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Math Help - Group homomorphism

  1. #1
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    Group homomorphism

    Let \phi :G\to H be a group homomorphism.
    (1) phi is injective iff ker phi is 1
    (2) |G:ker\phi |=|\phi (G)|

    proof of (1)
    \Rightarrow
    Suppose phi is injective. Let a\in ker\phi.
    Then \phi(a)=1
    \phi(a^{-1})=\phi^{-1}(a)=1^{-1}=1=\phi(a)
    Since \phi is injective, a=a^{-1}
    Therefore, 1_G\mapsto 1_H

    \Leftarrow
    Suppose ker phi is 1. If \phi(x)=\phi(y), then \phi^{-1}(y)\phi(x)=1
    \Rightarrow\phi(y^{-1}x)=1 So y^{-1}x\in ker\phi
    y^{-1}x=1\Rightarrow y=x

    Is the above correct?

    How do I prove part 2?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Group homomorphism

    This isn't quite right. What if H has more than one element of order two? But, you have just proven it in your second line. You know that \phi(1)=1 and if \phi(a)=1 then by the DEFINITION of injectivity a=1. Your converse is correct though, good job!


    Part two is just a restatement (in a sense) of the first isomorphism theorem. Know it?
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  3. #3
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    Re: Group homomorphism

    Quote Originally Posted by Drexel28 View Post
    This isn't quite right. What if H has more than one element of order two? But, you have just proven it in your second line. You know that \phi(1)=1 and if \phi(a)=1 then by the DEFINITION of injectivity a=1. Your converse is correct though, good job!


    Part two is just a restatement (in a sense) of the first isomorphism theorem. Know it?
    How do I do the first part of part one then?
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  4. #4
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    Re: Group homomorphism

    Quote Originally Posted by dwsmith View Post
    Let \phi :G\to H be a group homomorphism.
    (1) phi is injective iff ker phi is 1
    (2) |G:ker\phi |=|\phi (G)|

    proof of (1)
    \Rightarrow
    Suppose phi is injective. Let a\in ker\phi.
    Then \phi(a)=1
    as alex pointed out, you're essentially done here, by the injectivity of φ (because φ(1_G) = 1_H, right?)

    \phi(a^{-1})=\phi^{-1}(a)=1^{-1}=1=\phi(a)
    Since \phi is injective, a=a^{-1}
    Therefore, 1_G\mapsto 1_H
    this, on the other hand, is completely incomprehensible to me. in the first place φ^-1(a) doesn't exist, because a is in G, and has no pre-image under φ.

    i suspect you meant (φ(a))^-1, however, all that does is show that a is either 1, or of order 2.

    \Leftarrow
    Suppose ker phi is 1. If \phi(x)=\phi(y), then \phi^{-1}(y)\phi(x)=1
    \Rightarrow\phi(y^{-1}x)=1 So y^{-1}x\in ker\phi
    y^{-1}x=1\Rightarrow y=x

    Is the above correct?

    How do I prove part 2?
    is there an isomorphism between G/ker(φ) and φ(G) (perhaps mapping the coset g(ker(φ)) --> φ(g) would work)?
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