Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. Let $H$ be a Lie subgroup of $G$ and its Lie algebra $\mathfrak{h}$ is naturally a subalgebra of $\mathfrak{g}$. Let $G^\circ$ denote the identity component of $G$, $H^\circ$ denote the identity component of $H$. Assume now $\mathfrak{h}$ is an ideal of $\mathfrak{g}$, then we know $H^\circ$ is a normal subgroup of $G^\circ$. Is it true that $H^\circ$ is also a normal subgroup of $G$?
I think it is highly possible that this is not true, but I cannot find a counter example since one need to find a suitable unconnected Lie group here.
Suppose $G$ is a Lie group and $H$ is a normal subgroup of $G$. If $g$ is an element of the group, the map $x\in G\mapsto gxg^{-1}\in G$ restricts to a map $x\in H\mapsto gxg^{-1}\in H$. The latter is continuous and maps the identity element to the identity element, so it maps the connected component of $H$ of the identity element to itself.