# 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:

1. 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 $H$.
2. 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 $X \in \mathfrak{g}, Y \in \mathfrak{h}$.

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That's not what I asked :) Try to find an example of a subgroup $H \lt G$ such that $\mathfrak{h}$ is an ideal of $\mathfrak{g}$ but $H$ isn't normal in $G$. – t.b. Sep 5 '12 at 9:10
@t.b. I have an example here I think that $G = O_2$ and $H = I, \left(\begin{array}{cc} -1 & 0 \\ 0& 1 \end{array}\right)$ a subgroup of order 2. It is closed because it is a finite point set and it is not normal (for example conjugating by the rotation matrix of $45^\circ$ shows this). However, I don't know what the Lie algebra of $H$ is.... – user38268 Sep 5 '12 at 9:34
okay, any finite non-normal subgroup would do (it has Lie algebra $0$), showing that connectedness of $H$ is certainly necessary. On the other end of the spectrum just take a finite group $\Gamma$ and a non-normal subgroup $\Gamma_0$ and consider $G = K \times \Gamma$ and $G = K \times \Gamma_0$ for an example with $\mathfrak{g = h = k}$. An example with a proper ideal $0 \lneqq \mathfrak{h} \lneqq \mathfrak{g}$ would be more satisfactory, I guess: For this, you can use semi-direct products like $O(2) \ltimes \mathbb{R}^2$ and $\mathbb{Z}/2 \ltimes \mathbb{R}^2$ in your example. – t.b. Sep 5 '12 at 9:48
Do you know the fact that $Ad_{e^X} Y = e^{ad X} Y$? – Jason DeVito Sep 5 '12 at 17:22
@JasonDeVito Yes I do. This comes from a variation of the fact that $[X,Y] = \frac{d}{dt} e^{tX}Ye^{-tX}\big|_{t = 0}$ – user38268 Sep 5 '12 at 22:15

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.

$$\hspace{6in} \square$$

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In your final line of the second equation something more needs to be said. Since $\mathfrak{h}$ is an ideal, it's closed under finite sums, but why is it closed under infinite sums? The solution is that $\mathfrak{h}\subseteq \mathfrak{g}$ is a closed subset (since everything is finite dimensional). In infinite dimensional vector spaces, a subspace need not be closed. – Jason DeVito Sep 6 '12 at 13:29