Let $G$ be a complex connected and simply connected Lie group, with the Lie algebra $\mathfrak{g}$, and let $\mathfrak{k}\subset \mathfrak{g}$ be a real form of $\mathfrak{g}$.
Define the $\mathbb{R}$-linear map $\theta:\mathfrak{g}\rightarrow \mathfrak{g}$ by $\theta(x+iy)=x-iy,~~x,y\in \mathfrak{k}$. I can prove this is an automorphism and can be uniquely lift to an automorphism $\theta:G\rightarrow G$. (A little bit abuse the notations.)
Let $K=G^{\theta}$. Show this is a real Lie group with the Lie algbera $\mathfrak{k}$.
Attempt:
I cannot show $K$ is a group and an embedded submanifold (I try to show $K$ is a Lie subgroup so that $K$ is a real Lie group). And I only can show that $\mathfrak{k} \subset \mathsf{Lie}(K)$.
Since $\theta$ is an automorphism of $G$, its fixed points form a subgroup $K\subset G$ and since $\theta$ is continuous, $K$ is a closed subset of $G$. Now invoke the theorem that a closed subgroup of a Lie group is a Lie subgroup. This contains a description of the Lie algebra of $K$, which allows you to verify that $\mathfrak k$ is the Lie algebra of $K$.