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Math Help - Element of a Group Ring

  1. #1
    MHF Contributor Swlabr's Avatar
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    Element of a Group Ring

    Let o(a)=n, <a> \unlhd G. Then, the question is does \widehat{<a>} = 1+a+...+a^{n-1} \in Z(RG) hold, for R an arbitrary ring?

    Clearly, it is enough to show that \widehat{<a>}g = g\widehat{<a>} \forall g \in G, that is, g^{-1}\widehat{<a>}g = \widehat{<a>}.

    So, according to my notes, g(1+a+...+a^{n-1})g^{-1} = 1+a^g+...+(a^{n-1})^g, which I understand perfectly well. However, apparently this is just \widehat{<a>} because <a> \unlhd G. I don't get why that is true!

    My thought is perhaps that the action of g \in G on elements of a (finite) normal subgroup N perhaps induces a unique element of N? That is to say, for \{x_0, x_i,...,x_n\} = N \unlhd G, |N| \leq \infty, g \in G, then {x_i}^g \neq {x_j}^g for i \neq j?

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by Swlabr View Post
    Let o(a)=n, <a> \unlhd G. Then, the question is does \widehat{<a>} = 1+a+...+a^{n-1} \in Z(RG) hold, for R an arbitrary ring?

    Clearly, it is enough to show that \widehat{<a>}g = g\widehat{<a>} \forall g \in G, that is, g^{-1}\widehat{<a>}g = \widehat{<a>}.

    So, according to my notes, g(1+a+...+a^{n-1})g^{-1} = 1+a^g+...+(a^{n-1})^g, which I understand perfectly well. However, apparently this is just \widehat{<a>} because <a> \unlhd G. I don't get why that is true!

    My thought is perhaps that the action of g \in G on elements of a (finite) normal subgroup N perhaps induces a unique element of N? That is to say, for \{x_0, x_i,...,x_n\} = N \unlhd G, |N| \leq \infty, g \in G, then {x_i}^g \neq {x_j}^g for i \neq j?

    Thanks in advance!
    as you mentioned we have 1+a + \cdots + a^{n-1} \in Z(RG) if and only if 1+a+ \cdots + a^{n-1}=1+gag^{-1} + \cdots + ga^{n-1}g^{-1}, \ \ \forall g \in G. \ \ \ \ \ (1)

    now since o(a)=n, we have a^i \neq a^j, and hence ga^ig^{-1} \neq ga^jg^{-1}, for all 0 \leq i \neq j \leq n-1. thus, since RG is a free ring, every ga^ig^{-1}

    in the RHS of (1) must be equal to some a^j in the LHS of (1). this means <a> has to be a normal subgroup of G.

    conversely, if <a> is a normal subgroup of G, then \{ga^ig^{-1}: \ 0 \leq i \leq n-1 \}= g<a>g^{-1}=<a> and hence (1) holds.
    Last edited by NonCommAlg; May 24th 2009 at 10:32 AM. Reason: free ring not group! haha
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    as you mentioned we have 1+a + \cdots + a^{n-1} \in Z(RG) if and only if 1+a+ \cdots + a^{n-1}=1+gag^{-1} + \cdots + ga^{n-1}g^{-1}, \ \ \forall g \in G. \ \ \ \ \ (1)

    now since o(a)=n, we have a^i \neq a^j, and hence ga^ig^{-1} \neq ga^jg^{-1}, for all 0 \leq i \neq j \leq n-1. thus, since RG is a free ring, every ga^ig^{-1}

    in the RHS of (1) must be equal to some a^j in the LHS of (1). this means <a> has to be a normal subgroup of G.

    conversely, if <a> is a normal subgroup of G, then \{ga^ig^{-1}: \ 0 \leq i \leq n-1 \}= g<a>g^{-1}=<a> and hence (1) holds.

    That's much easier than I was anticipating. I do have one further question - why is RG a Free Ring? According to Wiki, Group Rings are Free Modules, things I know next to nothing about, but Free Rings are not mentioned...
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    Quote Originally Posted by Swlabr View Post

    why is RG a Free Ring? According to Wiki, Group Rings are Free Modules, things I know next to nothing about, but Free Rings are not mentioned...
    by "free ring" here i meant a free R-module which also has a structure of a ring. (it would be an algebra if R was commutative) so it's not anything official. i just made it up!
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