Let $U$ be the subgroup of $G = GL_n(F)$ of upper triangular matrices. Prove that the normalizer of $U$ in $G$ is $U$, i.e $N_G(U) = U$.
Two observations:
Let $L$ be the subgroup of $G$ of lower triangular matrices. This subgroup is conjugate to $U$ by the matrix $$ P = \begin{bmatrix} 0 & 0 & \dots & 1 \\ 0 & & 1 & 0 \\ \vdots & && \vdots \\ 1 & 0 & \dots & 0 \end{bmatrix}. $$ Considering the conjugation action of $G$ acting on $G/U$ we have $L \in O_U$, which is restating what's above. I feel like there is something very natural I should do at this point, but don't see or know.
More directly, we have $U \subset N_G(U)$. We need to show $N_G(U) \subset U$, so looking for a contradiction suppose there exists a matrix $B$ such that $B \in N_G(U)$ and $B \notin U$. That is $BUB^{-1} = U$ and $B$ is not upper triangular, thus in $B$ there exists some $b_{ij} \neq 0$ where $1 \le j < i \le n$. Now maybe break up the matrix $B = B_u + B_l$ where $B_u$ is the upper diagonal values of $B$ and $B_l$ is the lower diagonal values below the diagonal. So $(B_u + B_l)U = U(B_u + B_l)$, thus for any $T_1 \in U$, there exists $T_2 \in U$ such that $B_uT_1 + B_l T_1 = T_2 B_u + T_2 B_l$. Then $B_lT_1 - T_2 B_l = T_2 B_u - B_u T_1 \in U$. Now this seems to be getting a bit muddled.
Can someone provide some guidance for either proving the statement with using that fact that upper and lower matrices are conjugate or showing $N_G(U) \subset U$? Thanks.