Assume that $G$ is a connected Lie group and that $\alpha:G\rightarrow G$ is an automorphism of $G$. Furthermore let $\alpha_*:\mathfrak{g}\rightarrow\mathfrak{g}$ be the corresponding tangent map at the identity (arrow between the corresponding Lie algebra).
One can easily show that $\alpha_*$ is a Lie algebra automorphism since $\alpha$ is a (local diffeomorphism) which implies that $\alpha_*$ is bijective. Ready.
Now I want to show two things:
- $\alpha^2=Id_G$ $\Longleftrightarrow$ $\alpha_*^2=Id_{\mathfrak{g}}$
- If we assume that $\sigma^2=Id$ then $H=\{g\in G:\alpha(g)=g\}$ is a closed subgroup of $G$ and its Lie algebra $\mathfrak{h}$ is given by $\mathfrak{h}=\{X\in\mathfrak{g}:\alpha_*(X)=X\}$
For the first part I can only give a prove of the implication from the left to the right: Assume $\alpha^2=Id$, then it follows from the observation we made before that $\alpha_*^2=(\alpha^2)_*=(Id_G)_*=Id_{\mathfrak{g}}$ which proves the implication. For the other implication i think you have to use that $G$ is connected and that therefore the connected component of the identity is equal to the whole Lie group $G$, but i don't know how to use it precisely. Someone an idea?
For the second point one can easily prove that $H$ is a subgroup of $G$. To conclude that $H$ is closed it would be enough to show that $H$ is a submanifold of $G$ at the identity, but i don't now how to prove this, maybe by a local diffeomorphism? Or is there another way to conclude the result? Is the conclusion of how $\mathfrak{h}$ looks like a direct consequence?
Can someone help me with this problems? :)