Consider ? and ? as additive groups, the latter being a subgroup of the former. Prove ?/? ? (s)^1 { a+bi ∈ ? a? + b? 1} (Hint use the morphism f: ?→ s^1 x (maps to) e^ 2πix
Determine the identity element in S^1. Determine the kernel of the map f. Now use the first isomorphism theorem.
Let's first consider what the two groups look like. The group R/Z is an additive group, with cosets looking like r+Z, where r is in [0,1). Addition is like addition in the reals, modulo 1. You can and probably should prove that statement, by showing that if a+bc mod 1 then (a+Z) + (b+Z) (c+Z) in R/Z, and that this doesn't depend on the choice of a, b or z, so that each can be taken to be in [0,1). That's just the usual expression of numbers mod n as equivalence classes with the standard term used to express the classes as being in [0, n). The group s^1 (I'll just call it S) is a multiplicative group of the complex number with norm 1. Note that f(r) e^ir (with r a real number) is periodic with period 2π. So they expression e^2πix is just a minor change in variable making h(x) e^2πix periodic with period 1. We need to prove that h maps cosets of R/Z map to a unique point in S, and preserves the group operations. It should be pretty clear at this point that this is the case, but let's go through the motions with a more formal proof. It should be clear that the function h(x) e^2πix (which we wish to prove is an isomorphism) is bijective and preserves the group operations: 1. If x y+n (mod 1) so that x and y are in the same coset of R/Z, then h(x) e^2πix e^2πi(y+n) e^2πiy X e^2πin, e^2πiy since e^2πin cos(2πn) + isin(2πn) 1. By the way, this also illustrates that the kernel of H is 0+Z. The integers in R map to the multiplicative identity in S. So if x y in R/Z, h(x) h(y), showing that h is well-defined. You can reverse the math to show that it's 1-1. And if point s in S e^2πit, then h(t) s, so H is onto. 2. Since h(a) + h(b) e^2πia + e^2πib e^2πi(a+b) h(a+b), h preserves the group operations. Therefore h is an isomorphism of R/Z to S.
