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Thread: kernel of group

  1. #1
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    kernel of group

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

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

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

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