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Math Help - Groups- Proof using on the defination of a group

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    Groups- Proof using on the defination of a group

    Let G be a group and let a, b G. Using only the definition of a group show that the two equations xa = b and ay = b have unique solutions in G. Is it always true that x = y?
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    Quote Originally Posted by amm345 View Post
    Let G be a group and let a, b G. Using only the definition of a group show that the two equations xa = b and ay = b have unique solutions in G. Is it always true that x = y?

    all elements of a group have inverses and the inverse is commutative
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    Quote Originally Posted by amm345 View Post
    Let G be a group and let a, b G. Using only the definition of a group show that the two equations xa = b and ay = b have unique solutions in G. Is it always true that x = y?

    Multiply xa = b by the inverse of a from the right and ay = b by the inverse of a from the left and you'll get uniqueness (since x (y) is expressable by means of a,b)

    No, in general the solutions aren't equal (if the group is non-abelian).
    Take for example, in the symmetric group S_3, x(12) = (123), (12)y = (123).

    The first equation's solution is x = (123)(12) = (13), whereas the second one's is y = (12)(123) = (23).

    Tonio
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    I'm sorry, but I just don't see how multiplying by inverse of a on the right of the first and the left of the second proves uniqueness.. Please explain more?
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    Quote Originally Posted by amm345 View Post
    I'm sorry, but I just don't see how multiplying by inverse of a on the right of the first and the left of the second proves uniqueness.. Please explain more?

    xa=y so x^{-1}xa=x^{-1}y so 1\cdot a=x^{-1}y so a=x^{-1}y

    multiplying is well definited so the unique solution is x^{-1}y, the product cannot be two different elements of g
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    Quote Originally Posted by amm345 View Post
    I'm sorry, but I just don't see how multiplying by inverse of a on the right of the first and the left of the second proves uniqueness.. Please explain more?

    I already did: xa = b ==> xa*a^(-1) = b*a^(-1) ==> x = ba^(-1) and you're expressing the solution x as a function of a,b and this means x is unique since a, b are unique (arbitrary. but once they're given they're fixed and unique).
    The same is true with the other one.

    Tonio
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