Left-invariant vector fields form the Lie algebra

Let $G$ be a Lie group. A smooth vector field $X$ on $G$ is left-invariant if $(L_g)_*X = X$ for all $g\in G$ , where $L_g$ is left translation.

Lemma.

The space $\mathfrak X_L(G)$ of left-invariant vector fields is closed under the usual Lie bracket of vector fields, hence is a Lie algebra.
Evaluation at the identity defines a Lie algebra isomorphism $\mathfrak X_L(G)\;\xrightarrow{\ \cong\ }\; T_eG,$ where $T_eG$ is the Lie algebra of G. Explicitly, for $v\in T_eG$ , the corresponding left-invariant field is $X_v(g)=(dL_g)_e(v).$

Remarks

Idea of proof. Left-invariance is preserved by brackets because pushforward by a diffeomorphism commutes with the vector-field bracket. The map $v\mapsto X_v$ is inverse to evaluation at $e$ by construction, and the induced bracket on $T_eG$ matches the Lie algebra bracket (compare Lie bracket).

Context. This lemma is the conceptual bridge from global group structure to infinitesimal structure and underlies constructions such as the exponential map, where one exponentiates elements of $T_eG$ .