[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*}
[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]
[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]
