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Thread: 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; Apr 26th 2008 at 12:42 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 $\displaystyle (1, 0)$ 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 :

    $\displaystyle (p,q)*(r,s)=(r,s)$
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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 $\displaystyle (p',\,q')$ such that $\displaystyle (p,\,q)*(p',\,q')=(p',\,q')*(p,\,q)=(e_1,\,e_2)$ for all pairs $\displaystyle (p,\,q)$. (you can, for example, solve $\displaystyle (p',\,q')*(p,\,q)=(e_1,\,e_2)$ for $\displaystyle (p',\,q')$ and then deduce the equality $\displaystyle (p,\,q)*(p',\,q')=(p',\,q')*(p,\,q)$)

    ($\displaystyle (e_1,\,e_2)$ : identity element)
    Last edited by flyingsquirrel; Apr 26th 2008 at 01:05 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 $\displaystyle (p',\,q')$ such that $\displaystyle (p,\,q)*(p',\,q')=(p',\,q')*(p,\,q)=(e_1,\,e_2)$ for all pair $\displaystyle (p,\,q)$. (you can, for example, first solve $\displaystyle (p',\,q')*(p,\,q)=(e_1,\,e_2)$ for $\displaystyle (p'\,q')$ and the equality $\displaystyle (p,\,q)*(p',\,q')=(p',\,q')*(p,\,q)$ will come after)

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

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

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

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