Results 1 to 4 of 4

Thread: Element of a Group Ring

  1. #1
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176

    Element of a Group Ring

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

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

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

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

    Thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Swlabr View Post
    Let $\displaystyle o(a)=n, <a> \unlhd G$. Then, the question is does $\displaystyle \widehat{<a>} = 1+a+...+a^{n-1} \in Z(RG)$ hold, for $\displaystyle R$ an arbitrary ring?

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

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

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

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

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

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

    conversely, if $\displaystyle <a>$ is a normal subgroup of $\displaystyle G,$ then $\displaystyle \{ga^ig^{-1}: \ 0 \leq i \leq n-1 \}= g<a>g^{-1}=<a>$ and hence (1) holds.
    Last edited by NonCommAlg; May 24th 2009 at 09:32 AM. Reason: free ring not group! haha
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by NonCommAlg View Post
    as you mentioned we have $\displaystyle 1+a + \cdots + a^{n-1} \in Z(RG)$ if and only if $\displaystyle 1+a+ \cdots + a^{n-1}=1+gag^{-1} + \cdots + ga^{n-1}g^{-1}, \ \ \forall g \in G. \ \ \ \ \ (1)$

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

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

    conversely, if $\displaystyle <a>$ is a normal subgroup of $\displaystyle G,$ then $\displaystyle \{ga^ig^{-1}: \ 0 \leq i \leq n-1 \}= g<a>g^{-1}=<a>$ 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...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    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!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Nov 30th 2011, 07:10 PM
  2. Element of a Group
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Oct 6th 2010, 01:34 AM
  3. Nilpotent Element of a Ring
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Jul 23rd 2010, 01:15 PM
  4. Replies: 3
    Last Post: Mar 23rd 2010, 07:05 PM
  5. Ring with 0 only nilpotent element
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: Oct 27th 2008, 12:12 PM

Search Tags


/mathhelpforum @mathhelpforum