# Thread: subgroup(pG) of an abelian group(G)

1. ## subgroup(pG) of an abelian group(G)

Hi;

If G is an abelian group. prove that pG={px : x belongs to G} is a subgoup of G, where p is a prime number

2. Originally Posted by fuzzy topology
Hi;

If G is an abelian group. prove that pG={px : x belongs to G} is a subgoup of G, where p is a prime number

What do you think? This seems pretty standard unless I am misinterpreting or missing some subtlety.

3. Originally Posted by Drexel28
What do you think? This seems pretty standard unless I am misinterpreting or missing some subtlety.
How can this be so ? (Yes, its true in case of Z! )

Suppose G = < a >

Here pa is not in G for any prime p.

4. Originally Posted by ques
How can this be so ? (Yes, its true in case of Z! )

Suppose G = < a >

Here pa is not in G for any prime p.
See, that isn't how I interpreted the question. Perhaps we should wait and see what the OP has to say.

5. Originally Posted by ques
How can this be so ? (Yes, its true in case of Z! )

Suppose G = < a >

Here pa is not in G for any prime p.
It would be, as abelian groups are written additively!

$ = \{\ldots, -2a, -a, 0, a, 2a, \ldots\}$ so $p = \{\ldots -2pa, -pa, 0, a, 2a, \ldots\}$.

That said, I agree with Drexel. There is either nothing really to prove or I am missing something...

6. Originally Posted by Swlabr
It would be, as abelian groups are written additively!

$ = \{\ldots, -2a, -a, 0, a, 2a, \ldots\}$ so $p = \{\ldots -2pa, -pa, 0, a, 2a, \ldots\}$.

That said, I agree with Drexel. There is either nothing really to prove or I am missing something...
I guess that I get your point. pa in multiplication is a^p.

For n=/= p, (p,n ) = 1.

7. ## standard way

Thank you for your help, but in our university, they don't accept that the proof is trivial, we must do proofs for every statement. Like this statement, it seems very clear, but we should follow the steps in order to show a subset is a subgroup of a given group.

For example, let us proof our statement,
pG is not empty since pe=p is in pG
Second, we must proof the closure low is satisfied as well as the inverse of each elemet in pG, Suppose px and py are in pG we must show that (px)(py) is again in pG and (px)-1 also in pG
I got the following, (px).(py)=p^2(xy)=p(pxy) and pxy is in G since both px and y are in G. In fact I am streamly confusing about the last part of the proof it seems for me un acceptable. And I am like you think that there is nothing to proof.

Thank you very much

8. Originally Posted by fuzzy topology
Thank you for your help, but in our university, they don't accept that the proof is trivial, we must do proofs for every statement. Like this statement, it seems very clear, but we should follow the steps in order to show a subset is a subgroup of a given group.

For example, let us proof our statement,
pG is not empty since pe=p is in pG
Second, we must proof the closure low is satisfied as well as the inverse of each elemet in pG, Suppose px and py are in pG we must show that (px)(py) is again in pG and (px)-1 also in pG
I got the following, (px).(py)=p^2(xy)=p(pxy) and pxy is in G since both px and y are in G. In fact I am streamly confusing about the last part of the proof it seems for me un acceptable. And I am like you think that there is nothing to proof.

Thank you very much
Abelian groups are written additively. So what is mean by the element $px$ is $x+x+\ldots+x$ $p$-times.

So what you need to show is inverses, (if $a \in pG$ then $-a \in pG$) and closure ( $a + b \in pG$ for $a, b \in pG$). Everything is done additively, not multiplicatively as you were trying to do...

However, this holds for all $p \in \mathbb{Z}$, not just primes...

9. Thank you very much. I am get a little bet confusing because of the notation I use.

Thanks alot