Example of a non-connected Lie subgroup where the normalizer of Lie algebra is not the Lie algebra of the normalizer

Let $$G$$ be a Lie group and $$H\subset G$$ be a connected Lie subgroup, with Lie algebras $$\mathfrak{g}$$ and $$\mathfrak{h}$$, respectively. In this case we know that $$Lie(N(H)) = \mathfrak{n} (\mathfrak{h})$$ Where $$N(H)=\{ g\in G : gHg^{-1} \subseteq H \}$$ is the normalizer of $$H$$, $$Lie(N(H))$$ is the Lie algebra of such a closed subgroup and $$\mathfrak{n} (\mathfrak{h}) = \{ X\in \mathfrak{g} : ad(X)\mathfrak{h} \subset \mathfrak{h} \}$$. With $$ad(X)Y=[X,Y]$$ the Lie bracket of the corresponding vector fields.

However, I'm looking for an example on the case where $$H$$ is a non-connected subgroup and the above inequality does not hold. Any help would be greatly appreciated. At the moment I don't know the theory of solvable nor nilpotent Lie groups, so any example avoiding those is preferred.

An attempt

I want a non abelian group as I want at least one of the sets to be nontrivial.

I find computing normalizers rather hard, so if we have a discrete subgroup (which will be non-connected), we have that the Lie algebra $$\mathfrak{h}$$ is trivial and then $$\mathfrak{n} (\mathfrak{h}) = \mathfrak{g}$$. Now I want an element that is not in the Lie algebra of the normalizer. But I'm stuck at finding explicitly what is the Lie algebra of the Normalizer, even on a simple matrix group.

What about $$C_2 \times C_2 \simeq H = \{ \pmatrix{a&0\\0&d} : a,d \in \{\pm 1\}\} \subset G = GL_2(\mathbb R)$$?
• Thank you, sounds good for me. I found that the normalizer $N(H)$ is the set of diagonal matrices with $a,d\neq 0$, and then $Lie (N(H))$ is the set of diagonal matrices; additionaly, as the subgroup is discrete $\mathfrak{n}(\mathfrak{h})=\mathfrak{g}$. Thus, it suffices to pick any non diagonal matrix $A\in \mathfrak{g}$ but $A\not\in Lie(N(H))$. Is that so? Commented Jun 10, 2023 at 13:53