On the correspondence between normal Lie subgroup and ideal of Lie algebras.

Question

I have proved the following:

Proposition. Let $G$ be a Lie group and let $H$ be an embedded Lie subgroup of $G$. Let $\mathfrak{g}$, respectively $\mathfrak{h}$, be the Lie algebra of $G$, respectively of $H$.

$1.$ If $H$ is normal in $G$, then $\mathfrak{h}$ is an ideal of $\mathfrak{g}$.

Assume that both $G$ and $H$ are connected.

$2.$ If $\mathfrak{h}$ is an ideal of $\mathfrak{g}$, then $H$ is normal in $G$.

In the proof of $2$, I actually established without using the connectedness of $H$ and $G$ that for all $X\in\mathfrak{g}$ and $Y\in\mathfrak{h}$, one has $\exp(X)\exp(Y)\exp(X)^{-1}$, which is only a local information. This is why I suspect that the result does not hold anymore droping the connectedness assumption on either $H$ or $G$.

If $G$ is connected and $H$ is any non-central discrete subgroup of $G$, then we have a counterexample of $2$ when $H$ is disconnected. For an explicit counterexample, let $G=\textrm{GL}(n,\mathbb{C})$ and $H$ be the subgroup generated by a non-central matrix with finite order, e.g. a diagonal matrix with distinct $n$th roots of the unity on the diagonal.

Now I am looping in vain for the following setup:

$G$ disconnected, $H$ connected, $\mathfrak{h}$ is an ideal of $\mathfrak{g}$ and $H$ is not normal in $G$.

I have already investigated a lot of real matrix Lie groups.

Any enlightenment will be greatly appreciated!

Answer 1

The multiplication given by Tsemo is not associative. Here is an example given by my course instructor Prof. Rui Loja Fernandes.

Let $G=\mathbb{R}^3 \ltimes S_{3}$, where $S_{3}$ acts on $\mathbb{R}^3$ by $\sigma(a_{1},a_{2},a_{3})=(a_{\sigma(1)},a_{\sigma(2)},a_{\sigma(3)})$. The connected component $G^{0} $of $G$ containing the identity is $\mathbb{R}^3\times \{I\}$, which is abelian, so the Lie algebra $\mathfrak{g}$ is abelian.

Let $H=\{(x,0,0)\mid x\in \mathbb{R}\} \times \{I\}$ be a conncted Lie subgroup of $G$. The Lie algebra of $H$ is an ideal of $\mathfrak{g}$ since $\mathfrak{g}$ is abelian.

But $H$ is not a normal subgroup of $G$.

$$((0,-1,0),(1,3))\star ((1,0,0),I)\star((0,1,0),(1,3))=((,0,1),I)\notin H$$

Answer 2

Consider $G=\mathbb{R}^2\times \mathbb{Z}$ endowed with $(x,y)(x',y')=(x+A(x'),y+y')$, where $A\in G(2,\mathbb{R})$ is invertible and $A$ does not preserves a line $L$. $L=H$ is a normal subgroup of $\mathbb{R}^2$ but not of $G$