from 1) we have that:

cis(x)cis(y) = cis(x+y), hence we have closure under (complex) multiplication.

also cis(x)cis(-x) = cis(0) = 1+i0 = 1, so it is immediate that (cis(x))^{-1}= cis(-x), which is also in T, hence T contains all inverses. thus T is a group.

what (1) actually says is that f(x) = cis(x) is a homomorphism from (R,+) to (T,*). given cis(x) in T, it is clear that we have the pre-image x in R, so f is surjective.

thus T is isomorphic to R/ker(f), by the FHT.

what is ker(f)? the identity of T is 1 = 1+i0 (T is a subgroup of (C-{0},*) the multiplicative group of the field C), so if cis(x) = cos(x) + i sin(x) = 1+i0, then:

cos(x) = 1

sin(x) = 0

thus x = 2kπ, for some integer k. this shows ker(f) is contained in the subgroup of (R,+) generated by 2π.

on the other hand, if x is in <2π>, then cis(x) = cos(x) + i sin(x) = cos(2kπ) + i sin(2kπ) = 1+i0 = 1, so x is in ker(f).

hence the two sets are equal, so T is isomorphic to R/<2π> (f takes the real line and wraps it around the circle in such a way that the interval [0,2π) covers T exactly once).

there is really very little difference between 2 & 3 and 4 & 5....the inevitable conclusion is that R/<2π> is isomorphic to R/<1> = R/Z, one can think of this as the bijection between the unit interval [0,1], and the interval [0,2π].

(it is the RING properties of R which allow this to happen (R is a field, and thus automatically a ring), the "scaling map" x-->ax for any real number a is an additive homomorphism, because of the distributive law:

a(x+y) = ax+ay).