 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 $(d{L}_g{)}_e:T_{e}G\to T_{g}G$ 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, provingNo connectedness assumption on the Lie group is needed: the argument uses its left translations and its abelian Lie algebra.