[proofplan] We prove that $\exp(u[X,Y])$ belongs to $G$ for every real number $u$, because this is exactly the exponential characterization of membership in the [Lie algebra](/page/Lie%20Algebra). For $u\geq 0$, we write $\exp(u[X,Y])$ as a limit of powers of small group commutators built from $\exp(tX)$ and $\exp(tY)$. Each approximating product lies in $G$, and closedness of the matrix Lie group keeps the limit in $G$. For $u<0$, we exchange $X$ and $Y$ and use $[Y,X]=-[X,Y]$. [/proofplan]

[step:Build small group commutators from exponentials already lying in $G$] Fix $X,Y\in\mathfrak g$. Since $\mathfrak g$ is defined by one-parameter exponentials, for every $t\in\mathbb R$ we have \begin{align*} \exp(tX)\in G \end{align*} and \begin{align*} \exp(tY)\in G. \end{align*} For each $u\geq 0$ and each $m\in\mathbb N$, define \begin{align*} t_m:=\sqrt{\frac{u}{m}}\in\mathbb R. \end{align*} Define the matrix \begin{align*} C_m:=\exp(t_mX)\exp(t_mY)\exp(-t_mX)\exp(-t_mY)\in M(n,\mathbb C). \end{align*} Each factor in the product defining $C_m$ belongs to $G$, and $G$ is a subgroup of $GL(n,\mathbb C)$, so $C_m\in G$. Therefore \begin{align*} C_m^m\in G \end{align*} for every $m\in\mathbb N$. [/step]

[step:Prove the commutator exponential limit]We prove that, for every fixed $u\geq 0$, \begin{align*} \lim_{m\to\infty}C_m^m=\exp(u[X,Y]) \end{align*} in $M(n,\mathbb C)$. Choose any submultiplicative norm $\|\cdot\|_*$ on $M(n,\mathbb C)$. Define \begin{align*} A:=[X,Y]\in M(n,\mathbb C). \end{align*} The matrix exponential series gives, as $t\to 0$, \begin{align*}\exp(tX)=I+tX+\frac{t^2}{2}X^2+t^3R_X(t)\end{align*} and \begin{align*}\exp(tY)=I+tY+\frac{t^2}{2}Y^2+t^3R_Y(t),\end{align*} where $R_X$ and $R_Y$ are bounded on some interval $[-\delta,\delta]$. Applying the same expansion to $-X$ and $-Y$, multiplying the four expansions, and collecting terms up to degree $2$ gives \begin{align*}\exp(tX)\exp(tY)\exp(-tX)\exp(-tY)=I+t^2(XY-YX)+t^3R(t)\end{align*} for a matrix-valued function $R:(-\delta,\delta)\to M(n,\mathbb C)$ bounded near $0$. Let $K>0$ be a constant such that $\|R(t)\|_*\leq K$ whenever $|t|<\delta$. For all sufficiently large $m$, $t_m<\delta$, and hence \begin{align*} C_m=I+\frac{u}{m}A+E_m \end{align*} where \begin{align*} E_m:=t_m^3R(t_m) \end{align*} satisfies \begin{align*} \|E_m\|_*\leq K\left(\frac{u}{m}\right)^{3/2}. \end{align*} Thus \begin{align*} m\left(C_m-I\right)\to uA. \end{align*} The standard matrix exponential limit now gives \begin{align*} C_m^m\to \exp(uA)=\exp(u[X,Y]). \end{align*}[/step]

[guided]The key point is that the commutator product is designed so that all first-order terms cancel and the first nonzero term is the matrix commutator. We work in a fixed submultiplicative matrix norm $\|\cdot\|_*$ on $M(n,\mathbb C)$ so that convergence and product estimates are controlled in one finite-dimensional norm. Define \begin{align*} A:=[X,Y]=XY-YX. \end{align*} For $t$ near $0$, the matrix exponential series gives \begin{align*} \exp(tX)=I+tX+\frac{t^2}{2}X^2+t^3R_X(t), \end{align*} where $R_X(t)$ is bounded near $0$. This boundedness follows from absolute convergence of the exponential [power series](/page/Power%20Series) in the chosen submultiplicative norm: the tail beginning at degree $3$, divided by $t^3$, is locally bounded. The same statement holds with $Y$, $-X$, and $-Y$ in place of $X$. Multiplying the four second-order expansions gives \begin{align*} \exp(tX)\exp(tY)\exp(-tX)\exp(-tY)=I+t^2(XY-YX)+t^3R(t), \end{align*} where $R(t)$ is bounded near $0$. Let us indicate the cancellation. The degree-$1$ part is \begin{align*} tX+tY-tX-tY=0. \end{align*} The degree-$2$ part consists of the intrinsic second-order terms and the pairwise products of first-order terms. After collecting them, the terms $X^2/2$, $Y^2/2$, $X^2/2$, and $Y^2/2$ cancel against the corresponding products coming from inverse factors, and the surviving mixed contribution is exactly \begin{align*} XY-YX=[X,Y]. \end{align*} All remaining terms have degree at least $3$ in $t$, and boundedness of the exponential tails packages them as $t^3R(t)$ with $R$ bounded near $0$. Now fix $u\geq 0$ and put \begin{align*} t_m=\sqrt{\frac{u}{m}}. \end{align*} For all sufficiently large $m$, $t_m$ lies in the neighbourhood where $R$ is bounded. If $K>0$ satisfies $\|R(t)\|_*\leq K$ there, then \begin{align*} C_m=I+t_m^2A+t_m^3R(t_m)=I+\frac{u}{m}A+E_m, \end{align*} where \begin{align*} E_m:=t_m^3R(t_m) \end{align*} and \begin{align*} \|E_m\|_*\leq K\left(\frac{u}{m}\right)^{3/2}. \end{align*} Multiplying by $m$ gives \begin{align*} m(C_m-I)=uA+mE_m. \end{align*} Since \begin{align*} \|mE_m\|_*\leq Ku^{3/2}m^{-1/2}\to 0, \end{align*} we obtain \begin{align*} m(C_m-I)\to uA. \end{align*} The matrix exponential limit says that if $D_m\in M(n,\mathbb C)$ satisfies $m(D_m-I)\to B$, then $D_m^m\to\exp(B)$. Applying this with $D_m=C_m$ and $B=uA$ yields \begin{align*} C_m^m\to\exp(uA)=\exp(u[X,Y]). \end{align*}[/guided]

[step:Use closedness of the matrix Lie group for nonnegative parameters] Let $u\geq 0$. From the first step, $C_m^m\in G$ for every $m\in\mathbb N$. From the commutator exponential limit, \begin{align*} C_m^m\to\exp(u[X,Y]) \end{align*} in $M(n,\mathbb C)$. Since a matrix Lie group is a closed subgroup of $GL(n,\mathbb C)$, and convergence in $M(n,\mathbb C)$ to an invertible matrix preserves membership in a closed subset of $GL(n,\mathbb C)$, we obtain \begin{align*} \exp(u[X,Y])\in G. \end{align*} Thus $\exp(u[X,Y])\in G$ for every $u\geq 0$. [/step]

[step:Handle negative parameters by exchanging the two matrices] Let $u<0$. Then $-u>0$, and the already proved nonnegative case applied to the ordered pair $(Y,X)$ gives \begin{align*} \exp((-u)[Y,X])\in G. \end{align*} The matrix commutator is antisymmetric: \begin{align*} [Y,X]=YX-XY=-(XY-YX)=-[X,Y]. \end{align*} Therefore \begin{align*} (-u)[Y,X]=u[X,Y], \end{align*} and hence \begin{align*} \exp(u[X,Y])\in G. \end{align*} Combining this with the nonnegative case, we have proved that \begin{align*} \exp(u[X,Y])\in G \end{align*} for every $u\in\mathbb R$. [/step]

[step:Apply the exponential characterization of the Lie algebra] By the defining characterization \begin{align*} \mathfrak g=\{Z\in M(n,\mathbb C):\exp(tZ)\in G\text{ for every }t\in\mathbb R\}, \end{align*} a matrix $Z\in M(n,\mathbb C)$ lies in $\mathfrak g$ exactly when its full real one-parameter exponential curve remains in $G$. Applying this criterion to \begin{align*} Z=[X,Y], \end{align*} and using the conclusion of the previous step, we obtain \begin{align*} [X,Y]\in\mathfrak g. \end{align*} This proves that $\mathfrak g$ is closed under the matrix commutator. [/step]