Prove that the commutator is a Lie bracket in $\mathrm{Lie}(G)$
We need to check that ${\rm Lie}(G)$ is (1) a vector space, and (2) it is closed under the standard commutation operation. The commutator of matrices then always satisfies the axioms of a Lie bracket, and thus this is sufficient to show that ${\rm Lie}(G)$ is a Lie algebra.
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Let $X,Y\in{\rm Lie}(G)$. That $\lambda X\in{\rm Lie}(G)$ for all $\lambda\in\mathbb R$ is immediate from the definition. On the other hand, closure under addition, $X+Y\in{\rm Lie}(G)$, follows from the Lie product formula (remembering that $G$ is, by definition, closed): $$e^{X+Y} = \lim_{N\to \infty} \left(e^{X/N}e^{Y/N}\right)^N.$$ In this expression, we know that $X/N,Y/N\in\mathrm{Lie}(G)$, thus $e^{X/N}e^{Y/N}\in G$, and thus we get the limit of a sequence elements of $G\subseteq M(n,\mathbb C)$, which must converge in $G$ due to closedeness.
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To show that $[X,Y]\equiv XY-YX\in{\rm Lie}(G)$, observe that
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For any $g\in G$ and $Y\in{\rm Lie}(G)$, we have ${\rm Ad}(g)(Y)\equiv gYg^{-1}\in{\rm Lie}(G)$, and thus $e^{hX}Y e^{-hX}\in{\rm Lie}(G)$ for any $h\in\mathbb R$ and $X\in\mathrm{Lie}(G)$.
This result comes from observing how $e^{t (gYg^{-1})}=g e^{tY} g^{-1}\in G$.
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Finally, $$[X,Y] = \partial_h|_0 e^{hX}Y e^{-hX} \equiv \lim_{h\to 0}\frac{e^{hX}Y e^{-hX}-Y}{h}.$$ This implies $[X,Y]\in{\rm Lie}(G)$ because $e^{hX}\in G$, and we already proved that ${\rm Lie}(G)$ is a vector space.
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