# Is the identity component of a Lie subgroup normal (in original Lie group )?

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.

• What about $G=\Bbb C\rtimes \Bbb Z/4\Bbb Z$ wher ethe action of $\Bbb Z/4\Bbb Z$ on $\Bbb C$ is multiplicationj with $i$, and we let $H=\Bbb R\rtimes 1$? Then $H^0=H$ is nrmal in $G^0=\Bbb C$, but not in $G$. Nov 7, 2017 at 7:08
• I think you want $\mathfrak h$ is an ideal in $\mathfrak g$, not vice-versa. Nov 7, 2017 at 7:08
• Thank you for reminding me of the typo Robert, I have corrected them. And thank you for your suggestion Hagen, I am trying to figure out the meaning of the operator $\rtimes$ since I am a beginner for Lie groups. Nov 7, 2017 at 7:18
• That operator is not characteristic in any way to Lie groups; it is the semidirect product, and is there already when you consider just groups. Nov 7, 2017 at 7:33

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.
• Thanks Mariano. Here in the answer we suppose $H$ is a normal subgroup of $G$, but generally we cannot get this information only from $\mathfrak{h}$ is an ideal of $\mathfrak{g}$. Nov 7, 2017 at 17:16