Show Lie bracket left invariant I want to prove from the definition that the Lie bracket $[X,Y]$ of two left-invariant vector fields $X,Y: G \rightarrow TG$ where $G$ is a Lie group is again left-invariant.
Left-invariance basically means that for all $g \in G$ and $h \in G$ we have 
$$DL_{g}(h)(X(h)) = X(gh).$$ 
But somehow I am stuck when I try to apply $$DL_{g}(h)([X,Y](g)).$$
I just don't see how I can end up with $$[X,Y](hg).$$
There must be a trick to end up with this result. Does anybody have an idea?
 A: In general:
Let $\psi:M\to M_1$ a diffeomprohism. Then for $X,Y\in\mathfrak{X}(M)$ the relation $$[\psi_*X,\psi_*Y]=\psi_*[X,Y]$$ holds.
The proof are just calculations:
So let $f\in C^\infty(M)$.
\begin{align}
[\psi_*X,\psi_*Y](f)&=(\psi_*X)((\psi_*Y)(f))-(\psi_*Y)((\psi_*X)(f))\\
&=(\psi_*X)(Y(f\circ\psi)\circ\psi^{-1})-(\psi_*Y)(X(f\circ\psi)\circ\psi^{-1})\\
&=X(Y(f\circ\psi)\circ\psi^{-1}\circ\psi)\circ\psi^{-1}-Y(X(f\circ\psi)\circ \psi^{-1}\circ\psi)\circ\psi^{-1}\\
&=[X,Y](f\circ\psi)\psi^{-1}\\
&=\psi_*[X,Y](f)
\end{align}
Therefor if $X,Y$ are left-invariant, the commutator is left-invariant too.
Last we prove the following statement (as asked for in the comments):
$$(\psi_*X)(f)=X(f\circ\psi)\circ\psi^{-1}, \text{ where } f\in C^\infty(M_1,\mathbb{R})$$
Let $m\in M_1$. Actually we use the transport of a vector field (more general: the transport of a section, because a vector field is just a section of the tangent bundle):
\begin{align}
(\psi_*X)(f)(m)&=(\psi'_{\psi^{-1}(m)}X_{\psi^{-1}(m)})_{m}(f)\\
&=X_{\psi^{-1}(m)}(f\circ\psi)\\
&=X(f\circ\psi)\circ\psi^{-1}(m)
\end{align}
