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Math Help - FG-modules

  1. #1
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    FG-modules

    Hello, this is my first post.

    I was wondering how, if we have a finite group G and a field F and an FG-module FG, how we would show there is no FG-submodule complementary to V=\left\{\sum_{h\in G}x_h h\mid \sum_{h\in G} x_h=0\right\}? Obviously a proof by contradiction but how would I start?

    Thanks,

    John
    Last edited by JohnnyFrankston; October 12th 2009 at 12:24 AM.
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  2. #2
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    Quote Originally Posted by JohnnyFrankston View Post
    Hello, this is my first post.

    I was wondering how, if we have a finite group G and a field F and an FG-module FG, how we would show there is no FG-submodule complementary to V=\left\{\sum_{h\in G}x_h h\mid \sum_{h\in G} x_g=0\right\}? Obviously a proof by contradiction but how would I start?

    Thanks,

    John
    this is certainly not true! for example if F is a field of characteristic 0, then FG becomes a semisimple algebra (Maschke's theorem) and thus every ideal (FG submodule) of FG has a complement.
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    this is certainly not true! for example if F is a field of characteristic 0, then FG becomes a semisimple algebra (Maschke's theorem) and thus every ideal (FG submodule) of FG has a complement.
    Oh of course, I meant to say that |G|= 0\in F. Otherwise yes maschke comes to the rescue haha
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  4. #4
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    Quote Originally Posted by JohnnyFrankston View Post
    Oh of course, I meant to say that |G|= 0\in F. Otherwise yes maschke comes to the rescue haha
    suppose FG=V \oplus W and let e=\sum_{g \in G} g. see that ge=e for all g \in G and thus ve=0, for all v \in V. we have 1=v+w, for some v \in V, \ w \in W and so e=we. now suppose that e \in V.

    then since V,W are FG submodules, we'll have e=we \in V \cap W = \{0 \}, \color{red}^* which is nonsense. hence e \notin V, which means that |G|, as an element of F, is non-zero. Q.E.D.


    \color{red}*: note that V is an ideal of FG, i.e. it's both left and right FG sumodule. so if we assume that W is a right FG submodule and e \in V, then we \in V \cap W = \{ 0 \}.
    Last edited by NonCommAlg; October 12th 2009 at 03:12 AM.
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  5. #5
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    Quote Originally Posted by NonCommAlg View Post
    suppose FG=V \oplus W and let e=\sum_{g \in G} g. see that ge=e for all g \in G and thus ve=0, for all v \in V. we have 1=v+w, for some v \in V, \ w \in W and so e=we. now suppose that e \in V.

    then since V,W are FG submodules, we'll have e=we \in V \cap W = \{0 \}, \color{red}^* which is nonsense. hence e \notin V, which means that |G|, as an element of F, is non-zero. Q.E.D.


    \color{red}*: note that V is an ideal of FG, i.e. it's both left and right FG sumodule. so if we assume that W is a right FG submodule and e \in V, then we \in V \cap W = \{ 0 \}.
    Thanks a lot for that - the two bits I don't quite understand I've highlighted in red in the quote above.

    Essentially I don't understand why ge=e for all g \in G implies  ve=0 for all v\in V, nor do I understand why e\not\in V implies anything about the order of the group.
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  6. #6
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    Quote Originally Posted by JohnnyFrankston View Post
    Thanks a lot for that - the two bits I don't quite understand I've highlighted in red in the quote above.

    Essentially I don't understand why ge=e for all g \in G implies  ve=0 for all v\in V, nor do I understand why e\not\in V implies anything about the order of the group.
    if v=\sum_{g \in G}x_gg \in V, then \sum_{g \in G}x_g = 0 and thus ve=\sum_{g \in G} x_g ge=\sum_{g \in G}x_ge = \left(\sum_{g \in G}x_g \right)e = 0.

    we have e=\sum_{g \in G}x_g g, where x_g = 1, for all g \in G. now e \notin V means that |G|=\sum_{g \in G}1 =\sum_{g \in G} x_g \neq 0. (see the definition of V again!)
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