I want to prove that if $G$ is a connected Lie group, $H$ is a normal Lie subgroup of $G$, $\mathfrak{g}$ and $\mathfrak{h}$ their respective lie algebras, then $\mathfrak{h}$ is an ideal of $\mathfrak{g}$.
I try to proceed as follows:
Let $X\in\mathfrak{g}$, $Y\in\mathfrak{h}$, we want to compute $[X,Y]$ and show it is in $\mathfrak{h}$. Let $g:(-\varepsilon,\varepsilon)\rightarrow G$ and $h:(-\varepsilon,\varepsilon)\rightarrow H$ with $g(0) = h(0) = e$ be such that
$$\left.\frac{d}{dt}\right|_{t=0}g(t) = X$$
and
$$\left.\frac{d}{ds}\right|_{s=0}h(s) = Y$$
Then
$$[X,Y]=\left.\frac{d}{dt}\right|_{t=0}Ad_{g(t)}Y = \left.\frac{d}{dt}\right|_{t=0}\left.\frac{d}{ds}\right|_{s=0}g(t)h(s)g(t)^{-1}$$
Since $H$ is normal in $G$, we have that $g(t)h(s)g(t)^{-1}=\tilde{h}(s,t)\in H$ for all $t,s$. This means that differentiating $\tilde{h}$ once at the origin gives us an element of $\mathfrak{h}$, but here we have to differentiate it twice!
Am I doing something wrong or there is simply a detail I'm not getting?
Best Answer
Another quick way to see that this is true is to note that if $N \hookrightarrow G$ is a normal Lie subgroup, then it is the kernel of a Lie group homomorphism $f : G \to H$. Then all you need to do is check that $\frak{n}$ is the kernel of $f_* : \frak{g} \to \frak{h}$.