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I have an exercise of Lie group as follows: "Let $G,H$ be closed connected subgroup of $GL_n(\mathbb{R})$, and $H$ be subgoup of $G$. Suppose that $Lie(H)$ is an ideal of $Lie(G)$. Prove that $H$ is a normal subgroup of $G$." I get stuck to solve this problem. Also I have no idea to use the connectedness of $G$ and $H$. Some one can help me? Thanks a lot!

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1 Answer 1

This is essentially an application of the Lie subalgebra-subgroup correspondence:

Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. Suppose that $\mathfrak{h}$ is a Lie subalgebra of $\mathfrak{g}$. Then there is a unique connected immersed Lie subgroup $H\subseteq G$ whose Lie algebra corresponds to $\mathfrak{h}$.

You are given $H$ a closed connected Lie subgroup of the Lie group $G$. Choose $g\in G$, and let $H' = gHg^{-1}$. Then $H'$ is a Lie group with corresponding Lie algebra $Lie(H')\subseteq Lie(G)$. However, the assumption that $Lie(H)$ is an ideal of $Lie(G)$ says exactly that $Lie(H') = Lie(H)$. You then have that $H$ and $H'$ are two connected Lie subgroups of $G$ with the same Lie algebra. By the uniqueness in the above theorem, it follows that $H = H'$, and hence that $H$ is normal.

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Thank you so much. But I don't know about the theorem you said. Can you tell me the name of the theorem? By chance, if the Lie group $G$ is linear, i.e a closed subgroup of $GL_n{\mathbb{R}}$, do we have simple proof for this theorem? –  mapping Nov 3 '12 at 15:11
@mapping: I don't know if the theorem has a name. It is essentially the Frobenius Theorem‌​. The idea is as follows. Take a basis $X_1,\ldots, X_r$ for $\mathfrak{h}$. These are invariant vector fields. Since $\mathfrak{h}$ is a subalgebra, these vector fields involutive. Frobenius' Theorem says there is a unique connected integral submanifold $H\subseteq G$ through the identity element. This is the subgroup $H$. Of course, you have to prove it is a Lie subgroup. –  froggie Nov 3 '12 at 16:44
Thanks a lot! I will try to prove by this way. However I have trouble to prove $Lie(H')=Lie(H)$: If I take $X\in Lie{H'}$ then we have $\exp(tX)\in H'\forall t$. Then I can deduce that $g^{-1}Xg=Z\in Lie(H)$. So I need to prove $gZg^{-1}\in Lie(H)$. Using the condition $Lie(H)$ is an ideal of $Lie(G)$, we obtain $gZ-Zg\in Lie(H)$, hence $[gZ-Zg,g^{-1}]\in Lie(H)$. Finally I get $gZg^{-1}+g^{-1}Zg\in Lie(H)$. Until here, I don't know how to continue. Can you help me? –  mapping Nov 3 '12 at 17:53

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