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.
Thanks in advance!
Best Answer
What about $C_2 \times C_2 \simeq H = \{ \pmatrix{a&0\\0&d} : a,d \in \{\pm 1\}\} \subset G = GL_2(\mathbb R)$?