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Thread: Deducing a relation from others

  1. #1
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    Deducing a relation from others

    Hello, I'm trying to deduce the relation $\displaystyle xy^4 = y^4 x$ from $\displaystyle xy^2 = y^3 x$ and $\displaystyle yx^3 = x^2y$. So far I've gotten:

    $\displaystyle [x,y^4] = x^{-1} y^{-4} xy^4 = x^{-1}y^{-4}(xy^2)y^2 = x^{-1} y^{-1} (xy^2) = x^{-1}y^{-1} y^3 x = x^{-1}y^2x$

    However, everything I try from this point onwards seems to make it more complicated. Any help would be greatly appreciated!
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  2. #2
    Junior Member nimon's Avatar
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    Hmm...

    Suppose that $\displaystyle xy^{4}=y^{4}x$ is true and that $\displaystyle xy^{2} = y^{3}x$ and $\displaystyle yx^{3} = x^{2}y$. Then

    $\displaystyle xy^{4} = y^{6}x = y^{4}x \Rightarrow y^{2}= e \Rightarrow x = yx \Rightarrow y \,\,\text{is the identity}$

    But then $\displaystyle x^{3} = x^{2} \Rightarrow x \,\,\text{is the identity.}$ This would be a consequence of these relations; does this make sense in the group your considering? I only ask because its nice to know that it's plausible to derive the first relation from the other two before you begin. If $\displaystyle x,y$ are definitely not the identity element then you needn't waste your time!

    Please make sure I haven't made a silly mistake because I excel at those.
    Last edited by nimon; Apr 9th 2010 at 04:41 AM.
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  3. #3
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    Well the question asks for this group with presentation $\displaystyle \langle x, y \mid xy^2 = y^3 x, yx^3 = x^2 y \rangle$. No other relations are given and the 3rd relation was given as a hint. I arrived at the same conclusion as you, but I just can't seem to get the third relation.
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