Let $G$ be a Lie-group. Let $\mathfrak{g}\equiv \text{L}(G)$ be it's Lie-algebra: i.e. the set of left-invariant vector fields on $G$: $$ \text{L}(G):= \{ X \in \Gamma(TG)\; | \; {L_g*} X = X, \forall g\in G\}$$ where $L_{g*}$ - push-forward wrt. left translation s.t. $(L_{g*}X)_h = X_{gh}$.
Let $[G,G]$ be a closed commutator subgroup of $G$ (that is, also a Lie-group).
Statement: Prove that $\text{L}([G,G]) = [\mathfrak{g}, \mathfrak{g}]$, where $$[\mathfrak{g}, \mathfrak{g}] := span\{[X,Y] \in \mathfrak{g}\;|\; \forall X,Y \in \text{L}(G)\}$$
My attempt: By definition, we have: $$ \text{L}([G,G]) := \{Z\in \Gamma([G,G])\;|\; L_{[g,h]*}Z=Z, \forall g,h \in G\}$$ where $[g,h]:= g^{-1}h^{-1}gh$.
Then $(L_{[g,h]})_*=(L_{g^{-1}h^{-1}gh})_*=(L_{g^{-1}}\circ L_{h^{-1}}\circ L_g \circ L_h)_* = (L_{g^{-1}})_*\circ(L_{h^{-1}})_*\circ (L_g)_* \circ (L_h)_*$. That is: $$ (L_{[g,h]})_* Z= Z, \;\;\forall g,h \in G \\ \Rightarrow (L_g \circ L_h)_* Z = (L_h \circ L_g)_*Z, \;\; \forall g,h \in G$$ So, we have that Z remains the same under the commutator of two pushforwards along two arbitrary left translations. We also know that left translations are in correspondence with flows of right-invatiant vector fields: $L_g \leftrightarrow \phi_t^Q$ for some $Q \in \text{R}(G)$.
This is where I'm stuck. I think I somehow must show that equality above should lead to the conclusion that $Z$ has to be expressed as a linear combination of commutators from L$(G)$, but I don't know what exactly should I use to prove it. How do I proceed from this point?