I think it may be simpler than that.

Curvature is (or can be) defined as the change in a vector as it is parallel-transported around a loop (an infinitesimal parallelogram in particular, but for my purposes here it doesn't matter).

For there to be curvature, the parallel transport of a vector

V from a point

p to a point

q must

depend on the path from p to q.

Contrast this with the case of Euclidean space: there is a

global notion of 'parallel', such that given a vector

V at any point

p, you can immediately construct a parallel vector

V at any other point

q,

without consideration of the path between the points.

Euclidean space has no curvature, as an immediate consequence of this -

i.e. 'parallel transporting' any vector around any loop just leaves you with the same vector you started with.

Now the text says that in a Lie group:

Quote:

The infinitesimal group elements are to be pictured as particular vector fields on G (or, indeed, H). That is, we think of ‘moving G’ infinitesimally along the relevant vector field V on G, in order to express the transformation that corresponds to pre-multiplying each element of the group by the infinitesimal element represented by V.

and

Quote:

The relevant notion of ‘parallelism’ comes from the group action, supplying the needed notion of ‘parallel transport’...

(pages 312 and 313 respectively)

The first quote means that (just as in the Euclidean case), given a vector

V (say, at the origin), it can immediately be transferred to any other point on the manifold,

without consideration of the path taken.

The second quote confirms that this is indeed how 'parallel' (and hence 'parallel transport') is defined.

Thus, 'parallel transport' is path-independent in Lie groups, just like in Euclidean space, and it follows immediately from this that Lie groups have no curvature.