Math Help - Post subject: Quotient Group is isomorphic to the Circle Group

1. 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 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 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 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/<2>

4. Prove that g(x) = cis(2x) is a homomorphism from R onto T, with kernel Z

Let g: R -> T by g(x) = cis(2x).
g is subjective since ever element of T is of the from cis(a) = cis(2(a/2)) = g(a/2) for some a R. The kernel of g is the set of x R such that cis(2x) = 1. This equation only holds true if and only if 2x = 2k for some k Z. Divide by 2 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

2. Re: Post subject: Quotient Group is isomorphic to the Circle Group

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).