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Thread: Is this true for all groups

  1. #1
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    Is this true for all groups

    Let H be a group and $\displaystyle \gamma \in Aut(H)$ . Then $\displaystyle \gamma (Ker \theta) =ker \theta$ for a group G and $\displaystyle \theta \in Hom(H, G)$.

    This is true if H is a cyclic group. I want to know whether is this true for any other finite groups.
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  2. #2
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    Quote Originally Posted by dimuk View Post
    Let H be a group and $\displaystyle \gamma \in Aut(H)$ . Then $\displaystyle \gamma (ker \theta) =ker \theta$ for a group G and $\displaystyle \theta \in Hom(H, G)$.
    the answer is no. let $\displaystyle N$ be any non-characteristic normal subgroup of a group $\displaystyle H.$ let $\displaystyle G=H/N$ and $\displaystyle \theta: H \longrightarrow H/N$ be the natural homomorphism. obviously $\displaystyle \ker \theta = N.$

    since $\displaystyle N$ is not charactersitic, there exists $\displaystyle \gamma \in \text{Aut}(H)$ such that $\displaystyle \gamma(N) \neq N.$ thus $\displaystyle \gamma(\ker \theta) \neq \ker \theta. \ \ \Box$
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  3. #3
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    Is this true for all groups

    Then how about the finite abelian groups?
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  4. #4
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    Quote Originally Posted by dimuk View Post
    Then how about the finite abelian groups?
    If $\displaystyle G$ is a finite abelian group then $\displaystyle G\simeq C_1\times ... \times C_k$ where $\displaystyle C_1,...,C_k$ are cyclic. You proved this result for cyclic groups. Now it remains to prove (or disprove) that if it is true for $\displaystyle H_1,H_2$ then it is true for $\displaystyle H_1\times H_2$.
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  5. #5
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    Quote Originally Posted by dimuk View Post
    Then how about the finite abelian groups?
    i don't know if you understood the idea in my previous post or not. as long as $\displaystyle H$ has a normal subgroup which is not characteristic, the claim in your problem would be false.

    here's a simple example of finite abelian groups: let $\displaystyle H=\mathbb{Z}_2 \oplus \mathbb{Z}_2.$ define $\displaystyle \gamma: H \longrightarrow H$ by: $\displaystyle \gamma(a,b)=(b,a).$ then $\displaystyle \gamma \in \text{Aut}(H).$ let $\displaystyle N=\{(0,0),(1,0)\}$ and $\displaystyle G=\mathbb{Z}_2$ and define

    $\displaystyle \theta: H \longrightarrow G$ by: $\displaystyle \theta(a,b)=b.$ clearly $\displaystyle \theta \in \text{Hom}(H,G),$ and $\displaystyle \ker \theta=\{(0,0),(1,0) \}=N \neq \{(0,0),(0,1)\}=\gamma(N)=\gamma(\ker \theta).$
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