# $G$ is the semidirect product of $U_n$ & $D_n$

Let $GL_n \Bbb R$ be the set of all invertible matrices of size n. Let $U_n$ be the set of upper triangular matrices with $1$'s on the diagonal and $D_n$ be the set of diagonal matrices with non-zero entries. Let G be the subgroup generated by $U_n$ & $D_n$. Prove that $G$ is the semidirect product of $U_n$ & $D_n$.

It is clear that $U_n$ is normal in $G$. But what is the order of Aut($U_n$)? and What will be the map? Please help.

• $U_n$ isn't a normal subgroup. Geometrically, it's the group that stabilises a complete flag of subspaces, conjugation of such an element stabilises a different flag. And dimensionwise the semidirect product can't fill up all of $GL_n$... Also, why tag this question "finite groups"? – Olivier Bégassat Sep 22 '14 at 2:57
• @ Olivier Bégassat, I have edited it. Sorry for the mistake. – Ri-Li Sep 22 '14 at 3:10
• @user152715, you haven't edited the biggest mistake: $\;U_n\;$ is not normal in $\;G\;$ ...and nor is $\;D_n\;$ , so how are you going to form a semidirect product with them?. – Timbuc Sep 22 '14 at 3:29
• @Timbuc Can you tell me why $U_n$ is not normal in G? – Ri-Li Sep 22 '14 at 3:48
• @OlivierBégassat, are you sure that $U_n$ is not normal in the subgroup of $GL_n$ generated by $U_n$ and $D_n$? – Mariano Suárez-Álvarez Sep 22 '14 at 6:33