Here is the exercise from Lee's Introduction to smooth manifold 8-25
Prove that if $G$ is an abelian Lie group, then $Lie(G)$ is abelian. [Hint: show that the inversion map $i:G\rightarrow G$ is a group homomorphism, and use $di_e: T_eG\rightarrow T_eG$ is given by $di_e(X)=-X$.]
where $Lie(G)$ is defined as all left-invariant vector fields. I don't know how to get started on this. Does anyone know why the hint helps?
Thanks in advance!
Best Answer
Since $G$ abelian, then $i$ is a Lie group homomorphism and hence $i_* : \text{Lie}(G) \to \text{Lie}(G)$ is a Lie algebra homomorphism, where for each $X \in \text{Lie}(G)$, $i_*X \in \text{Lie}(G)$ is a $i$-related vector field to $X$. By the second hint, we can show that for any $X \in \text{Lie}(G)$,
$$ i_*X = -X. $$ Hence for any $X,Y \in \text{Lie}(G)$, $$ -[X,Y] = i_*[X,Y] =\cdots= [X,Y] \implies [X,Y] = 0. $$