Lie bracket of Invariant Vector fields  Let $G$ be a Lie group and let $\xi.\eta$ be left invariant vector fields. We can now construct right invariant vector fields $X_\xi$ and $X_\eta$ by defining $X_\xi(e)=\xi(e)$ and $X_\eta(e)=\eta(e)$. For $GL_n$, it is true that $[X_\xi,X_\eta]=X_{[\eta,\xi]}$. Is it true for any Lie group?
 A: For a left-invariant vector field $X$ on $G$, denote by $X^R$ the right-invariant 
vector field with the same value at the identity. From $\iota_*X=\iota_*L_{g*}X=R_{g^{-1}*}\iota_*X$, we see that $\iota_*X$ is right-invariant, where $\iota$ is the inversion map $g\mapsto g^{-1}$. Since $\iota_*X(1)=-X(1)$, we get $X^R=-\iota_*X$. Finally, 
$[X^R,Y^R]=[-\iota_*X,-\iota_*Y]=\iota_*[X,Y]=-[X,Y]^R$.   
A: The above statement ist true for GL(n). As Deane pointed out, this neceassrily implies that the statement must be true for any subgoup of GL(n) that is a Lie group as well, i.e., any subgriup of GL(n).
It can be shown, however, that every Lie-Group can be considered as subgroups of some GL(n), where one uses representation theory. The above statement is really sloppy, but the essential point is that you can choose group representations that make the Lie group resemble some subgroup of GL(n). Consider for instanve the Spin Group which is either defined via the Clifford algebra or as the universal cover of SO(n).
EDIT: As I was fortunately pointed out, I mixed the terms Lie groups and Lie algebras. The theorem of Ado, which I was referring to, is valid for Lie algebras.
Although it may be applied, the proof given above is more elementary and gives more "computational material". I am indebted to Ben whose comment raised my awareness to the mistake in my answer.
