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Thread: Generators and relations

  1. #1
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    Generators and relations

    Use generators and relations to show that if $\displaystyle x\in\text{D}_{2n}$ which is not a power of r, then $\displaystyle rx=xr^{-1}$

    $\displaystyle \text{D}_{2n}=<r,s:r^n=s^2=1, \ rs=sr^{-1}>$

    Basically, x in this case is a flip. I am not sure what needs to proved. x is just a flip so by definition it satisfy $\displaystyle rx=xr^{-1}$
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  2. #2
    MHF Contributor Amer's Avatar
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    Re: Generators and relations

    since x is not a power of r it will be a multiplication of power of r and s say

    $\displaystyle x = r^k s $

    we have to show that $\displaystyle r r^k s = r^k s r^{-1} $
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  3. #3
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    Re: Generators and relations

    Quote Originally Posted by Amer View Post
    since x is not a power of r it will be a multiplication of power of r and s say

    $\displaystyle x = r^k s $

    we have to show that $\displaystyle r r^k s = r^k s r^{-1} $
    Can I do this:

    Since $\displaystyle rs=sr^{-1}$, $\displaystyle r^ksr^{-1}=r^krs=r^{k+1}s\equiv rr^ks=r^{k+1}s$
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Re: Generators and relations

    Quote Originally Posted by dwsmith View Post
    Can I do this:

    Since $\displaystyle rs=sr^{-1}$, $\displaystyle r^ksr^{-1}=r^krs=r^{k+1}s\equiv rr^ks=r^{k+1}s$
    Yup, that works.
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