The Galois Correspondence at Work

Any σ ∈ Gal(L1L2/K) restricted to L1 or L2 is an automorphism since L1 and L2 are both Galois over K. So we get a function R : Gal(L1L2/K)→ Gal(L1/K)×Gal(L2/K) by R(σ) = (σ|L1 , σ|L2). We will show R is an injective homomorphism. To show R is a homomorphism, it suffices to check the separate restriction maps σ 7→ σ|L1 and σ 7→ σ|L2 are each homomorphisms from Gal(L1L2/K) to Gal(L1/K) and Gal(L2/K). For σ and τ in Gal(L1L2/K) and any α ∈ L1, (στ)|L1(α) = (στ)(α) = σ(τ(α)), and τ(α) ∈ L1 since L1/K is Galois, so σ(τ(α)) = σ|L1(τ |L1(α)) = (σ|L1 ◦ τ |L1)(α). Thus (στ)|L1(α) = (σ|L1 ◦ τ |L1)(α) for all α ∈ L1, so (στ)|L1 = σ|L1 ◦ τ |L1 . The proof that (στ)|L2 = σ|L2 ◦ τ |L2 is the same. The kernel of R is trivial, since if σ is the identity on L1 and L2 then it is the identity on L1L2. Thus R embeds Gal(L1L2/K) into the direct product of the Galois groups of L1 and L2 over K. If the groups Gal(L1/K) and Gal(L2/K) are abelian, their direct product is abelian. Therefore the embedded subgroup Gal(L1L2/K) is abelian.