Question

How do you prove a group is cyclic? I'm having trouble with how the proof structure should look like.

Answer 1

use function

Answer 2

To show a group \(G\) is cyclic, show that \(\exists \ a \in \ G \) s.t. \(G= <a> :=\{ \ a^n \ |\ \ n\in \mathbb{Z} \ \}\).

Answer 3

If it is hard to show such an \(a\) exists then sometimes is easier to show it is isomorphic to something that is known to be cyclic like \(\mathbb{Z}_n\)

Answer 4

What if G is defined as \[\left\{ fn:n \in \mathbb{Z} \right\}, where fn =x+n./} What is the generator \in this fi=unction? What is the generator for a function?

Answer 5

\[\left\{ fn:n \in \mathbb{Z} \right\}\]

Answer 6

fn = x+n

Answer 7

Do I let n=a, where a is an element of integers? But what is x? Is x an element of real numbers?

Answer 8

Can I really generate this whole group?

Answer 9

sec didnt see you were here

Answer 10

still here?

Answer 11

Yes

Answer 12

\(\{x+n \ | \ n \in \mathbb{Z}\}\)
?

Answer 13

Yes, but is x an integer?

Answer 14

your asking me?

Answer 15

I have no idea what x is...this is confusing.

Answer 16

It says n is, but not it says nothing about x?

Answer 17

What question is this?

Answer 18

let me go find that book

Answer 19

D4 chapter 7

Answer 20

k found it. one sec

Answer 21

yes, x must be real as it permutes the reals

Answer 22

weird question

Answer 23

What do you mean by permute?

Answer 24

read questions 1-3

Answer 25

So what is the generator for this set? I am suppose to get all reals or all integers. I'm confuse what set I'm trying to generate, and what the generator is for this?

Answer 26

G is a set of functions, the functions are of the form x+n, so like

x+0
x+1
x+2
x+3

the group operation is composition,

\(f_n*f_m = f_n(f_m)= f_m+n = x+m+n = f_{n+m}\)

Answer 27

we want to show that one function generates them all, and I think it must be \(f_1=x+1\)

Answer 28

Generators can be functions?

Answer 29

if the set is a set of functions

Answer 30

Let \(f_n\in G\) then \(f_n=x+n\) where \(n\in \mathbb{Z}\)
then \(f_n = (f_1)^n=f_1\cdot f_1\cdot f_1 \cdot .......\cdot f_1\) (n times)

\(= x+n*1 = x+n\) thus \(x+1\) generates \(G\)

Answer 31

That's all for the proof?

Answer 32

example \(n=3\)

\((f_1)^3=(f_1(f_1(f_1))) = f_1(f_1(x+1))=f_1((x+1)+1)=f_1(x+2)=(x+2)+1=\\x+3\)

Answer 33

Why \[f1^n\]

Answer 34

we have shown that \(G = \langle x+1\rangle\)

Answer 35

I see, so that really is all for the proof?

Answer 36

Because f1 composed with itself generates the all group. Right?

Answer 37

We use \(f_1\) because that is \(x+1\) and remember that our operation is composition so \((f_n)^2=f_n*f_n = f_n(f_n(x))\)

Answer 38

because for any element x+n we can compose x+1 with itself n times and we get x+n

Answer 39

He should have told you guys to do parts 1,2 and 3 also. It would have helped you understand part 4.

Answer 40

if our operation is addition \(3^5 = 3+3+3+3+3 = 5*3\)
if our operation is multiplication \(3^5 = 3*3*3*3*3\)
if our operation is function composition \(f^3 = f(f(f(x)))\)

Answer 41

Okay.

Answer 42

Can show another way to write the proof, just to get a feel for verbalizing these things into paper?

Answer 43

im not sure what you mean. I only know the one way to prove it. which part is giving you problems?

Answer 44

The writing proof part. I don't know how to correctly write it.

Answer 45

i picked an arbitrary element of the group and called it \(f_n\) then I showed you can obtain \(f_n\) by taking repeated multiples of the element \(f_1 = x+1\). In fact I showed how many times we must multiply it to get our element (n times).
So I showed any element in G can be obtained through one element.

Answer 46

Okay.

Answer 47

I have other homework problems that I'm stuck on. Can I send you text letting you know when I'm open study?

Answer 48

sure

Answer 49

Marak gave us a sample exam.

Answer 50

I have my group theory midterm the Monday after this upcoming one. It covers chapters 1-9 of the book.

Answer 51

Let \(g\in G:=\{ \ x+n \ | \ n\in \mathbb{Z}\}\). Then \(g=f_m\) for some \(m \in \mathbb{Z}\).
I claim that \(f_m = (x+1)^m=(f_1)^m\).
Indeed \((f_1)^m = \underbrace{f_1*f_1*....*f_1}_\text{m times}=f_{\underbrace{1+1+1+....+1}_\text{m times}}=f_m\)

Answer 52

that last line is by part 2.

Answer 53

remember

\(f_1*f_1*....*f_1 = f_1(f_1(f_1(....f_1(x)))))))))\)

Answer 54

This is for D4? Since you use m instead of n times. Why is that?

Answer 55

I picked an arbitrary element in G. I could have used m,n,h,k,.... anything. I chose m because we use n to define the set. I did this in hopes to make it less confusing.

Answer 56

Okay.

Answer 57

Well I'll show how I written my proofs on Monday. Thanks, good nite.

Answer 58

npz gn

Answer 59

you dont have to write them as short as I do. you can explain more if you want. but remember the more you explain the more rope you are giving to hang yourself with. so be careful.