I want to show that isomoprhism by defining a bijective linear map $f:T_e G\to \mathfrak{g}$ that sends $a\mapsto X_a$. First, let's define how we obtain $X_a$ from $a$. Take $$X_a(g)=(L_g)_*(a)\text{ for all }g\in G$$ where $L_g$ is a left action of $G$ on itself. We want to show that
- $X_a$ is a left invariant vector field.
- $X_a$ is a smooth vector field.
- $f$ is a bijective linear map.
Proofs:
(1) We need to check that $(L_h)_*(X_a) = X_a$ for all $h\in G$. It follows from the fact that $L_{gh}=L_gL_h$ for all $h,g\in G$ i.e. $(L_h)_*(X_a(g))=(L_h)_*(L_g)_*(a)=(L_hL_g)_*(a)=(L_{hg})_*(a)=X_a(hg).$
(2) It follows from the fact that $L_g:G\to G$ is a smooth map.
(3) We can see that $f$ is linear as $(L_g)_*$ is linear.
Let's check that $f$ is injective. We have that $$f(a)(g)=f(b)(g)\Rightarrow (L_g)_*(a)=(L_g)_*(b).$$ Since $(L_g)_*$ is a linear isomorphism as $L_g$ is a diffeomorphism, then $a=b$.
Let's show that $f$ is surjective. Take any $Y\in \mathfrak{g}$, then we can obtain a vector $Y_e$ by letting $Y_e=(L_{g^{-1}})_*(Y_g).$ We want to show that $f(Y_e)=Y$. Indeed, check that $$f(Y_e)(g)=Y_g\text{ for all g}\in G.$$ We have that $f(Y_e)(g)=X_{Y_e}(g)=(L_g)_*(Y_e)=(L_{g})_*(L_{g^{-1}})_*(Y_g)=Y_g$.
As 1, 2, and 3 hold, then $\mathfrak{g}\cong T_e G$.