Proposition 1

— Let E be a Galois extension of K and Γ its Galois group. Let A be the derived group of Γ (I, p. 10, Def. 4) and F the field of invariants of A. For a subextension L to E to be abelian it is necessary and sufficient that it should be contained in F.

Firstly we note that F is also the field of invariants of the closure $\bar{\Delta}$ of A in Γ, and that $\bar{\Delta}$ is a closed normal subgroup of Γ. By V, p. 68, Cor. 4, F is thus a Galois extension of K. Further, the Galois group of F over K is isomorphic to $\Gamma / \bar{\Delta}$ and hence is commutative. Hence every subextension of F is abelian. Conversely let L be an abelian extension of K contained in E, and let Π be the Galois group of E over L. Since L is Galois, Π is a normal subgroup of Γ and the Galois group of L over K is isomorphic to $\Gamma / \Pi$ (V, p. 68, Cor. 4). Therefore $\Gamma / \Pi$ is commutative and Π contains A, whence $L \subset F$.