Suppose G ⊆ Sym(X) is transitive and abelian. Suppose (x)g = (x)h for g, h ∈ G. Let a = gh−1. Then (x)a = x. Claim (y)a = y for every y ∈ X. To see this, choose b ∈ G such that (x)b = y using transitivity. Then (y)a = (x)ba = (x)ab = (x)b = y, using that G is abelian. This proves the claim. But then a =idX, so gh−1 = idX and g = h. Thus G is regular.