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    Need help on aiii

    what does it mean by conjugate? or do I just power all of y by a?

    Also can someone tell me if I got these right please.

    ai) a b^-1 a^2 b a b a^-1

    aii) b^-3 a^-1 b^-1 a
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    (Original post by cooldudeman)
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    Need help on aiii

    what does it mean by conjugate? or do I just power all of y by a?

    Also can someone tell me if I got these right please.

    ai) a b^-1 a^2 b a b a^-1

    aii) b^-3 a^-1 b^-1 a
    Holy moly. Is that uni stuff?
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    (Original post by Andy98)
    Holy moly. Is that uni stuff?
    yep
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    (Original post by cooldudeman)
    yep
    No wonder I don't get it :lol:
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    (Original post by cooldudeman)
    ai) a b^-1 a^2 b a b a^-1

    aii) b^-3 a^-1 b^-1 a
    Can't help with aiii - don't even know what y^a means.

    aii however is incorrect, try forming yy^{-1} using that.

    Recall (ab)^{-1}=b^{-1}a^{-1}
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    (Original post by cooldudeman)
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    Need help on aiii
    From a bit of digging I suspect you are looking for the element: aya^{-1}

    You should have a definition of the symbology in your notes somewhere.
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    (Original post by ghostwalker)
    From a bit of digging I suspect you are looking for the element: aya^{-1}

    You should have a definition of the symbology in your notes somewhere.
    so aii is a b^-1 a^-1 b^-3 right?

    and thanks for aiii, but I will ask my lecturer just in case.

    For b, what does it mean by conjugate? I cannot for the life of me find the definition in my notes!
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    (Original post by Andy98)
    Holy moly. Is that uni stuff?
    Tots get you man When a question barely makes sense :laugh:
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    (Original post by cooldudeman)
    so aii is a b^-1 a^-1 b^-3 right?

    and thanks for aiii, but I will ask my lecturer just in case.

    For b, what does it mean by conjugate? I cannot for the life of me find the definition in my notes!
    There's always Google.

    Two elements x,y are conjugate if there exists an element, say a, such that ax=ya or putting it another way axa^{-1}=y, and I suspect with the notation used in your photo, that's y=x^a
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    (Original post by ghostwalker)
    There's always Google.

    Two elements x,y are conjugate if there exists an element, say a, such that ax=ya or putting it another way axa^{-1}=y, and I suspect with the notation used in your photo, that's y=x^a
    so is it like aza^-1=z' = a^2 b^2 ab^-1 ab^-1 ?

    Also for 3c, is it like we find aza^-1=z' and a^-1 za=z' and bzb^-1=z' and b^-1 zb=z'
    ??
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    (Original post by cooldudeman)
    so is it like aza^-1=z' = a^2 b^2 ab^-1 ab^-1 ?
    Well, at the end z has a^-1 already, so you'd end up with a^-2 at the end

    aza^{-1}=z' = a^2 b^2 ab^{-1} ab^{-1}a^{-2}

    Also for 3c, is it like we find aza^-1=z' and a^-1 za=z' and bzb^-1=z' and b^-1 zb=z'
    ??
    I suspect so. Those are certainly conjugate, but:

    Conjugacy is an equivalence relation, and you could do (ab)z(ab)^{-1} etc, but then there would be an infinite number of possibilities.

    But since you're expected to list them all, that can't be the case.
    I don't know what's expected there - I'd need to see some of the background material, or someone else may be familiar with this and more able to help.
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    (Original post by ghostwalker)
    Well, at the end z has a^-1 already, so you'd end up with a^-2 at the end

    aza^{-1}=z' = a^2 b^2 ab^{-1} ab^{-1}a^{-2}



    I suspect so. Those are certainly conjugate, but:

    Conjugacy is an equivalence relation, and you could do (ab)z(ab)^{-1} etc, but then there would be an infinite number of possibilities.

    But since you're expected to list them all, that can't be the case.
    I don't know what's expected there - I'd need to see some of the background material, or someone else may be familiar with this and more able to help.
    Thanks at least for helping. I will ask around uni.

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