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Math Help - Group Theory - newcomer needs help

  1. #1
    Newbie Alborg's Avatar
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    Group Theory - newcomer needs help

    Given that A = {}

    Define * on A by (p,q) * (r,s) = (pr+qs, ps+qr)

    Does anyone know what an identity element for * on A might be? And how do I show that (A,*) is a group other than having shown that it is associative and is closed under *?

    Thank you in advance,
    Aalborg
    Last edited by CaptainBlack; April 26th 2008 at 01:42 PM.
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  2. #2
    MHF Contributor
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    Quote Originally Posted by Alborg View Post
    Given that A = {}

    Define * on A by (p,q) * (r,s) = (pr+qs, ps+qr)

    Does anyone know what an identity element for * on A might be? And how do I show that (A,*) is a group other than having shown that it is associative and is closed under *?

    Thank you in advance,
    Aalborg
    I think (1, 0) is an identity element for * on A.
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  3. #3
    Moo
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    Hello,

    For the identity element, I think that you have to solve for p and q in :

    (p,q)*(r,s)=(r,s)
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Hi

    And how do I show that (A,*) is a group other than having shown that it is associative and is closed under *?
    You also need show that there exists (p',\,q') such that (p,\,q)*(p',\,q')=(p',\,q')*(p,\,q)=(e_1,\,e_2) for all pairs (p,\,q). (you can, for example, solve (p',\,q')*(p,\,q)=(e_1,\,e_2) for (p',\,q') and then deduce the equality (p,\,q)*(p',\,q')=(p',\,q')*(p,\,q))

    ( (e_1,\,e_2) : identity element)
    Last edited by flyingsquirrel; April 26th 2008 at 02:05 PM.
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  5. #5
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    Quote Originally Posted by flyingsquirrel View Post
    Hi

    You also need show that there exists (p',\,q') such that (p,\,q)*(p',\,q')=(p',\,q')*(p,\,q)=(e_1,\,e_2) for all pair (p,\,q). (you can, for example, first solve (p',\,q')*(p,\,q)=(e_1,\,e_2) for (p'\,q') and the equality (p,\,q)*(p',\,q')=(p',\,q')*(p,\,q) will come after)

    ( (e_1,\,e_2) : identity element)
    Given (p, q) \in A let (r, s) \in A be defined as such:

    If |p| > |q|, then take r = kp, s = -kq such that k(p^2 - q^2) = 1.

    If |q| > |p|, then take r = -kp, s = kq such that k(q^2 - p^2) = 1.

    This shows that every element x \in (A, *) has an inverse under the identity (1, 0).
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