# Lie algebra of normalizer or normal subgroup

I'm struggling with following problem from book by Olver "equivalence, invariance and symmetry". $G$ is Lie group with Lie algebra $\mathfrak g$ and $H$ its Lie subgroup with Lie algebra $\mathfrak h \subseteq \mathfrak g$. Normalizer subgroup is defined as $G_H=\{g \in G | gHg^{-1} \subseteq H \}$. I am to show that its Lie algebra is $\mathfrak g _ {\mathfrak h} = \{ v \in \mathfrak g | \forall w \in \mathfrak h [v,w] \in \mathfrak h \}$. I need some hints. I want to avoid using powerful results like Campbell-Baker-Hausdorff formula because I don't know precise statements for those and I'm learning the basics now. Also, my background in abstract algebra is pretty nonexistent.

Edit:

Since there was no answers so far I add similar problem. If someone can help me with one of them then hopefully I will understand how to solve the other. I'm to show that ideals in $\mathfrak g$ are in one to one correspondence with connected normal Lie subgroups of G. However it would be most elegant to start with solving the first problem because from that one can directly deduce this result.

Another edit:

Ok now I know that for left invariant vectors fields $X,Y$ we have

$\frac{d}{dt}|_0 Ad_{e^{tX}}Y=[X,Y]$

For right invariant vector fields similar formula holds, except for a minus sign. From that it is easy to see that Lie algebra of normal subgroup is an ideal and that Lie algebra of normalizer subgroup $G_H$ is contained in $\mathfrak g_{\mathfrak h}$. However I don't see how to prove the converses:

For first problem that $\mathfrak g_{\mathfrak h}$ is contained in Lie algebra of $G_H$.

For second problem that for every ideal in $\mathfrak g$, corresponding subgroup is normal.

I think you have to assume that $H$ is connected for the result to be true. I'just give a hint how to prove the result you are looking for, let me know if you want me to add details.
Under the assumption, that $H$ is connected, you first prove that $G_H=G_{\mathfrak h}:=\{g\in G:Ad(g)(\mathfrak h)\subset\mathfrak h\}$. Second, you can use $Ad(exp(X))(Y)=e^{ad(X)}(Y)$, where the exponential in the right hand side is in $L(\mathfrak g,\mathfrak g)$, to show that for $X\in\mathfrak g_{\mathfrak h}$ and $t\in\mathbb R$, you have $\exp(tX)\in G_{\mathfrak h}=G_H$. This implies that $\mathfrak g_{\mathfrak h}$ is contained in the Lie algebra of $G_H$.
For the problem of ideals vs. normal subgroups, one also has to assume that $G$ is connected and then you can proceed similarly. First show that a connected Lie subgroup $H$ is normal if and only if its Lie algebra satisfies $Ad(g)(\mathfrak h)\subset\mathfrak h$ for all $g\in G$. Then use $Ad(exp(X))(Y)=e^{ad(X)}(Y)$ to show that the latter property follows from $ad(X)(\mathfrak h)\subset\mathfrak h$ for all $X\in\mathfrak G$.