Exhibit an element in the center of a group ring

Let R be a ring with 1 \neq 0, and let G = \{g_1, \ldots, g_n \} be a finite group. Prove that the element N = \sum_{i=1}^n g_i is in the center of the group ring R[G].

Let M = \sum_{i=1}^n r_i g_i be an element of R[G]. Note that for each g_i \in G, the action of g_i on G by conjugation permutes the subscripts. Then we have the following.

NM  =  \left( \displaystyle\sum_{i=1}^n g_i \right) \left( \displaystyle\sum_{j=1}^n r_jg_j \right)
 =  \displaystyle\sum_{j=1}^n \displaystyle\sum_{i=1}^n r_jg_ig_j
 =  \displaystyle\sum_{j=1}^n \displaystyle\sum_{i=1}^n r_j g_j g_j^{-1} g_ig_j
 =  \displaystyle\sum_{j=1}^n r_j g_j \left( \displaystyle\sum_{i=1}^n g_j^{-1}g_ig_j \right)
 =  \displaystyle\sum_{j=1}^n r_j g_j \left( \displaystyle\sum_{i=1}^n g_i \right)
 =  \left( \displaystyle\sum_{j=1}^n r_jg_j \right) \left( \displaystyle\sum_{i=1}^n g_i \right)
 =  MN.

Thus N \in Z(R[G]).

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  • Bobby Brown  On November 5, 2010 at 3:45 pm

    indices i and j are mixed up, and got quite confusing, but I think I sorted it out. Nice proof.

  • Will  On November 15, 2011 at 2:36 am


    I might be totally lost, but I think that the g_i at the very end of the third to last line should be deleted, no?

    Definitely a nice proof.

    • nbloomf  On November 15, 2011 at 11:32 am

      You’re right; I’m not sure how it got there. Thanks!

  • Ali  On May 10, 2012 at 8:06 am

    In the step where you go from (the summation of) gj^(-1)gigj to just gi, don’t you have to assume the elements gi,gj, etc. of G are commutative? And we don’t know that the group is commutative, so can we make that step?

    • nbloomf  On August 17, 2012 at 11:45 pm

      This step is tricky. Note that the element \sum_{i=1}^n g_j^{-1}g_ig_j is the sum over all the elements g_i in the group after they have been conjugated by g_j. But conjugation by g_j merely permutes the elements of the group, so this sum is precisely \sum_{i=1}^n g_i. No commutativity is required.

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