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Math Help - Groups

  1. #1
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    Groups

    Hi. Hope someone can help with my problem.

    Let G be a group with identity e. If A and B are subgroups of G then, from the Direct Product of Groups Theory, AxB is a subgroup of GxG. Use this result througout.

    a) Show that, if A and B are two normal subgroups of G, then AxB is normal subgroup of GxG (Not sure if I am to prove that A and B are two normal subgroups as well as proving AxB is a normal subgroup of GxG).

    b) Let H be the subset

    H={(g,g):g E G}

    Show that H is a subgroup of GxG

    Normally I can do these type of questions, it's just the Gx G and AxB bit has confused me a bit.

    Thanx
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  2. #2
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    Quote Originally Posted by bex23 View Post
    a) Show that, if A and B are two normal subgroups of G, then AxB is normal subgroup of GxG (Not sure if I am to prove that A and B are two normal subgroups as well as proving AxB is a normal subgroup of GxG).
    You need to prove (g_1,g_2)(a,b)(g_1^{-1},g_2^{-1})\in A\times B thus (g_1ag_1^{-1},g_2bg_2^{-1})\in A\times B because g_1ag_1^{-1}\in A and g_2bg_2^{-1} \in B.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    You need to prove (g_1,g_2)(a,b)(g_1^{-1},g_2^{-1})\in A\times B thus (g_1ag_1^{-1},g_2bg_2^{-1})\in A\times B because g_1ag_1^{-1}\in A and g_2bg_2^{-1} \in B.
    I seem to be having a brain dead day, as I can't even remember how to do that
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  4. #4
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    Groups

    Right, I think I have figured out part a) it's just part b) that is giving me problems. I know that to prove that H is a subgroup of GxG, I need to prove the three subgroup axioms Closure, Identity and Inverse. So far I have only gone so far as Closure and even then I am not sure that this is correct any help will be greatly appreciated.

    Closure:

    Let (g_1, g_2) and (g_3, g_4) be any two elments of GxG. By the definition of GxG, we know that g_1, g_2, g_3 and g_4 are elemnts of G.

    Since G is a group, it is closed under the operation o and we have

    (g_1,g_2)o(g_3,g_4)=(g_1 o g_3, g_2 o g_4)
    which belongs to H

    So the closue axiom holds.

    Is this correct and if it is (which I don't think it is) can you help me get started with the other two axioms.

    Thanx

    Bex
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  5. #5
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    It is correct.

    For identity element consider (e,e), show it is identity element.
    For inverse, note (g_1,g_2)^{-1} = (g_1^{-1},g_2^{-1}).
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  6. #6
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    Groups

    Identity:
    To prove the identity axiom we use the identity element (e,e) which belongs to GxG and the arbitrary element
    (g_1,g_2) which also belongs to GxG, Thus

    (e,e) o (g_1,g_2)= (e o g_1, e o g_2)
    =(g_1,g_2) which belongs to H

    Inverse:

    For this I have two methods (neither which I am very confidant on).

    Method1:


    To prove the inverse axiom, let (g_1,g_2) be and element of GxG where g_1^-1 and g_2^-1 are the inverses respectively. Proving that (g_1^-1, g_2^-1) is the inverse of (g_1,g_2) in GxG this will prove the inverse axiom (I hope!!).

    We have

    (g_1,g_2)o(g_1^-1, g_2^-1)=(g_1og_1^-1, g_2og_2^-1)
    =(e,e)

    Also

    (g_1^-1, g_2^-1)o(g_1,g_2)=(e,e)

    Hence, (g_1^-1,g_2^-1) is the inverse of (g_1,g_2)

    Method 2:

    Let (g_1,g_2) be an element of GxG then the inverse is
    (g_1^-1,g_2^-1), which belongs to H as it is of the form (g,g) for g E G.

    I think method 2 is the more wrong of the two but put in in here just in case.

    Have I gone completly off the mark here. I was surprised when you said that the first part was right, and so now the rest will probably be all wrong. Lol!! Can you just tell me if I have gone in the right direction, and if not point me in the right one.

    Thanx

    Bex
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