Prove that the identification of $T_e G$ with the space of right-invariant vector fields induces the opposite Lie algebra structure. Let $G$ be a Lie group and denote by $T_e G$ the tangent space of $G$ at the identity element $e$. 

How do I show that the identification of $T_e G$ with the space of right- invariant vector fields induces the opposite Lie algebra structure ?

I am familiar with the fact that $T_e G$ is isomorphic to the space of left- invariant vector fields on $G$ and I guess that the same holds for right-invariant vector fields. 
There is a hint saying that one should consider the Lie group $\bar{G}$ equal to $G$ as a smooth manifold with the multiplication $\bar{gh} = hg$. Moreover, one should use the following lemma : 

Let $G$ and $H$ be two Lie groups and let $\varphi : G \rightarrow H$ be a Lie group morphism. Then the differential $d_e \varphi : T_e G \rightarrow T_e H$ is a Lie algebra morphism. 

So, if I apply this lemma to $\varphi : G \rightarrow \bar{G}, g \mapsto g^{-1}$, then $d_e \varphi : T_e G \rightarrow T_e \bar{G}$ is a Lie algebra morphism. 
Unfortunately, I am stuck and would be pleased if someone can help me to solve this problem.
Thanks. 
 A: The sketch you give is almost complete, one has only to conclude from this the result. Let me explain:
You start with the Lie group $G$ and for obvious reasons also the opposite group (what you called $\overline{G}$) is a Lie group (with multiplication in the opposite order, whence opposite group). We now have the following facts mentioned already by you:


*

*$\iota \colon G \rightarrow \overline{G} , x \mapsto x^{-1}$ is a morphism of Lie groups. 
Computing $T_e \iota$ (what you denote $d_e\iota$) one obtains the well known (i.e. contained in many introductory Lie theory texts) formula $T_e \iota(v) = - v$

*As mentioned, $T_e G$ is isomorphic (as Lie algebra!) to the Lie algebra of left invariant vector fields on $G$ and $T_e \overline{G}$ is isomorphic (as Lie algebra) to the left invariant vector fields on $\overline{G}$.

*Using that $\iota \colon G \rightarrow \overline{G}$ is a Lie group morphism, $T_e \iota$ is a Lie algebra morphism and we have by bilinearity and antisymmetry of the Lie bracket.
$$-[x,y]_G = T_e \iota ([x,y]_G) = [T_e \iota (x),T_e \iota (y)]_{\overline{G}} = [x,y]_{\overline{G}} = [y,x]_G$$ (here subscripts denote Lie brackets with respect to the group structure of $G$ or $\overline{G}$ respectively). 


Observe now that due to the opposite composition in $\overline{G}$ left invariant vector fields on $\overline{G}$ correspond exactly to right invariant vector fields on $G$. Thus the Lie bracket on $T_e \overline{G} = T_e G$ induced by the right invariant vector fields is the Lie bracket induced by the opposite group $\overline{G}$, which is the negative of the Lie bracket induced by $G$ (due to 3.). 
We conclude that the right-invariant vector fields induce the opposite Lie algebra structure.
