# Upper Triangular Matrix with determinant 1

Considering the group of invertible triangular matrices $$T(n,K)\subset GL(n,K)$$. Then the set $$ST(n,K):= \{M\in T(n,K), \det(M)=1\}$$ forms a normal subgroup of $$T(n,K)$$. Also, it is known that the group of upper triangular matrices with all diagonal entries being 1 is a normal subgroup of $$T(n,K)$$, call it $$U(n,K)$$. However, take any $$P\in ST(n,K), A\in U(n,K), PAP^{-1}$$ is an upper triangular matrix with its diagonal entries being the product of the diagonal entries of $$P$$ and $$P^{-1}$$ in the respective position, which may not be 1. So it's not a normal subgroup of $$ST(n,K)$$. I tested other common types of matrices (some are modified to be upper triangular), but could not find a normal subgroup for $$ST(n,K)$$. Does it have a normal subgroup? If so, what does it look like?

• $K$ is a field? Commented Oct 7, 2020 at 13:43
• Yes. K is a field. Commented Oct 7, 2020 at 13:51

For $$n=3$$ for simplicity, isn't it:
$$\begin{pmatrix} a \ * \ * \\ 0 \ b \ * \\ 0 \ 0 \ c \end{pmatrix} \begin{pmatrix} 1 \ * \ * \\ 0 \ 1 \ * \\ 0\ 0 \ 1 \end{pmatrix}\begin{pmatrix} a \ * \ * \\ 0 \ b \ * \\ 0 \ 0 \ c \end{pmatrix}^{-1}=\begin{pmatrix} 1 \ * \ * \\ 0 \ 1 \ * \\ 0 \ 0 \ 1 \end{pmatrix}$$