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Math Help - Prove c is an isomorphism iff G is abelian

  1. #1
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    Prove c is an isomorphism iff G is abelian

    Let G be any group. Define c:G-->G by c(x)=x^-1 for all x in G. prove that c is an isomorphism iff G is abelian.


    Let c be an isomorphism. We want to show G is abelian.
    Since c is an isomorphism, c is 1-1, onto, and a homomorphism.
    We have c(a)=c(b) implies a=b by 1-1
    c(a)=a^-1
    c(b)=b^-1
    a^-1=b^-1
    a^-1a=b^-1a
    e=b^-1a
    be=bb^-1a
    b=a
    We have c(x)=y by onto
    y=c(x)=x^-1
    y=x^-1
    yx=x^-1x
    yx=e
    y^-1yx=y^-1e
    x=y^-1 onto
    Now c(ab)=c(a)c(b)
    c(a)c(b)=a^-1b^-1
    Since c(ab)=a^-1b^-1
    G must be abelian since (ab)^-1=b^-1a^-1

    Now suppose G is abelian. We want to show this implies c is an isomorphism.
    Since G is abelian, we know ab=ba
    1-1 and onto follow by the same ideas as before
    c(a)c(b)
    a^-1b^-1
    Since c is abelian (ab)^-1=a^-1b^-1
    c(a)c(b)=c(ab)

    I am really not sure if I did any of those correct
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kathrynmath View Post
    Let G be any group. Define c:G-->G by c(x)=x^-1 for all x in G. prove that c is an isomorphism iff G is abelian.


    Let c be an isomorphism. We want to show G is abelian.
    Since c is an isomorphism, c is 1-1, onto, and a homomorphism.
    We have c(a)=c(b) implies a=b by 1-1
    c(a)=a^-1
    c(b)=b^-1
    a^-1=b^-1
    a^-1a=b^-1a
    e=b^-1a
    be=bb^-1a
    b=a
    We have c(x)=y by onto
    y=c(x)=x^-1
    y=x^-1
    yx=x^-1x
    yx=e
    y^-1yx=y^-1e
    x=y^-1 onto
    Now c(ab)=c(a)c(b)
    c(a)c(b)=a^-1b^-1
    Since c(ab)=a^-1b^-1
    G must be abelian since (ab)^-1=b^-1a^-1

    Now suppose G is abelian. We want to show this implies c is an isomorphism.
    Since G is abelian, we know ab=ba
    1-1 and onto follow by the same ideas as before
    c(a)c(b)
    a^-1b^-1
    Since c is abelian (ab)^-1=a^-1b^-1
    c(a)c(b)=c(ab)

    I am really not sure if I did any of those correct
    I think you're making this harder than it need be.

    If c is an isomorphism than for any x,y\in G there are x',y' which map to them. Then, xy=c(x')c(y')=c(x'y')=(x'y')^{-1}=(y')^{-1}(x')^{-1} but c(x')=x\implies (x')^{-1}=x and simillarly for y so that the above states that xy=yx

    For your second argument I believe the fact that it's a homomorphism from what you said, now show it's bijective.
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  3. #3
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    bijective, we have 1-1 , onto
    1-1
    Assume c(x)=c(y)
    Then x^-1=y^-1
    xx^-1=xy^-1
    e=xy^-1
    ey=xy^-1y
    y=x

    onto:
    y=c(x)=x^-1
    y=x^-1 solve for x
    xy=xx^-1
    xy=e
    x=y^-1
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kathrynmath View Post
    bijective, we have 1-1 , onto
    1-1
    Assume c(x)=c(y)
    Then x^-1=y^-1
    xx^-1=xy^-1
    e=xy^-1
    ey=xy^-1y
    y=x

    onto:
    y=c(x)=x^-1
    y=x^-1 solve for x
    xy=xx^-1
    xy=e
    x=y^-1
    Correct!
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