The Lie algebra of the normalizer of a closed subgroup of a linear algebraic group

Question

Let $G$ be a connected algebraic group over an algebraically closed field of characteristic zero, and $H \subseteq G$ a closed connected subgroup. In Humphreys Linear algebraic groups Exercise 13.1 we are asked to show that the Lie algebra of $N=N_G(H)=\{x\in G : xHx^{-1}=H\}$, denoted $\mathscr{L}(N)$, is equal to $\mathfrak{n}=\mathfrak{n}_\mathfrak{g}(\mathfrak{h})=\{X \in \mathfrak{g} : [X,\mathfrak{h}] \subseteq \mathfrak{h}\}$.

One inclusion, $\mathscr{L}(N) \subseteq \mathfrak{n}$, is easy to see: Note that the adjoint representation satisfies $\text{Ad}x(\mathfrak{h})=\mathfrak{h}$ for all $x \in N$, so $\text{Ad} x$ can be written in the form $\begin{bmatrix} * && * \\ 0 && * \end{bmatrix}$, so $\text{ad} X$ inherits this form for all $X \in \mathscr{L}(N)$, so $\text{ad}X(\mathfrak{h})=[X,\mathfrak{h}]\subseteq \mathfrak{h}$, i.e. $\mathscr{L}(N)\subseteq \mathfrak{n}$.

How do we get the other inclusion?

Edit I can show that if this statement holds for $G=\mathrm{GL}(n,k)$ then it holds for a general connected algebraic group. This follows easily from the fact that $N=N_{\mathrm{GL}(n,k)}(H)\cap G$, so $\mathfrak{n}=\mathscr{L}(N_{\mathrm{GL}(n,k)}(H))\cap \mathfrak{g}$. So now I only need help proving other inclusion for $G=\mathrm{GL}(n,k)$.

Answer 1

Here's a cleaner proof from the beginning.

Consider the algebraic group $K=\{x \in G : \mathrm{Ad}_x (\mathfrak{h})=\mathfrak{h}\}$. Note that $x \in N \iff \mathrm{Int}_x (H)=H \iff \mathrm{Ad}_x (\mathfrak{h})=\mathscr{L}(\mathrm{Int}_x(H))=\mathfrak{h}\iff x \in K$. (These implications are all straightforward except $\mathrm{Int}_x (H)=H \Leftarrow \mathrm{Ad}_x (\mathfrak{h})=\mathscr{L}(\mathrm{Int}_x(H))$, which follows from the fact that $H$ and $\mathrm{Int}_x(H)$ are connected, and have the same Lie algebra, and hence must be equal by Theorem 13.1 in Humphreys.) So $N=K$, and hence $\mathscr{L}(N)=\mathscr{L}(K)=\mathfrak{n}$, where the last equality follows from Theorem 13.2 in Humphreys.