# Why does the maximal compact subgroup of a Lie group inject into the compact form?

I've seen multiple sources state the following (without proof or reference), but I don't see why it's true.

Let $G$ be a Lie group, and $G_u$ be a compact connected Lie group such that the complexification of their Lie algebras are isomorphic ($\mathfrak g \otimes \Bbb C \cong \mathfrak g_u \otimes \Bbb C$). Then the maximal compact subgroup $K$ of $G$ naturally includes into $G_u$.

What is a proof or reference for the above claim?

In the semisimple case, the argument goes as follows: the conjugation $\theta$ of a complex semisimple $\mathfrak{g}^\mathbb{C}$ with respect to a compact real form $\mathfrak{g}_u$ is a Cartan involution, that is, $-B(X,\theta Y)$ is a positive definite bilinear form on $\mathfrak{g}^\mathbb{C}$. If $\theta$ commutes with the conjugation $\bar{\cdot}$ of $\mathfrak{g}^\mathbb{C}$ with respect to $\mathfrak{g}$, then it restricts to an involution of $\mathfrak{g}$, which is the definition of a Cartan involution of a real semisimple Lie algebra. EDIT: It's not actually a further assumption that $\theta$ commutes with $\bar{\cdot}$. $\mathfrak{g}_u$ can be conjugated around by inner automorphisms to guarantee this.
An involution of a real vector space is diagonalizable with eigenvalues $\pm 1$. The corresponding decomposition of $\mathfrak{g}$ under the restriction of $\theta$ is $\mathfrak{g}=\mathfrak{k}\oplus \mathfrak{p}$. $\mathfrak{g}_u$ is then $\mathfrak{k}\oplus i\mathfrak{p}$, and so $K$ the immersed subgroup of $G$ corresponding to $\mathfrak{k}$ embeds naturally in $G_u$. In case the center of $G$ is finite, $K$ is compact and in fact maximal compact.