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    If \forall x,y \in G, xy=x^{-1}y^{-1}, prove that G must be abelian.

    My proof goes something like this:
    xy = x^{-1}y^{-1} \iff x^2y^2 = e, but note also that yx = y^{-1}x^{-1} \iff y^2x^2 = e. So that x^2y^2 = y^2x^2.

    But I have a sinking feeling in my stomach that doesn't prove that G is abelian. It needs to reduce down to xy = yx, doesn't it?

    How can I go about this proof?
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    (Original post by Zacken)
    If \forall x,y \in G, xy=x^{-1}y^{-1}, prove that G must be abelian.

    My proof goes something like this:
    xy = x^{-1}y^{-1} \iff x^2y^2 = e, but note also that yx = y^{-1}x^{-1} \iff y^2x^2 = e. So that x^2y^2 = y^2x^2.

    But I have a sinking feeling in my stomach that doesn't prove that G is abelian. It needs to reduce down to xy = yx, doesn't it?

    How can I go about this proof?
    You're quite right: you've shown that products of squares must commute, but not that products of all elements must commute.

    In fact, this kind of group is *very* specific. The first thing you should do when seeing rules like this is to check some special cases. What happens when x=y? When x=y^{-1}? When x=e?
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    (Original post by Smaug123)
    You're quite right: you've shown that products of squares must commute, but not that products of all elements must commute.

    In fact, this kind of group is *very* specific. The first thing you should do when seeing rules like this is to check some special cases. What happens when x=y? When x=y^{-1}? When x=e?
    x=y \Rightarrow x^2 = x^{-2}

    x=y^{-1} \Rightarrow y^{-1}y = yy^{-1} = e

    x=e \Rightarrow y = y^{-1} \Rightarrow y^2 = e

    I'm not really seeing how that gives any intuitive help.
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    (Original post by Zacken)
    x=y \Rightarrow x^2 = x^{-2}

    x=y^{-1} \Rightarrow y^{-1}y = yy^{-1} = e

    x=e \Rightarrow y = y^{-1} \Rightarrow y^2 = e

    I'm not really seeing how that gives any intuitive help.
    Indeed, the first two do not help. But the third is a really really strong condition on G: every element is self-inverse. (That is, G is a "boolean group".) Among other things, if the group is finite, it is therefore of order a power of 2.

    I'm trying to think of a non-magic way to show that the group is abelian. The proof follows in one line, but the correct expression to consider is a bit unmotivated. Give me a minute
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    (Original post by Zacken)
    x=y \Rightarrow x^2 = x^{-2}

    x=y^{-1} \Rightarrow y^{-1}y = yy^{-1} = e

    x=e \Rightarrow y = y^{-1} \Rightarrow y^2 = e

    I'm not really seeing how that gives any intuitive help.
    OK, motivation: we'll go for a contradiction. Suppose xy \not = y x. What can we deduce?
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    (Original post by Smaug123)
    OK, motivation: we'll go for a contradiction. Suppose xy \not = y x. What can we deduce?
    We can deduce that y^{-1}xyx^{-1} \neq e?
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    (Original post by Zacken)
    We can deduce that y^{-1}xyx^{-1} \neq e?
    And given that x, y are self-inverse…?
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    (Original post by Smaug123)
    And given that x, y are self-inverse…?
    Then y^{-1}x^{-1}y^{-1}x^{-1} = yxyx = (yx)^2 \neq e?
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    (Original post by Zacken)
    Then y^{-1}x^{-1}y^{-1}x^{-1} = yxyx = (yx)^2 \neq e?
    Precisely. That's a contradiction.

    More neatly stated (without contradiction): x y = (xy)^{-1} = y^{-1} x^{-1} = y x.
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    (Original post by Smaug123)
    Precisely. That's a contradiction.

    More neatly stated (without contradiction): x y = (xy)^{-1} = y^{-1} x^{-1} = y x.
    Oh my god! I'm so dense. That was so obvious. Thank you! :eek:
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    (Original post by Zacken)
    Oh my god! I'm so dense. That was so obvious. Thank you! :eek:
    No problem - it's really simple if you spot the right expression to consider, but that takes a bit of intuition. Contradiction often reduces the amount of intuition you need, but tends to produce a less neat proof.
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    (Original post by Smaug123)
    No problem - it's really simple if you spot the right expression to consider, but that takes a bit of intuition. Contradiction often reduces the amount of intuition you need, but tends to produce a less neat proof.
    Wait a second, re-reading your proof, I can't see why xy = (xy)^{-1}. We're given that xy = x^{-1}y^{-1} but (xy)^{-1} = y^{-1}x^{-1} \neq x^{-1}y^{-1} unless we assume G is abelian? :confused:
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    (Original post by Zacken)
    Wait a second, re-reading your proof, I can't see why xy = (xy)^{-1}. We're given that xy = x^{-1}y^{-1} but (xy)^{-1} = y^{-1}x^{-1} \neq x^{-1}y^{-1} unless we assume G is abelian? :confused:
    x y \in G, and we've already shown that for all g \in G, g = g^{-1}.
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    (Original post by Smaug123)
    x y \in G, and we've already shown that for all g \in G, g = g^{-1}.
    Fair enough (xy)^{-1} \in G but where does the equality of xy = (xy)^{-1} come from?
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    (Original post by Zacken)
    Fair enough (xy)^{-1} \in G but where does the equality of xy = (xy)^{-1} come from?
    Let g = xy. Then e g = e^{-1} g^{-1} by the original condition in the question (letting x = e, y = g), so g = g^{-1}. That is, (xy) = (xy)^{-1}.
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    (Original post by Smaug123)
    Let g = xy. Then e g = e^{-1} g^{-1} by the original condition in the question (letting x = e, y = g), so g = g^{-1}. That is, (xy) = (xy)^{-1}.
    *wipes sweat off forehead* This abstractness is hard to swallow. Thanks again, it makes a load more sense now.

    +Rep
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    (Original post by Zacken)
    *wipes sweat off forehead* This abstractness is hard to swallow. Thanks again, it makes a load more sense now.

    +Rep
    Thanks sorry, do tell me if I'm being too abstract. It takes a bit of time to get the right frame of mind to just swap variables in and out for each other like that. A big part of becoming a mathematician is, in my opinion, gaining the ability to detach your understanding of a thing (eg. the element xy as a member of G) from its name ("xy", which is not helpful when it comes to deriving the fact that xy = (xy)^{-1}).
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    (Original post by Smaug123)
    Thanks sorry, do tell me if I'm being too abstract. It takes a bit of time to get the right frame of mind to just swap variables in and out for each other like that. A big part of becoming a mathematician is, in my opinion, gaining the ability to detach your understanding of a thing (eg. the element xy as a member of G) from its name ("xy", which is not helpful when it comes to deriving the fact that xy = (xy)^{-1}).
    No, no, you make way more sense than my textbook! - quite a good teacher, really.

    I'm still struggling to get used to that, trying to detach my pre-conceived notions, but I always get scared that detaching them might lead to mistakes somewhere, so I juggle the two in my head. I'll get the hang of it someday.
 
 
 
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