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Math Help - Subgroup Problem

  1. #1
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    Subgroup Problem

    Let G be an abelian group. Let H be the subset of G consisting of the identity e together with all elements of G of order 2. Show that H is a subgroup of G.

    I'm not sure how to go about showing closure. Can it be assumed that any operation on elements of order 2 will produce another element of order 2? Doesn't seem right...
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by GoldendoodleMom View Post
    Can it be assumed that any operation on elements of order 2 will produce another element of order 2?
    In fact there is only one operation possible : *

    For the closure : let x,\,y \in H,
    (x*y)^2=(x*y)*(x*y)=x*y*x*y=x*x*y*y thanks to associativity and commutativity
    Then, x*x*y*y=(x*x)*(y*y)=e*e=e because x and y have order 2
    Hence, x*y also has order 2 and lies in H.
    Last edited by flyingsquirrel; April 16th 2008 at 10:04 AM. Reason: "are of order two" -> "have order two"
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  3. #3
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    Quote Originally Posted by flyingsquirrel View Post
    In fact there is only one operation possible : *

    Then, x*x*y*y=(x*x)*(y*y)=e*e=e because x and y are of order 2
    Hence, x*y is of order 2 and lies in H.

    Thanks for your help flyingsquirrel. I'm just stuck on why x*x or y*y = e?
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by GoldendoodleMom View Post
    Thanks for your help flyingsquirrel. I'm just stuck on why x*x or y*y = e?
    By data x,y \in H, which means they have order 2. Saying x has order 2 implies x*x = e.
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