# Quotient Group is isomorphic to the Circle Group

by spotsymaj
Tags: circle, isomorphic, quotient
 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 $$\in$$ 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$$\pi$$x) is a homomorphism from R onto T, with kernel Z Let g: R -> T by g(x) = cis(2$$\pi$$x). 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$$\pi$$x) = 1. This equation only holds true if and only if 2$$\pi$$x = 2$$\pi$$k 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
 P: 16 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.
 P: 16 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 $$< 2 \pi >$$ do you know how to do this?
P: 4

## Quotient Group is isomorphic to the Circle Group

 Quote by Eric Wright Do you know how to do this?
Not really!! I have taught myself this whole course
P: 4
 Quote by Eric Wright ...its 1( =cis(0) ) !
What does this mean?
P: 16
 Quote by spotsymaj 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
P: 16
 Quote by spotsymaj Not really!! I have taught myself this whole course
Take an arbitrary element of $$<2 \pi >$$ and show it is in the kernel.

Next take an arbitrary element of the kernel and show it is in $$<2 \pi >$$

This is a technique called double containment and it is how we normally show that sets are equal.
 Mentor P: 16,561 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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