Hi;
please help me in proving this statement;
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
Thank you in advance
It would be, as abelian groups are written additively!
$\displaystyle <a> = \{\ldots, -2a, -a, 0, a, 2a, \ldots\}$ so $\displaystyle p<a> = \{\ldots -2pa, -pa, 0, a, 2a, \ldots\}$.
That said, I agree with Drexel. There is either nothing really to prove or I am missing something...
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 $\displaystyle px$ is $\displaystyle x+x+\ldots+x$ $\displaystyle p$-times.
So what you need to show is inverses, (if $\displaystyle a \in pG$ then $\displaystyle -a \in pG$) and closure ($\displaystyle a + b \in pG$ for $\displaystyle a, b \in pG$). Everything is done additively, not multiplicatively as you were trying to do...
However, this holds for all $\displaystyle p \in \mathbb{Z}$, not just primes...