let fx(z) = xz
let θ : Q* → Aut(Q)
with θ(x) = fx
Is θ an isomorphism?
Surjectivity and injectivity
The Attempt at a Solution
θ is a homomorphism because
θ(xy) = fxy = fx * fy = θ(x)θ(y)
θ is injective because the neutral element of Aut(Q) is f1 and
θ-1(f1) = 1 the neutral element of Q*
Prooving the surjectivity is where I am uncertain...
∀fx ∈ Aut(Q) , There is a unique x ∈ Q* for which
θ(x) = fx
Is the codomain = Aut(Q) ?