Post subject: Quotient Group is isomorphic to the Circle Group

A portion of a homework problem was given me to solve for practice. I have solved some but not all of the homework problem and I hope you all can help.

Here is the problem:

1. For each x http://sosmath.com/CBB/latexrender/p...58e3fcf6fd.png R it is conventional to write cis(x) = cos(x) + i sin(x). Prove that cis(x+y) = cis(x) cis(y).

Let x, y http://sosmath.com/CBB/latexrender/p...58e3fcf6fd.png R. We want to prove that cis(x+y) = cis(x) cis(y). Thus,
cis(x+y) = cos(x+y) + i sin(x+y)
= (cos(x) cos(y) - sin(x)sin(y)) + i(cos(x)sin(y) + sin(x)cos(y))
= cos(x) cos(y) - sin(x) sin(y) + i sin(x) cos(y) + i sin(y) cos(x)
= (cos(x)+ i sin(x))(cos(y) + i sin(y))
= cis(x) cis(y)

2. Let T designate the set {cis(x) : x http://sosmath.com/CBB/latexrender/p...58e3fcf6fd.png R}, that is, the set of all the complex numbers lying on the unit circle, with the operation of multiplication. Use part 1 to prove that T is a group.

3. Use the FHT to conclude that T isomorphic R/<2http://sosmath.com/CBB/latexrender/p...2c7a9dac6d.png>

4. Prove that g(x) = cis(2http://sosmath.com/CBB/latexrender/p...2c7a9dac6d.pngx) is a homomorphism from R onto T, with kernel Z

Let g: R -> T by g(x) = cis(2http://sosmath.com/CBB/latexrender/p...2c7a9dac6d.pngx).
g is subjective since ever element of T is of the from cis(a) = cis(2http://sosmath.com/CBB/latexrender/p...2c7a9dac6d.png(a/2http://sosmath.com/CBB/latexrender/p...2c7a9dac6d.png)) = g(a/2http://sosmath.com/CBB/latexrender/p...2c7a9dac6d.png) for some a http://sosmath.com/CBB/latexrender/p...58e3fcf6fd.png R. The kernel of g is the set of x http://sosmath.com/CBB/latexrender/p...58e3fcf6fd.png R such that cis(2http://sosmath.com/CBB/latexrender/p...2c7a9dac6d.pngx) = 1. This equation only holds true if and only if 2http://sosmath.com/CBB/latexrender/p...2c7a9dac6d.pngx = 2http://sosmath.com/CBB/latexrender/p...2c7a9dac6d.pngk for some k http://sosmath.com/CBB/latexrender/p...58e3fcf6fd.png Z. Divide by 2http://sosmath.com/CBB/latexrender/p...2c7a9dac6d.png then you get ker(g) = Z. Hence, g is a homomorphism with a kernel of Z

5. Conclude that T is isomorphic R/Z

By the FHT g: R -> T is a homomorphism and R is subjective to T. Since Z is the kernel of g then H is isomorphic to R/Z


I really have no clue how to do 2 and 3 so any help would be great on that. Also if you can verify that my other three are correct that would be great. Thanks for the help in advance

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).