Let $B$ be the subgroup of $G = GL_n(\mathbb C)$ of invertible upper triangular matrices, and let $U \subset B$ be the set of upper triangular matrices with diagonal entries 1. Prove that $B = N(U)$ and that $B = N(B)$.

Direct computation is so complicated...So I tried to prove $B \subset N(B)$ and $N(B) \subset B$. Since $B \subset N(B)$ is trivial, I have only to show $N(B) \subset B$. But it is so difficult to me. Is this method correct? I need help.

  • $\begingroup$ The normalizer where? Do you mean the normalizer of $U$ in $B$? Or the normalizer of $U$ in $G$? $\endgroup$ Commented May 21, 2017 at 17:55
  • $\begingroup$ I want to know the normalizer of $B$ in $G$ and $U$ in $G$. $\endgroup$
    – Sonatina
    Commented May 22, 2017 at 0:06

1 Answer 1


you may have confused normalizer with centralizer


you may think $U$ as a single element $u$ just like this, $Z(u)=\left\{g∈G:gu=ug\right\}$

actually, there may be many elements in $U$

for example, in $GL_2(\mathbb{C})$, $B=\left\{\begin{pmatrix}a&b\\&d\end{pmatrix}:ad≠0;a, b, d∈\mathbb{C}\right\}$


you may think there's only $e$ in $U$, but it's not true

take any $\begin{pmatrix}a&b\\c&d\end{pmatrix}∈GL_2(\mathbb{C})$


that is, $\begin{pmatrix}a&ae_{1}+b\\c&ce_{1}+d\end{pmatrix}=\begin{pmatrix}a+ce_{2}&b+de_{2}\\c&d\end{pmatrix}$

we have, $ce_{1}=ce_{2}=0, ae_{1}=de_{2}$

due to the arbitrariness of $e_{1},e_{2}$

we have, $c=0$, that is, $N(U)=B$


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