The commutator of two tangent vectors is again tangent

Step 1 (setup). Let $X \in \mathfrak{g}$, $Y \in \mathfrak{g}$ come from curves $\alpha(t), \beta(t) \in G$ with $\alpha(0)=\beta(0)=I$, $\alpha'(0)=X$, $\beta'(0)=Y$; to first order $\alpha(t) \approx I + tX$ and $\beta(t) \approx I + tY$.

Step 2 (the commutator curve). Define $\gamma(t) = e^{tX} e^{tY} e^{-tX} e^{-tY}$, which lies in $G$ because $G$ is a group and $\gamma(0) = I$. Expanding each factor to second order in $t$ (using $e^{sZ} \approx I + sZ + \tfrac12 s^2 Z^2$) and multiplying out gives $\gamma(t) = e^{tX} e^{tY} e^{-tX} e^{-tY} = I + t^2[X,Y] + O(t^3)$; the linear terms in $t$ cancel exactly because it is a commutator $g h g^{-1} h^{-1}$, leaving a quadratic leading term equal to $XY - YX$.

Step 3 (extracting the tangent vector). Reparametrize by $s = t^2$ and set $\sigma(s) = \gamma(\sqrt{s})$ for $s \ge 0$; this is a smooth curve in $G$ with $\sigma(0) = I$ and $\sigma'(0) = [X,Y]$ by Step 2. Since $\sigma$ is a curve through the identity of $G$, its velocity vector $[X,Y]$ lies in $T_e G = \mathfrak{g}$ by definition of the tangent space. Hence $[X,Y] \in \mathfrak{g}$.

Step 4 (Jacobi identity). Direct algebraic expansion of $[X,[Y,Z]] + [Y,[Z,X]] + [Z,[X,Y]] = 0$ using $[X,Y] = XY - YX$ shows every term of the form $XYZ$ appears exactly twice with opposite signs and cancels, so the identity holds automatically for any associative matrix product — no extra geometric input is needed once the bracket is $XY-YX$. Together with bilinearity and antisymmetry $[X,Y]=-[Y,X]$ (immediate from the formula), this confirms $(\mathfrak{g}, [\cdot,\cdot])$ is a Lie algebra.