronBuie Maybe someone would down vote for a false proof. This is only an example. There are other confusing factors. As for your last comment, I didn't get any reps. Anyway, what's wrong with separate answers?2012-11-08
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    @DonAntonio I opened a meta thread on multiple answers. http://meta.math.stackexchange.com/questions/6525/whats-wrong-with-a-user-posting-different-and-independent-answers-to-a-question2012-11-08
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    @CameronBuie Just to make it clear, what exactly is reputation-farming?2012-11-08
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    @DonAntonio What's wrong with multiple answers?2012-11-08
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    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.2012-11-08
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    @CameronBuie Since I'm not rep hungry, I decline your proposal. Please answer what's wrong with posting several answers.2012-11-08
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    I answered that. Read the first two sentences of my prior comment.2012-11-08
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    @CameronBuie I'm not trying to get more reputation by posting several answers. So what's wrong with posting several answers?2012-11-08
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    I'm not claiming that you are. I merely stated that *looks like* you *might* be, and that puts people off.2012-11-08
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    Anyway, I've made my proposal, and you've declined. I support your right to make your own choice. I'll leave it to others to try to convince you in the meta post, if they like.2012-11-08
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    @CameronBuie As I wrote, merging several answers to one can cause confusions.2012-11-08
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    @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.2012-11-08
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    @WillieWong The three answers are based on different ideas. Please explain if you think otherwise.2012-11-08
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    @Willie And I strongly disagree with you. The answers are logically independent and, as such, there are good reasons for drawing sharp boundaries between them - see my [answer to the associated meta question.](http://meta.math.stackexchange.com/q/6525/242)2012-11-08
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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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    Because $G$ is finitely generated and $G/pG$ is a torsion group(i.e. every element has finite order).2012-11-08
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    Consider generators $x_1, \dots, x_n$ of $G/pG$.2012-11-08
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    A finitely generated torsion abelian group is finite, @Jack. It's an easy proof, try it.2012-11-08
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    @Jack If every element of a group has finite order, it is called a torsion group.2012-11-08