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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?

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  • $\begingroup$ $K$ is a field? $\endgroup$
    – 1123581321
    Commented Oct 7, 2020 at 13:43
  • $\begingroup$ Yes. K is a field. $\endgroup$
    – Divide1918
    Commented Oct 7, 2020 at 13:51

1 Answer 1

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For $n=3$ for simplicity, isn't it:

$\begin{pmatrix} a \ * \ * \\ 0 \ b \ * \\ 0 \ 0 \ c \end{pmatrix} \begin{pmatrix} 1 \ * \ * \\ 0 \ 1 \ * \\ 0\ 0 \ 1 \end{pmatrix}\begin{pmatrix} a \ * \ * \\ 0 \ b \ * \\ 0 \ 0 \ c \end{pmatrix}^{-1}=\begin{pmatrix} 1 \ * \ * \\ 0 \ 1 \ * \\ 0 \ 0 \ 1 \end{pmatrix}$

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  • $\begingroup$ Did you meant to write all diagonal entries 1 in the middle matrix? $\endgroup$
    – Divide1918
    Commented Oct 7, 2020 at 14:02
  • $\begingroup$ Yes, thanks. I fixed it $\endgroup$
    – 1123581321
    Commented Oct 7, 2020 at 14:03

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