Correct. No simplification needed.

Remember that h takes an element of Q to anautomorphismof K. In this case, acts on the element . And because it is an automorphism, it is multiplicative. Therefore . Also, . Thus

You can't write h(q) on its own here; h(q) is an automorphism, and it has to act on an element of K. In this case, it acts on the identity element of K, and an automorphism always takes the identity to the identity. Therefore h(q)(1) = 1, and (k.h(q)(1),q.1) = (k.1,q.1) = (k,q), as required.

Careful here! The inverse of (k,q) has to be an element (j,p) such that (j,p)*(k,q) = (1,1). But (j,p)*(k,q) = (j.h(p)k, pq). If this is equal to (1,1) then pq = 1 (in Q), and so . But j.h(p)k = 1 (in K), and so (because an automorphism takes inverses to inverses). Therefore . (In other words, the inverse of (k,q) isnot.)

Afterthought: I think you would find this whole calculation easier to follow if you change the notation for an automorphism. Instead of writing , write it as . That will remind you that is an operator, that has to act on an element of K. It satisfies the conditions , , and .