Characterize the center of a group ring

Let \mathcal{K} = \{k_1, \ldots, k_m \} be a conjugacy class in the finite group G. Let R be a ring with 1.

  1. Prove that the element K = \sum_{i=1}^m k_i is in the center of the group ring R[G]. [Hint: Check that g^{-1}Kg = K for all g \in G.]
  2. Let \mathcal{K}_1, \ldots, \mathcal{K}_n be the conjugacy classes of G, and for each i, let K_i be the sum of the elements in \mathcal{K}_i (as described in part 1). Prove that an element \alpha \in R[G] is in the center if and only if \alpha = \sum_{i=1}^n a_iK_i for some elements a_i \in Z(R).

  1. Let g \in G. Note that conjugation by g permutes the elements of \mathcal{K}, so that (as an element of R[G]) we have g^{-1}Kg = g^{-1} \left( \sum_{i=1}^m k_i \right) g = \sum_{i=1}^m g^{-1}k_ig = \sum_{i=1}^m k_i = K. Thus gK = Kg for all g \in G. Then for all M = \sum_{i=1}^t r_i g_i \in R[G], we see the following.
    KM  =  \left( \displaystyle\sum_{i=1}^m k_i \right) \left( \displaystyle\sum_{j=1}^t r_j g_j \right)
     =  \displaystyle\sum_{i=1}^m \displaystyle\sum_{j=1}^t r_j k_i g_j
     =  \displaystyle\sum_{j=1}^t \displaystyle\sum_{i=1}^m r_j k_i g_j
     =  \displaystyle\sum_{j=1}^t r_j \left( \displaystyle\sum_{i=1}^m k_i \right) g_j
     =  \displaystyle\sum_{j=1}^t r_j K g_j
     =  \displaystyle\sum_{j=1}^t r_j g_j K
     =  \left( \displaystyle\sum_{j=1}^t r_j g_j \right) K
     =  MK

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

  2. First we show that N = \sum_{i=1}^n a_i K_i \in Z(R[G]). Let M = \sum_{j=1}^t r_j g_j \in R[G]. Then we have the following.
    NM  =  \left( \displaystyle\sum_{i=1}^n a_i K_i \right) M
     =  \displaystyle\sum_{i=1}^n a_i K_i M
     =  \displaystyle\sum_{i=1}^n a_i M K_i
     =  \displaystyle\sum_{i=1}^n a_i \left( \displaystyle\sum_{j=1}^t r_j g_j \right) K_i
     =  \displaystyle\sum_{i=1}^n \left( \displaystyle\sum_{j=1}^t a_i r_j g_j \right) K_i
     =  \displaystyle\sum_{i=1}^n \left( \displaystyle\sum_{j=1}^t r_j a_i g_j \right) K_i
     =  \displaystyle\sum_{i=1}^n \left( \displaystyle\sum_{j=1}^t r_j g_j \right) a_i K_i
     =  \left(\displaystyle\sum_{j=1}^t r_j g_j \right) \left( \displaystyle\sum_{i=1}^n a_i K_i \right)
     =  MN

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

    Now let M = \sum_{i=1}^t r_i g_i \in Z(R[G]). First, let s \in R be arbitrary. By examining each coefficient of Ms = sM, we see that r_i \in Z(R) for all i. Now recall that G acts transitively (by conjugation) on each of its conjugacy classes. If K is a conjugacy class of G and g_1,g_2 \in K, then we have h^{-1}g_1h = g_2 for some h \in G. The coefficient of g_2 in M = g^{-1}Mg is r_2 on one hand and r_1 on the other, so that in fact r_1 = r_2. In fact the coefficient of each g_i \in K is r_1, and we have M = \sum_{i=1}^m r_i K_i.

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  • Kelley  On March 3, 2011 at 11:43 pm

    In a group ring RG, R is a commutative ring so you need not show that r_i \in Z(R), since we know that r_i \in R = Z(R)

    • nbloomf  On March 4, 2011 at 6:53 am

      The ring R doesn’t have to be commutative.

  • RD Reese  On November 19, 2013 at 7:49 pm

    In the Dummit and Foote context, R was commutative, but in general this does not need to hold.

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