When I was reading Brian C Hall's book: Lie Groups, Lie Algebras, and Representations, it said
It is a straightforward exercise to verify that the normalizer $$N=\{A\in SU(3)~|~\mathrm{Ad}(g)\mathfrak{h}\subset \mathfrak{h}\}$$ is a subgroup of $SU(3)$, where $\mathfrak{h}$ is the Cartan subalgebra of $SU(3)$.
To show $N$ is the group, everything is trivial to me except that the inverse is in $N$, that is,
if $g\in SU(3)$ such that $$\mathrm{Ad}(g)\mathfrak{h}=g\mathfrak{h}g^{-1}\subset \mathfrak{h},$$ then we have $g^{-1}\mathfrak{h}g\subset \mathfrak{h}$.
Since $\mathfrak{h}$ is generated by two elements $$H_1=\begin{pmatrix}1&0&0\\0&-1&0\\0 &0 &0 \end{pmatrix},\ \ \ H_2=\begin{pmatrix}0&0&0\\0&1&0\\0 &0 &-1 \end{pmatrix},$$ suppose $A=(a_{ij})\in SU(3)$ and $A^{-1}=(\bar{a}_{ji})$, then we can do the direct and brutal computation, which is too much work. So I was wondering if there is any simple and elegant method?
Is it true that $$N=N':=\{A\in SU(3)~|~\mathrm{Ad}(g)H=H\},$$ where $H$ is the corresponding connected Lie group of $\mathfrak{h}$? If so, then $N'$ is obviously a group and we are done.
I can only prove that $N=N'':=\{A\in SU(3)~|~\mathrm{Ad}(g)H\subset H\}$.
Hint (expanding Tobias Kildetoft's comment): This has nothing to do with Lie groups and Lie algebras, let alone specific matrix calculations. Rather, let $\rho: G \rightarrow GL(V)$ be any group homomorphism (i.e. representation of an arbitrary group $G$ on an arbitrary vector space $V$), and let $W$ be a finite-dimensional subspace of $V$. Show that
$$G_W := \{g \in G: \rho(g) (W) \subseteq W \}$$
is a subgroup of $G$, by using that (by a well-known linear algebra result about injectivity and surjectivity of endomorphisms of finite-dimensional vector spaces) the inclusion $\subseteq$ in the definition can be replaced by an equality $=$.