Exhibit the set of all invertible diagonal matrices with diagonal entries equal as a direct product

Let G = \{ [a_{i,j}] \in GL_n(F) \ |\ a_{i,j} = 0\ \mathrm{if}\ i > j, a_{i,i} = a_{j,j}\ \mathrm{for\ all}\ i,j \}, where F is a field, be the group of upper triangular matrices all of whose diagonal entries are equal. Prove that G \cong D \times U, where D is the group of nonzero multiples of the identity matrix and U the group of strictly upper triangular matrices.

Recall that \overline{SL}_n(F) is normal in GL_n(F), where \overline{SL}_n(F) denotes the upper triangular matrices with only 1 on the diagonal. Then U = G \cap \overline{SL}_n(F) is normal in G. Now if A \in D, we have A = kI for some nonzero field element k. Then for all M \in GL_n(F), we have AM = kIM = kM = Mk = MkI = MA, so that D \leq Z(GL_n(F)); more specifically, D is normal in G.

We can see that D \cap U is trivial, since every element in U has only 1s on the main diagonal. Finally, if A \in G has the element k on the main diagonal, then k^{-1}A \in U, and in fact A = kI \cdot k^{-1}A \in DU. Thus G = DU; by the recognition theorem, we have G \cong D \times U.

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  • Bobby Brown  On October 20, 2010 at 3:12 pm

    U is NOT SLn(F) intersect G.

    • nbloomf  On October 21, 2010 at 9:02 am

      True- it should be the strictly upper triangular elements in SL_n(F). Should be fixed now- thanks!

  • Gobi Ree  On January 4, 2012 at 12:05 am

    Where can I recall that \overline{SL}_n(F) is normal in GL_N(F)? Was it in the D&F?

    • nbloomf  On June 12, 2012 at 12:24 pm

      Bah- I thought we showed that, but now I can’t find it. I’ll have to mark this incomplete and come back to it.

      We did show previously that SL_n(F) is normal in GL_n(F), that might play a role in the fix.

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