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Thread: Group homomorphism

  1. #1
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    Group homomorphism

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

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

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

    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 $\displaystyle H$ has more than one element of order two? But, you have just proven it in your second line. You know that $\displaystyle \phi(1)=1$ and if $\displaystyle \phi(a)=1$ then by the DEFINITION of injectivity $\displaystyle a=1$. 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 $\displaystyle H$ has more than one element of order two? But, you have just proven it in your second line. You know that $\displaystyle \phi(1)=1$ and if $\displaystyle \phi(a)=1$ then by the DEFINITION of injectivity $\displaystyle a=1$. 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 $\displaystyle \phi :G\to H$ be a group homomorphism.
    (1) phi is injective iff ker phi is 1
    (2) $\displaystyle |G:ker\phi |=|\phi (G)|$

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

    $\displaystyle \phi(a^{-1})=\phi^{-1}(a)=1^{-1}=1=\phi(a)$
    Since $\displaystyle \phi$ is injective, $\displaystyle a=a^{-1}$
    Therefore, $\displaystyle 1_G\mapsto 1_H$
    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.

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

    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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