Results 1 to 4 of 4

Math Help - Finite Abelian Groups

  1. #1
    Junior Member
    Joined
    Jun 2011
    Posts
    34

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,932
    Thanks
    782

    Re: Finite Abelian Groups

    Quote Originally Posted by jcir2826 View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jun 2011
    Posts
    34

    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,932
    Thanks
    782

    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?
    Last edited by SlipEternal; September 29th 2013 at 12:04 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. finite abelian groups
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: April 10th 2013, 10:23 AM
  2. finite abelian groups question
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 9th 2013, 08:04 PM
  3. finite abelian groups
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 2nd 2010, 09:00 PM
  4. Finite Abelian Groups
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: November 8th 2009, 04:12 PM
  5. finite abelian groups, tensor product
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 13th 2009, 07:27 PM

Search Tags


/mathhelpforum @mathhelpforum