# The normalizer of invertible upper triangular matrices

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.

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

you may have confused normalizer with centralizer

$$N(U)=\left\{g∈G:gU=Ug\right\}$$

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

$$U=\left\{\begin{pmatrix}1&e\\&1\end{pmatrix}:e∈\mathbb{C}\right\}$$

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

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

$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1&e_1\\&1\end{pmatrix}=\begin{pmatrix}1&e_2\\&1\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}$$

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, $$N(U)=B$$