= Solution
The correspondence in part (a) respects <Lie brackets>. If the <Lie algebra> is abelian, any two <left-invariant vector fields> satisfy $[X,Y]=0$ everywhere, not just at the identity. Part (b) therefore gives $d\omega(X,Y)=0$ for every such pair.
At a point $g$, the map $(dL_g)_e:T_eG\to T_gG$ is an isomorphism. Thus every <tangent vector> at $g$ is the value of a <left-invariant vector field>. The two-form $d\omega$ consequently vanishes on every pair of <tangent vectors> at every point, proving
$$
\boxed{d\omega=0.}
$$
No connectedness assumption on the <Lie group> is needed: the argument uses its left translations and its <abelian Lie algebra>.
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