Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Jack is trying to prove:

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

(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.

share|improve this question
    
How is the operation of $n \in Z$ on $g \in G$ defined? –  Herp Derpington Nov 8 '12 at 0:00
1  
$$\forall\,g\in G\;\;,\;pg\in pG\Longrightarrow p(g+pG)=pG\Longrightarrow$$ 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. –  DonAntonio Nov 8 '12 at 2:32
    
@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 $n<0$. –  Cameron Buie Nov 8 '12 at 8:37
    
@DonAntonio: Why not make that an answer? –  Cameron Buie Nov 9 '12 at 6:53
    
@CameronBuie, I will. It's just that there were already several answers... –  DonAntonio Nov 9 '12 at 9:07

4 Answers 4

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.

share|improve this answer

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

share|improve this answer

$G/pG$ 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 $p$.

share|improve this answer
8  
Three answers, Makoto? Why don't you just combine them into one, and just give them as three ways to see it? –  Cameron Buie Nov 8 '12 at 0:22
5  
Why would anyone downvote extra effort? By contrast, when one posts multiple answers, it looks like reputation-farming. –  Cameron Buie Nov 8 '12 at 3:47
8  
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. –  Cameron Buie Nov 8 '12 at 8:14
7  
@CameronBuie Since I'm not rep hungry, I decline your proposal. Please answer what's wrong with posting several answers. –  Makoto Kato Nov 8 '12 at 8:26
12  
@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. –  Willie Wong Nov 8 '12 at 9:01

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.

share|improve this answer
    
Because $G$ is finitely generated and $G/pG$ is a torsion group(i.e. every element has finite order). –  Makoto Kato Nov 8 '12 at 0:21
    
Consider generators $x_1, \dots, x_n$ of $G/pG$. –  Makoto Kato Nov 8 '12 at 0:27
    
A finitely generated torsion abelian group is finite, @Jack. It's an easy proof, try it. –  DonAntonio Nov 8 '12 at 3:57
1  
@Jack If every element of a group has finite order, it is called a torsion group. –  Makoto Kato Nov 8 '12 at 6:51

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.