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Math Help - Intersection of abelian subgroups is a normal subgroup

  1. #1
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    Intersection of abelian subgroups is a normal subgroup

    "Prove that if H and K are abelian subgroups of a group G, then the intersection of H and K is a normal subgroup of <H,K>"

    Just to be clear a normal subgroup of G is a subgroup where its elements multiplied by an element of G is an element in the normal subgroup, and <H,K> is the group where its elements are linear combinations of the elements of H and the elements of K, correct? I ask this just to be sure I understand the question, but I still don't know how to do this proof?
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    Re: Intersection of abelian subgroups is a normal subgroup

    Hi,
    Your description of normal subgroups is faulty. A subgroup H of G is a normal subgroup iff for any x in G x-1Hx = H. Also, <H,K> is not "linear combinations" of elements of H and K. Here's a solution with a few details left for you:

    Intersection of abelian subgroups is a normal subgroup-mhfgroups3.png
    Thanks from HowDoIMath
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    Re: Intersection of abelian subgroups is a normal subgroup

    Ok it looks like my definition of a normal subgroup was too much like the definition of an ideal under a ring. So far I've only worked with cyclic groups with only one generating element like <H> or <K>, but never <H,K> so that's how I got confused. I don't have a very firm grasp of group theory, but to make sure I understand this, "if x_1.x_2.x_3....x_n where each x_i is in H or K" means that we are working with multiplicative groups and NOT additive groups, correct? And N_G means "normal group of"?
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    Re: Intersection of abelian subgroups is a normal subgroup

    Yes, the convention is that the group operation is "multiplication" unless specified otherwise. Now for G a group and H a subgroup of G, NG(H) is called the normalizer (in G) of H. N_G(H)=\{x\in G\,:\,x^{-1}Hx=x\}. This is easily shown to be a subgroup of G that contains H. H is a normal subgroup of G iff NG(H) = G.
    Last edited by johng; April 6th 2013 at 09:30 AM.
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