# Proof that the tangent space at the neutral element of a Lie group carries the structure of a Lie algebra

I am currently dealing with Lie groups. To show that the tangent space at the neutral element of a Lie group carries the structure of a Lie algebra I tried to prove the following Lemma:
Let G be a Lie group of dimension n.
(i) For every tanget vector $$\xi \in T_{e}G$$ there exists exactly one left-translation-invariant vector field $$X^{\xi} \in \Gamma^{\infty}(TG)$$ with $$X^{\xi}(e)=\xi$$. It is given by $$$$X^{\xi}(g) = d_{e}L_{g}(\xi)$$$$ where $$L_{g}:G \rightarrow G$$ is defined as $$L_{g}(x)=gx$$. In particular, the mapping $$\xi \mapsto X^{\xi}$$ is linear.
(ii) The Lie bracket of two left-translation-invariant vector fields is again left-translation-invariant.

For left-translation-invariant vector fields we have the following definition:

Let $$G$$ be a Lie group. A vector field $$X \in \Gamma^{\infty}(TG)$$ is said to be left-translation-invariant, if $$$$d_{p}L_{g}X(p) = X(L_{g}(p)) \quad \text{for all }p,g \in G$$$$ In other words: $$X$$ is $$L_{g}$$-related to itself.

I came up with the following proof:
(i) To show the existence of such a vector field, it is sufficient to show that $$X^{\xi}(g)=d_{e}L_{g}(\xi)$$ is a left-translation-invariant vector field with the property $$X^{\xi}(e)=\xi$$. It is $$$$X^{\xi}(e) = d_{e}L_{e}(\xi)=\xi$$$$ Furthermore it holds \begin{align} d_{p}L_{g}X^{\xi}(p) &= d_{p}L_{g}(d_{e}L_{p}(\xi)) \\ &= d_{e}(L_{g}\circ L_{p})(\xi) \\ &= d_{e}L_{gp}(\xi) \\ &= X^{\xi}(gp) \\ &= X^{\xi}(L_{g}(p)) \end{align} So $$X^{\xi}$$ is such a vector field. To show the uniqueness, let $$Y$$ be another left- translation-invariant vector field with $$Y(e)=\xi$$. Consider \begin{align} X^{\xi}(p)-Y(p) &= d_{e}L_{p}(X^{\xi}(e))-d_{e}L_{p}(Y(e)) \\ &= d_{e}L_{p}(X^{\xi}(e)-Y(e)) \\ &= d_{e}L_{p}(\xi-\xi) \\ &= d_{e}L_{p}(0) \\ &= 0 \end{align} The linearity of the map $$\xi \mapsto X^{\xi}$$ follows directly from the linearity of the differential: $$$$X^{\lambda\xi+\eta}(g) = d_{e}L_{g}(\lambda\xi+\eta) = \lambda d_{e}L_{g}(\xi) + d_{e}L_{g}(\eta) = \lambda X^{\xi}(g)+X^{\eta}(g)$$$$ This proves (i).
(ii) follows directly from following statement on Lie brackets

Let $$F:M \rightarrow N$$ be a smooth map between manifolds and let $$X_{1}, X_{2} \in \Gamma^{\infty}(TM)$$ and $$Y_{1}, Y_{2} \in \Gamma^{\infty}(TN)$$ be vector fields such that $$X_{i}$$ is F-related to $$Y_{i}$$ for $$i=1,2$$. Then $$[X_{1},X_{2}]$$ is F-related to $$[Y_{1},Y_{2}]$$.

Although I think the proof is correct, I would be grateful if someone would look at it again.