= Solution
A <left-invariant vector field> $X$ on a <Lie group> satisfies
$$
(dL_g)_h X_h=X_{gh}\qquad(g,h\in G),
$$
or equivalently $(L_g)_*X=X$ for every $g$. For $\xi\in T_eG$, define
$$
X^\xi_g=(dL_g)_e\xi.
$$
This is smooth because multiplication is smooth and its differential depends smoothly on the base points. The identity $L_g\circ L_h=L_{gh}$ gives $(dL_g)_hX^\xi_h=X^\xi_{gh}$, so $X^\xi$ is a <left-invariant vector field>. Conversely, left invariance forces any such field to have this form with $\xi=X_e$. Evaluation at $e$ and the construction above are inverse <linear maps>. Hence
$$
\boxed{T_eG\cong\{\text{left-invariant vector fields on }G\}.}
$$
The <Lie bracket of vector fields> is their <commutator> as <derivations> on <smooth functions>: $[X,Y]f=X(Yf)-Y(Xf)$. It is another vector field because the cross terms cancel in the Leibniz rule for its action on a product. The bracket is preserved by a <diffeomorphism>: conjugating the <derivation> operators by <pullback of a smooth function> preserves their <commutator>. Therefore
$$
(L_g)_*[X,Y]=[(L_g)_*X,(L_g)_*Y]=[X,Y]
$$
for left-invariant $X,Y$. These fields are consequently closed under the bracket.
Transport the bracket to the <tangent space> by
$$
\boxed{[\xi,\eta]_{\mathfrak g}=[X^\xi,X^\eta]_e.}
$$
Bilinearity and antisymmetry come from the <commutator>. The <Jacobi identity> follows by expanding the three nested <commutators> of <derivation> operators: their six terms cancel in pairs. Thus $T_eG$ is a <Lie algebra>, and the displayed correspondence is a <Lie algebra isomorphism> onto the algebra of <left-invariant vector fields>.
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