I am studying GTM218 and found an unproven theorem that the author left for exercise.
Here is the theorem:
Theorem 7.35 (Characterization of Semidirect Products). Suppose $G$ is a Lie group, and $N;H \subseteq G$ are closed Lie subgroups such that $N$ is normal, $N\cap H = \{e\}$, and $N+H =G$. Then the map $N\rtimes H \rightarrow G$ is a Lie group isomorphism between $N\rtimes H$ and G, where $\theta : H\rtimes N \rightarrow N$ is the action by conjugation: $\theta_{h}(n) = hnh^{-1}$
I think the condition '$H$ is closed' is not needed. Here is my proof of this theorem without that condition:
Consider the immersion $H\rightarrow G$ and $N\rightarrow G$, we combine them to get the immersion $H\times N \rightarrow G\times G$. By calculating its derivative we know that the combined map is an immersion too. Then we compose this function with the map $H\times N \rightarrow G: (h,n) \rightarrow hnh^{-1}$. Hence we construct a smooth map from $H \times N$ to $G$.
Because $N$ is a closed Lie subgroup of $G$, we know it is also an embedded manifold of $G$. So we can conclude that the map $\theta: H\times N \rightarrow N: (h,n) \rightarrow hnh^{-1}$ is smooth.
By considering group theory(we can find the proof in some algebra books) $\theta$ is a group action by automorphisms. So $\theta$ is a smooth group action of $H$ on $N$ by automorphisms.
We then can construct the semiproduct of Liegroups $N \rtimes H$. Define the map $\varphi : N\rtimes H \rightarrow G$ by $\varphi(n,h):= nh$. It can be identified by composing the immersion map $N\times H \rightarrow G\times G$ and $G\times G \rightarrow G:(n,h)\rightarrow nh$. Because $\varphi$ is the composition of two smooth maps we find $\varphi$ is a smooth map. By algebra consideration, $\varphi$ is a group isomorphism. Hence $\varphi$ is a bijective smooth homomorphism of two Lie groups.
But we know Lie group homeomorphism must have constant rank, and a bijective smooth map between two manifolds must be a Lie group isomorphism. Hence we get the conclusion
I don't find anywhere that the condition '$H$ is closed' be used. It looks like a consequence of my proof.
You're right. See the note about page 169 on my list of corrections.