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Quotient Group is isomorphic to the Circle Group 
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#1
Dec212, 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 


#2
Dec212, 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. 


#3
Dec212, 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? 


#4
Dec212, 12:42 PM

P: 4

Quotient Group is isomorphic to the Circle Group



#5
Dec212, 08:03 PM

P: 4




#6
Dec212, 08:33 PM

P: 16

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 


#7
Dec212, 08:35 PM

P: 16

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. 


#8
Dec212, 08:48 PM

Mentor
P: 18,279

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: http://www.physicsforums.com/showpos...17&postcount=3 


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