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Math Help - Is this true for all groups

  1. #1
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    Is this true for all groups

    Let H be a group and \gamma \in Aut(H) . Then \gamma (Ker \theta) =ker \theta for a group G and \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 \gamma \in Aut(H) . Then \gamma (ker \theta) =ker \theta for a group G and \theta \in Hom(H, G).
    the answer is no. let N be any non-characteristic normal subgroup of a group H. let G=H/N and \theta: H \longrightarrow H/N be the natural homomorphism. obviously \ker \theta = N.

    since N is not charactersitic, there exists \gamma \in \text{Aut}(H) such that \gamma(N) \neq N. thus \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 G is a finite abelian group then G\simeq C_1\times ... \times C_k where 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 H_1,H_2 then it is true for 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 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 H=\mathbb{Z}_2 \oplus \mathbb{Z}_2. define \gamma: H \longrightarrow H by: \gamma(a,b)=(b,a). then \gamma \in \text{Aut}(H). let N=\{(0,0),(1,0)\} and G=\mathbb{Z}_2 and define

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