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Thread: Solvable Group

  1. #1
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    Solvable Group

    Show $\displaystyle B=\left\{ \left. \left(\begin{array}{cc}x&z\\0&y\end{array}\right) \;\; \right| \;\; x,y\in \mathbb{F}_p^\times,\; z\in\mathbb{F}_p \right\} $ is a solvable group.

    The group operation is matrix multiplication.
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  2. #2
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    Quote Originally Posted by mathman88 View Post
    Show $\displaystyle B=\left\{ \left. \left(\begin{array}{cc}x&z\\0&y\end{array}\right) \;\; \right| \;\; x,y\in \mathbb{F}_p^\times,\; z\in\mathbb{F}_p \right\} $ is a solvable group.

    The group operation is matrix multiplication.

    Check the commutator group: it then should be "almost" obvious that it is abelian so...

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    Check the commutator group: it then should be "almost" obvious that it is abelian so...

    Tonio
    $\displaystyle [B,B] = \left\{ \left. \left(\begin{array}{cc}1&z\\0&1\end{array}\right) \;\; \right| \;\; z\in\mathbb{F}_p \right\} $

    And it easy to see $\displaystyle [B,B] $ is abelian.

    We then have $\displaystyle \{1\} \triangleleft [B,B] \triangleleft B $ and $\displaystyle [B,B]/\{1\} = [B,B] $ is cyclic, but how is $\displaystyle B/[B,B] $ cyclic?
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    Quote Originally Posted by mathman88 View Post
    $\displaystyle [B,B] = \left\{ \left. \left(\begin{array}{cc}1&z\\0&1\end{array}\right) \;\; \right| \;\; z\in\mathbb{F}_p \right\} $

    And it easy to see $\displaystyle [B,B] $ is abelian.

    How does this show $\displaystyle B $ is solvable?

    (!!) Because then the derived series is finite, of course: $\displaystyle G\supseteq G'\supseteq G''=1$ and this an abelian series for the group...what definition of "solvable" do you have?

    Tonio
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  5. #5
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    Quote Originally Posted by tonio View Post
    (!!) Because then the derived series is finite, of course: $\displaystyle G\supseteq G'\supseteq G''=1$ and this an abelian series for the group...what definition of "solvable" do you have?

    Tonio
    $\displaystyle G $ is solvable $\displaystyle \iff \{1\} \triangleleft G_i \triangleleft \ldots G_1 \triangleleft G_0=G $ and $\displaystyle \forall \; j, \; G_j/G_{j+1} $ is cyclic.
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  6. #6
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    Quote Originally Posted by mathman88 View Post
    $\displaystyle G $ is solvable $\displaystyle \iff \{1\} \triangleleft G_i \triangleleft \ldots G_1 \triangleleft G_0=G $ and $\displaystyle \forall \; j, \; G_j/G_{j+1} $ is cyclic.

    This is not the usual definition, and it's a rather misleading and even incorrect one (imo, of course. See following note), but it never minds: since the group $\displaystyle B$ is finitely generated then $\displaystyle B\slash [B,B]$ is finitely generated and thus can be decomposed in a direct product of cyclic groups...

    Note: for your definition to work it MUST be that G is fin. generated, or at least that all its abelian factor groups are, otherwise it fails: an infinitely generated abelian group wouldn't be solvable according to your definition which, of course, is absurd.

    Tonio
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