An ideal integrates to a connected immersed normal subgroup

Fix $X\in \mathfrak{g}$. By [F2], ${\mathrm{ad}}_X$ restricts to an endomorphism of $\mathfrak{h}$. For $Y\in \mathfrak{h}$, solve $u^{\mathrm{\prime }}={\mathrm{ad}}_{X}u$, $u(0)=Y$ inside the finite-dimensional space $\mathfrak{h}$. Its inclusion into $\mathfrak{g}$ solves the same ambient initial-value problem, so uniqueness in [F4] gives $e^{t{\mathrm{ad}}_X}Y\in \mathfrak{h}$. Applying this with $-t$ proves equality $e^{t{\mathrm{ad}}_X}\mathfrak{h}=\mathfrak{h}$.

By [F3] and step 1.1, ${\mathrm{Ad}}_{\exp X}\mathfrak{h}=\mathfrak{h}$ for every $X\in \mathfrak{g}$. Define $K=\{g\in G:{\mathrm{Ad}}_g\mathfrak{h}=\mathfrak{h}\}\mathrm{.}$ The representation law in [F3] makes $K$ a subgroup. The local exponential neighborhood in [F4] lies in $K$, so $K$ is open; every other coset is open as well, and therefore $K$ is also closed. Since $K$ contains $e$, connectedness puts the identity component $G^0$ inside $K$.

For $g\in K$, the composite $C_g\circ i:H\to G$ is an injectively immersed homomorphism with connected source, and [F5] says that its identity tangent image is ${\mathrm{Ad}}_g\mathfrak{h}=\mathfrak{h}$. Uniqueness in [F1] therefore identifies its image $gH{g}^{-1}$ with $H$. Thus every $g\in K$ normalizes $H$, and step 2.1 gives $H⊴G^0$.

If $\mathfrak{h}$ is invariant under every ${\mathrm{Ad}}_g$, then $K=G$ by definition, and step 3.1 gives $H⊴G$. The cases $\mathfrak{h}=0$ and $\mathfrak{h}=\mathfrak{g}$ are included: their connected integral subgroups are respectively $\{e\}$ and $G^0$. Disconnected $G$ is allowed, and the stronger global conclusion uses exactly the separately stated full $\mathrm{Ad}$-invariance. No closedness conclusion follows. The only choice use is [A1], inherited through [F1], [F3], and [F4].