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Math Help - kernel of group

  1. #1
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    kernel of group

    Let G=Sym(n). Suppose that \theta \in Hom(C, G), where C is a cyclic group. Then how to show that \gamma (Ker \theta)=Ker \theta for \gamma \in Aut(C).
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  2. #2
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    Quote Originally Posted by dimuk View Post

    Let G=Sym(n). Suppose that \theta \in Hom(C, G), where C is a cyclic group. Then how to show that \gamma (Ker \theta)=Ker \theta for \gamma \in Aut(C).
    the claim is true for any group G. let C=<x> and \ker \theta = <x^m>. let \gamma(x)=x^r and x = \gamma(x^s). then for any integer k:

    \subseteq: \ \ \theta(\gamma(x^{km}))=(\theta(x^{km}))^r = 1_G.

    \supseteq: \ \ x^{km}=\gamma(x^{skm}) \in \gamma(\ker \theta).
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  3. #3
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    kernel of group

    Thanks.

    But I can't understand what you mean.
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  4. #4
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    Quote Originally Posted by dimuk View Post
    Thanks.

    But I can't understand what you mean.
    forget about that proof! it can be done much easier and in general case:

    let N=<x^m> be any subgroup of C=<x> and \gamma \in \text{Aut}(C). let \gamma(x)=x^r. then \gamma(x^{km})=x^{rkm} \in N. thus \gamma(N) \subseteq N. replacing \gamma with \gamma^{-1} gives us N \subseteq \gamma(N). thus \gamma(N)=N.
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