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Quotient Group is isomorphic to the Circle Group

by spotsymaj
Tags: circle, isomorphic, quotient
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Dec2-12, 12:09 PM
P: 4
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 [tex]\in[/tex] 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 [tex]\in[/tex] 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 [tex]\in[/tex] 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[tex]\pi[/tex]>

4. Prove that g(x) = cis(2[tex]\pi[/tex]x) is a homomorphism from R onto T, with kernel Z

Let g: R -> T by g(x) = cis(2[tex]\pi[/tex]x).
g is subjective since ever element of T is of the from cis(a) = cis(2[tex]\pi[/tex](a/2[tex]\pi[/tex])) = g(a/2[tex]\pi[/tex]) for some a [tex]\in[/tex] R. The kernel of g is the set of x [tex]\in[/tex] R such that cis(2[tex]\pi[/tex]x) = 1. This equation only holds true if and only if 2[tex]\pi[/tex]x = 2[tex]\pi[/tex]k for some k [tex]\in[/tex] Z. Divide by 2[tex]\pi[/tex] 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
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Eric Wright
Dec2-12, 12:31 PM
P: 16
for #2 you need to verify the group axioms.

pick [tex] z,w,v \in T [/tex]

you need to show:

1. [tex] zw \in T [/tex]

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

[tex] zw=cis(x)cis(y)=cis(x+y) \in T [/tex]

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

2. [tex] (zw)v=z(wv) [/tex]

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

[tex] z \in T \Rightarrow |z|=1 [/tex]

use [tex] \frac{1}{z} = \frac{\bar z}{ |z|^2} [/tex]

and |z|=1 to show that 1/z is in T.
Eric Wright
Dec2-12, 12:40 PM
P: 16
For #3 define:

Note that R here is treated as an additive group!

[tex] f: R \rightarrow T [/tex]

by [tex] x \rightarrow cis(x) [/tex]

problem 1 shows its an additive homomorphism

you need to show that the kernel of f is [tex] < 2 \pi > [/tex]

do you know how to do this?

Dec2-12, 12:42 PM
P: 4
Quotient Group is isomorphic to the Circle Group

Quote Quote by Eric Wright View Post
Do you know how to do this?
Not really!! I have taught myself this whole course
Dec2-12, 08:03 PM
P: 4
Quote Quote by Eric Wright View Post

...its 1( =cis(0) ) !
What does this mean?
Eric Wright
Dec2-12, 08:33 PM
P: 16
Quote Quote by spotsymaj View Post
What does this mean?
[tex] 1=cis(0) \in T [/tex]

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
Dec2-12, 08:35 PM
P: 16
Quote Quote by spotsymaj View Post
Not really!! I have taught myself this whole course
Take an arbitrary element of [tex] <2 \pi > [/tex] and show it is in the kernel.

Next take an arbitrary element of the kernel and show it is in [tex] <2 \pi > [/tex]

This is a technique called double containment and it is how we normally show that sets are equal.
Dec2-12, 08:48 PM
micromass's Avatar
P: 18,331
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.

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