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Math Help - Cross Cancellation implies commutativity?

  1. #1
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    Cross Cancellation implies commutativity?

    Hello!

    If we have a group G: if a,b,c \in G
    and ab=ca implies b=c

    then prove G is Albelian.

    To me, it seems trivial, that if we assume ab = ca and since b=c, ab = ba  \forall a,b \in G

    I have a feeling this proof is not valid, but I am not sure how? Does anyone else see an error in this logic, or have a suggestion for a method to prove this properly? Or is this legit?

    Thank you!
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    Quote Originally Posted by matt.qmar View Post
    Hello!

    If we have a group G: if a,b,c \in G
    and ab=ca implies b=c

    then prove G is Albelian.

    To me, it seems trivial, that if we assume ab = ca and since b=c, ab = ba  \forall a,b \in G

    I have a feeling this proof is not valid, but I am not sure how? Does anyone else see an error in this logic, or have a suggestion for a method to prove this properly? Or is this legit?

    Thank you!
    You have not proven that, for all a,b in G, there exists c in G such that ab=ca.

    (Whether or not we want to show this, or whether it's true or easy, I'm not sure. I'm not an expert on group theory, just happened to spot the error.)
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    Well I thought perhaps that given three arbitrary elements of G (a,b, and c)
    to show the group is albelian, show ab = ba for two arbitrary elements.

    So I thought we could start by assuming ab = ca. Then by the assumption of G, we know b =c, so we substitute b for c in the RHS, and get ab = ba.

    I think I see the problem, though : when we say ab = ca, then a,b, and c are no longer arbitrary?

    I think the way to do it is to start with any a and b,
    then show ab must be = ba...

    Sorry, hope this was clear! Thanks for the reply!
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    Oh! just re-read your comment.

    In the assumption, we are saying G is a group with the property that,

    IF ab=ca for some a,b,c in G

    THEN b=c.
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  5. #5
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    Quote Originally Posted by matt.qmar View Post
    So I thought we could start by assuming ab = ca.
    By doing this, you assume that such c exists, for arbitrary a and b. If there is no such element c in G for any particular a and b, then your proof falls apart.

    Like I said, I'm not sure how to do the actual proof, just explaining how your reasoning has a gap in it.

    Edit: We posted at the same time and I didn't see this until after submitting:

    Quote Originally Posted by matt.qmar View Post
    Oh! just re-read your comment.

    In the assumption, we are saying G is a group with the property that,

    IF ab=ca for some a,b,c in G

    THEN b=c.
    I'm still not sure we're on the same page, so consider this. IF we can prove that there always exists such a c, then we can write a full proof as such.

    Let a, b in G. Then, by magical lemma, we know there exists c in G such that ab=ca. This implies that b=c, hence ab=ba, hence G is abelian. QED.

    But if there are groups in which we are not guaranteed c exists (that is, we do not have our magical lemma to rely upon), then this proof does not work and we have to find another way.
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    Aha! I see how you mean.

    Ok, I just thought...

    start with ab = ab.
    Then cabc = cabc (left and right multiply by some c in G)
    then by the property of G, abc = cab
    is this sufficent? since ab is in G (G is closed)
    then call ab = d
    so we have dc = cd, so G is commutative?


    but now we have the condition d need be a product of elements, although each element in a group is a product of two other elements
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    Quote Originally Posted by matt.qmar View Post
    Aha! I see how you mean.

    Ok, I just thought...

    start with ab = ab.
    Then cabc = cabc (left and right multiply by some c in G)
    then by the property of G, abc = cab
    is this sufficent? since ab is in G (G is closed)
    then call ab = d
    so we have dc = cd, so G is commutative?


    but now we have the condition d need be a product of elements, although each element in a group is a product of two other elements
    Hmm I'm not sure... I got a little lost at the end.

    But I think I just realized something obvious. I couldn't remember (!) whether we had inverses at our disposal when working in a group. But checking the axioms -- we do! So going back to the magical lemma proof in post #5, we can just let c=aba^{-1}. I think we're done then right?

    Edit: Looking at your latest proof (post #6) again, I believe it is valid, but has confusing lettering and presentation. If you did some re-lettering and added parentheses in a few places I think it would be easier to follow. Although I think it's cleaner to do it the "magical lemma" way, which is in fact very close to what you had in your post #1 anyway.
    Last edited by undefined; October 5th 2010 at 07:17 PM.
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    That is quite clear to me now! Thank you!
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    So this is how it looks...

    let e be the identity of G, let a, b \in G
    ab = ab(e) = ab(a^{-1}a)
    then by properties of G and commutativity
    a(b) = (aba^{-1})a implies b = aba^{-1}
    right multiply by a, and commute
    ba = ab(a^{-1}a) = ab(e) = ab
    and so arbitrary a and b in G commute so it's Albelian.

    I'm convinced. Thanks!!
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  10. #10
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    Quote Originally Posted by matt.qmar View Post
    So this is how it looks...

    let e be the identity of G, let a, b \in G
    ab = ab(e) = ab(a^{-1}a)
    then by properties of G and commutativity
    a(b) = (aba^{-1})a implies b = aba^{-1}
    right multiply by a, and commute
    ba = ab(a^{-1}a) = ab(e) = ab
    and so arbitrary a and b in G commute so it's Albelian.

    I'm convinced. Thanks!!
    You're welcome!

    I think you meant associativity where you wrote commutativity -- commutativity is what we want to prove!

    Also once you have that b=aba^{-1} you can simply substitute to get ab=ba.

    If you don't mind, an alternate way of stating the proof (I drop in and out of LaTeX just for ease of typing):

    Let a,b\in G (a and b not necessarily distinct), and let e be the identity element. Then

    ab=abe=aba^{-1}a=(aba^{-1})a.

    By the property of G, we have that b=aba^{-1}. Thus we substitute to get

    ab=ba.

    Thus G is abelian. QED.
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