I'm learning about Lie groups from Knapp's book 'Lie groups', 2ed, and I was confused by a comment on 364 right above the example.
Let $\mathfrak{g}$ be a real semisimple Lie algebra with a Cartan involution $\theta$ and inner product $B_\theta(\cdot, \cdot):= -B(\cdot, \theta(\cdot))$ induced from the Killing form $B$. We have a decomposition $\mathfrak{g} = \mathfrak{k}\oplus \mathfrak{p}$ into $\pm1$ eigenspaces. We also have the mapping $ad_{\mathfrak{g}}: \mathfrak{g} \to \operatorname{Der}(\mathfrak{g})$. We define $\operatorname{Int}_{\mathfrak{g}}(\mathfrak{k})$ to be the connected subgroup of $\operatorname{Aut}(\mathfrak{g})$ corresponding to $ad_{\mathfrak{g}}(\mathfrak{k})$. Then
Claim: The subgroup $\operatorname{Int}_{\mathfrak{g}}(\mathfrak{k})$ is compact.
Question: How does one justify this?
Attempt: Note that $\operatorname{Aut}(\mathfrak{g})^o$ has Lie algebra $\operatorname{Der}(\mathfrak{g})=ad_{\mathfrak{g}}(\mathfrak{g})$ which is itself semisimple with involution $\tilde{\theta}M:= -M^*$, where the adjoint $*$ is taken with respect to the inner product $B_\theta$. So $\tilde{\theta}$ becomes another Cartan involution giving the decomposition $ad(\mathfrak{g})= ad_{\mathfrak{g}}(\mathfrak{k}) \oplus ad_{\mathfrak{g}}(\mathfrak{p})$. Now the global theorem on Cartan Decomposition (see Theorem 6.31 in the book) says that $K := \operatorname{Int}_{\mathfrak{g}}(\mathfrak{k})$ is compact if and only if $\operatorname{Aut}(\mathfrak{g})^o$ has finite center. Finally, proposition $7.9$ from the book states that any connected group of matrices with semisimple Lie algebra must have finite center. This establishes the claim.
Question: I wonder if there's an easier way to see that $\operatorname{Int}_{\mathfrak{g}}(\mathfrak{k})$ is compact. I don't want to use proposition $7.9$ since it occurs much later in the book.
Hint: Check if it is a closed subgroup of some group which is known to be compact, like some $SO$ w.r.t. a real inner product.