Let $G$ be the $\mathbb R$-points of an algebraic group in $\operatorname{GL}_n(\mathbb C)$ which is defined over $\mathbb R$. Let $\mathfrak g$ be the Lie algebra of $G$. Assume $G$ is topologically connected, and closed under the conjugate transpose $ g \mapsto g^{\ast} = \space ^t\overline{g}$. Let $$\mathfrak h = \{ X \in \mathfrak{gl}_n(\mathbb C) : \space ^t\overline{X} = X\}$$ be the space of Hermitian matrices, and let $U(n) = \{g \in \operatorname{GL}_n(\mathbb C) : gg^{\ast} = 1 \}$ be the unitary group. Let $K = G \cap U(n)$. By the polar decomposition theorem, we have that $$U(n) \times \mathfrak h \rightarrow \operatorname{GL}_n(\mathbb C)$$ $$(u,X) \mapsto u \exp(X) $$ is a homeomorphism. If we set $\mathfrak p = \mathfrak h \cap \mathfrak g$, I want to prove the Cartan decomposition for $G$, which states that polar decomposition restricts to a homeomorphism
$$K \times \mathfrak p \rightarrow G$$
For $g \in G$, uniquely write $g = k \exp(X)$ for $k \in U(n)$ and $X$ Hermitian. I know how to show that $k$ actually lies in $K$, and therefore that $\exp(X)$ lies in $G$. However, I don't see why this implies $X \in \mathfrak p$.
I think my difficulty comes down to the following question: if $X$ is Hermitian, and $\exp(X)$ is in $G$, do we have $X \in \mathfrak g$?