>>13899155For part (ii) just construct the isomorphism.
f: V?(n)SO(2) V?(-n)SO(2)
(v,r) (v*,r) where (x,y)* = (x,-y) (inspired by how conjugation reverses phase of complex numbers)
To check that this is an iso:
f((v1, r1)*(v2,r2)) (group multiplication inside f is for V?(n)SO(2))
= ([v1 + (r1^n)v2]*, r1r2) = (v1* + (r1^(-n))v2*, r1r2) (the conjugation on r1^n gives (-r1)^n = r1^(-n))
= (v1*,r1)*(v2*,r2) = f((v1, r1))*f((v2,r2)) (group multiplication outside f is for V?(-n)SO(2))
so it is homo.
To check bijective, it is easy to see that conjugation is a bijection from V to V.
bijective + homo = iso
