Finite Abelian Groups

• Sep 28th 2013, 10:06 PM
jcir2826
Finite Abelian Groups
Let G be an abelian group, define a subgroup S of G to be pure if for all m in Z. S intersection mG = mS.
prove that if G is p-primary abelian group. then S is pure iff S intersection P^n S for all n >=0
• Sep 29th 2013, 06:48 AM
SlipEternal
Re: Finite Abelian Groups
Quote:

Originally Posted by jcir2826
Let G be an abelian group, define a subgroup S of G to be pure if for all m in Z. S intersection mG = mS.
prove that if G is p-primary abelian group. then S is pure iff S intersection P^n S for all n >=0

Please reread what you wrote. Something seems to be missing because "S intersection P^n S for all n >=0" is not a boolean expression. Also, what is P?
• Sep 29th 2013, 10:02 AM
jcir2826
Re: Finite Abelian Groups
Let G be an abelian group, not necessarily primary. Define a subgroup S of G to be pure subgroup if, for all m in Z, S intersection mG = mS.
Prove that if G is a p-primary abelian group, then a subgroup S of G is pure if and only if S intersection p^nS for all n>= 0.

p is prime number.
• Sep 29th 2013, 10:53 AM
SlipEternal
Re: Finite Abelian Groups
I don't know how to test if the expression "S intersection p^nS for all n>=0" is true or false. That is not a boolean expression. That expression describes a collection of sets. It does not suggest that the sets need to exist. It does not suggest that the sets need to be equal to other sets. It is just a collection of sets. The intersection operator is not a comparison operator. Neither is the ^ operator. You are not comparing S intersection p^nS to anything. So, since that set is neither true nor false, the expression "If G is a p-primary abelian group, then a subgroup S of G is pure if and only if S intersection p^nS for all n>=0." is not a sentence (it can not be evaluated to true or false).

Edit: I can guess at what you were trying to say. If $G$ is a $p$-primary abelian group, then a subgroup $S\le G$ is pure if and only if $S\cap p^n G = p^n S$ for all $n\ge 0$.

If that is the case, then the argument seems pretty self-evident. The first direction of the implication $\Longrightarrow$ is by definition. Next, to show the reverse implication, assume you have $S\cap p^n G = p^n S$ for all $n\ge 0$ and show that $S\cap m G = m S$ for all $m\in \mathbb{Z}$. That seems like a straightforward argument to me. Where are you running into issues?