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Thread: Simple group

  1. #1
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    Simple group

    Hi,

    Problem: $\displaystyle G$ - simple $\displaystyle \Rightarrow$ has $\displaystyle 8$ subgroups of order $\displaystyle 7$ and we could embedding it into $\displaystyle S_8$.

    Solution: First part is easy to show from Sylow Theorem.
    Suppose that we have $\displaystyle H_1,...,H_8$ - Sylow $\displaystyle 7$-subgroups and consider following mapping:
    $\displaystyle \phi : G \to S(\{H_1,...,H_8\})\cong S_8$
    $\displaystyle \phi(g)(H_i)=gH_ig^{-1}$, $\displaystyle g \in G$ and $\displaystyle i=1,...,8$.
    $\displaystyle \phi$ is homomorphism (it is easy to show that $\displaystyle \phi(gh)(H_i)=(\phi(g)\circ\phi(h))(H_i)$)

    $\displaystyle \mbox{Ker}\phi=\{g\in G : \forall_{i=1,...,8} \phi(g)(H_i)=H_i\}=\{g\in G : \forall_{i=1,...,8} gH_ig^{-1}=H_i\}$ <--- may I describe kernel such that? If yes then:

    $\displaystyle \mbox{Ker}\phi=\{1\}$ or $\displaystyle \mbox{Ker}\phi=G$ - because $\displaystyle G$ is simple.

    If $\displaystyle \mbox{Ker}\phi=G$, then $\displaystyle gH_i=H_ig$, so $\displaystyle H_1,...,H_8$ are simple - bad choice.

    So then $\displaystyle \mbox{Ker}\phi=\{1\}$ $\displaystyle \rightarrow$ $\displaystyle \phi$ is monomorphis. [qed]

    Is this solution good? Thanks for any advices.
    Last edited by Arczi1984; Dec 5th 2010 at 09:16 PM.
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  2. #2
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    Quote Originally Posted by Arczi1984 View Post
    Hi,

    Problem: $\displaystyle G$ - simple $\displaystyle \rightarrow$ has $\displaystyle 8$ subgroups of order $\displaystyle 7$ and we could embedding it into $\displaystyle S_8$.

    Solution: First part is easy to show from Sylow Theorem.
    Suppose that we have $\displaystyle H_1,...,H_8$ - Sylow $\displaystyle 7$-subgroups and consider following mapping:
    $\displaystyle \phi : G \to S(\{H_1,...,H_8\})\cong S_8$
    $\displaystyle \phi(g)(H_i)=gH_ig^{-1}$, $\displaystyle g \in G$ and $\displaystyle i=1,...,8$.
    $\displaystyle \phi$ is homomorphism (it is easy to show that $\displaystyle \phi(gh)(H_i)=(\phi(g)\circ\phi(h))(H_i)$)

    $\displaystyle \mbox{Ker}\phi=\{g\in G : \forall_{i=1,...,8} \phi(g)(H_i)=H_i\}=\{g\in G : \forall_{i=1,...,8} gH_ig^{-1}=H_i\}$ <--- may I describe kernel such that? If yes then:

    $\displaystyle \mbox{Ker}=\{1\}$ or $\displaystyle \mbox{Ker}=G$ - because $\displaystyle G$ is simple.

    If $\displaystyle \mbox{Ker}=G$, then $\displaystyle gH_i=H_ig$, so $\displaystyle H_1,...,H_8$ are simple - bad choice.

    So then $\displaystyle \mbox{Ker}=\{1\}$ $\displaystyle \rightarrow$ $\displaystyle \phi$ is monomorphis. [qed]

    Is this solution good? Thanks for any advices.
    A ""tiny"" omission in your post : you don't say what group G is!!

    Tonio
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  3. #3
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    You're right

    Let $\displaystyle G$ be a group of order $\displaystyle 168$.

    Like You wrote it was "tiny" omission My eye, You punched me
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  4. #4
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    Quote Originally Posted by tonio View Post
    A ""tiny"" omission in your post : you don't say what group G is!!

    Tonio


    As $\displaystyle \ker\phi=N_G(H)\,,\,\,N_G(H)=\ker\phi=G\Longleftri ghtarrow H\triangleleft G$ , which is absurd as we're assuming, I hope,

    that G is simple...

    Tonio
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