Quotient Group is isomorphic to the Circle Group

[spotsymaj]

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 \in 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 \in 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 \in 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\pi>

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

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

 

[Eric Wright]

for #2 you need to verify the group axioms.

pick z,w,v \in T

you need to show:

1. zw \in T

by the first problem you have proved... we have

zw=cis(x)cis(y)=cis(x+y) \in T

cis(x+y) is in T by definition since x+y is in R

2. (zw)v=z(wv)

not hard to show... (since T is a subset of the complex numbers)

3. there exists a multiplicative identity in T

...its 1( =cis(0) ) !

4. there exists an inverse for z in T

z \in T \Rightarrow |z|=1

use \frac{1}{z} = \frac{\bar z}{ |z|^2}

and |z|=1 to show that 1/z is in T.

 

[Eric Wright]

For #3 define:

Note that R here is treated as an additive group!

f: R \rightarrow T

by x \rightarrow cis(x)

problem 1 shows its an additive homomorphism

you need to show that the kernel of f is &lt; 2 \pi &gt;

do you know how to do this?

 

[spotsymaj]

Do you know how to do this?

Not really! I have taught myself this whole course

 

[spotsymaj]

...its 1( =cis(0) ) !

What does this mean?

 

[Eric Wright]

What does this mean?

1=cis(0) \in T

the multiplicative identity in T is the number 1 which is e^0=cos0+isin0=cis(0)
1 is in T since 1=cis(0) and 0 is a real number

 

[Eric Wright]

Not really! I have taught myself this whole course

Take an arbitrary element of &lt;2 \pi &gt; and show it is in the kernel.

Next take an arbitrary element of the kernel and show it is in &lt;2 \pi &gt;

This is a technique called double containment and it is how we normally show that sets are equal.

 

[micromass]

Spotsymaj, I appreciate that you're using LaTeX here to make your posts readable. As a hint: it would be better to use [itеx] ... [/itеx] instead of [tеx] ... [/tеx]. The latter automatically place everything on a separate line.

Here is an FAQ: https://www.physicsforums.com/showpost.php?p=3977517&postcount=3