Normal Subgroups of Lie groups I am trying to prove the following. Suppose that $G$ is a matrix Lie group and $H$ a subgroup of $G$. Let $\mathfrak{h},\mathfrak{g}$ be the Lie algebras respectively of $G$ and $H$ with $\mathfrak{h}$ and ideal of $\mathfrak{g}$. Suppose that $G,H$ are connected. Then $H \triangleleft G$. 
Now I know that the converse of this is true without the assumption that $G,H$ are connected (I have proven this). I am interested in proving this direction and I know that:


*

*Connectedness of $G$ means that every element $A \in G$ can be written as $A = e^{X_1}\ldots e^{X_n}$ with each $X_i \in \mathfrak{g}$. One can deduce similar things about $H$.

*One can realise $H$ as the kernel of some Lie group homomorphism. The problem is that a Lie algebra homomorphism $\phi : \mathfrak{g} \to \mathfrak{h}$ does not always give a unique $\Phi : G \to H$.


Now to prove normality I want to show given a $y \in H$ and any $x \in G$ that $xyx^{-1} \in H$. However even in the simplest case that 
$$x = e^{X}, y = e^{Y}$$
for some $X \in \mathfrak{g}, Y \in\mathfrak{h}$ I am not able to see why $xyx^{-1} \in H$. What am I missing here? Please do not post complete solutions.
Thanks.
Edit: To prove normality, I think it suffices in the proof to just consider conjugating an element in $H$ of the form $e^{Y}$ for $Y \in\mathfrak{h}$ and not a general product of elements of this form. Indeed by induction, it suffices to prove that any element of the form $e^{X}e^Ye^{-X} \in H$ for $X \in \mathfrak{g}, Y \in \mathfrak{h}$.
 A: I think I know how to show that given any $X \in \mathfrak{g}, Y \in \mathfrak{h}$ that $e^Xe^Ye^{-X} \in H$. Now 
$$\begin{eqnarray*} e^{X}e^Ye^{-X} &=& 1 + e^XYe^{-X} + e^{X}\frac{Y^2}{2!}e^{-X} + \ldots\\
&=& 1 + e^{\textrm{ad}_X}(Y) + \frac{1}{2!}(e^XYe^{-X})(e^{X}Ye^{X}) + \ldots \\
&=& 1 + e^{\textrm{ad}_X}(Y) + \frac{1}{2!}\left(e^{\textrm{ad}_X}Y\right)^2 + \ldots \\
&=& e^{\left(e^{\textrm{ad}_X}Y\right)}
\end{eqnarray*}$$
from which it suffices to prove that $e^{\textrm{ad}_X}Y$ is in the Lie algebra $\mathfrak{h}$. Now we can expand $e^{\textrm{ad}_X}$ out as a power series to get
$$\begin{eqnarray*}e^{\textrm{ad}_X}(Y) &=& \bigg(1 + \textrm{ad}_X + \frac{\textrm{ad}^2_X}{2!} + \ldots \bigg) (Y) \\ 
&=& Y + \textrm{ad}_X(Y) + \frac{\textrm{ad}_X^2(Y)}{2!} + \ldots \\
&=& \text{something in $\mathfrak{h}$}
\end{eqnarray*}$$
by definition of $\mathfrak{h}$ being an ideal and the fact $\mathfrak{h}$ is a subspace of $\mathfrak{g}$ that is finite dimensional, hence is closed. We now have that $e^{\left(e^{\textrm{ad}_X}\right)}$ is in $H$, so that $e^{-X}e^Ye^{X} \in H$ proving that $H$ is normal.
$$\hspace{6in} \square$$
