## 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 $G \cong D \times U$.

• 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.