Proving Cyclic Group Generators: An Exploration

In summary, the problem asks for a generator for a cyclic group with more than two elements, and you show that a^(-1)=a^2.
  • #1
Daveyboy
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0

Homework Statement



Prove any cyclic group with more than two elements has at least two different generators.

Homework Equations



A group G is cyclic if there exists a g in G s.t. <g> = G. i.e all elements of G can be written in the form g^n for some n in Z.


The Attempt at a Solution



Z has 1 and -1.
<i> where i = (-1)^1/2 so i and -i work

now I consider G={e, a, a^2}
All I can think of is a^3 could generate this aside from a, but that is pretty lame. Am I missing somethig?
 
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  • #2


In G={e,a,a^2}, a^3 doesn't generate. a^(3)=e. But a^2=a^(-1) does. If a is a generator, isn't it's inverse also a generator?
 
  • #3


An infinite cyclic group is isomorphic to Z, and you showed the two generators for it.
A finite cyclic group is isomorphic to [tex]Z_{n}[/tex], so an element "a" in [tex]Z_{n}[/tex] with gcd(a, n) = 1 generates the whole group.

Now for a finite cyclic group G with more than two elements, we can count how many elements in G can generate the whole group.

Hint) Use the Euler phi function.
 
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  • #4


Hi, I just found this post and noticed this was a homework problem for me.


I'm not really sure I follow this:
In G={e,a,a^2}, a^3 doesn't generate. a^(3)=e. But a^2=a^(-1) does. If a is a generator, isn't it's inverse also a generator?

I follow that a^3 doesn't generate and a^3=e, but afterwards I'm lost
 
  • #5


kathrynag said:
Hi, I just found this post and noticed this was a homework problem for me.I'm not really sure I follow this:
In G={e,a,a^2}, a^3 doesn't generate. a^(3)=e. But a^2=a^(-1) does. If a is a generator, isn't it's inverse also a generator?

I follow that a^3 doesn't generate and a^3=e, but afterwards I'm lost

Well, what's a^(-1) in this group? Does it generate the group? You can't be that lost in such a simple group.
 
  • #6


We want a * a^-1=a^-1*a=e
Can't be e. That'll give a.
Can't be a. That'll give a^2.
Then it must be a^2.
 
  • #7


kathrynag said:
We want a * a^-1=a^-1*a=e
Can't be e. That'll give a.
Can't be a. That'll give a^2.
Then it must be a^2.

A method like that could take a lot of time if you are dealing with Z100. But, yes, a^(-1)=a^2. Does a^2 generate the group?
 
  • #8


If you know x is a generator for G all you need to show is for some n (x-1)^n = x to get x-1 is also a generator. Note this isn't sufficient for your claim, you need to show something else. To see what consider 2 in Z4, 2 = 2-1. So you need to prove why your your generator can't be idempotent.

Also you need to consider infinite groups, so ask your self can an infinite group be generated by a single element?
 
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  • #9


Yes, a^2 generates the group.
Sorry, I've never heard the word idempotent yet.
 
  • #10


kathrynag said:
Yes, a^2 generates the group.
Sorry, I've never heard the word idempotent yet.

Ok then can you figure out how this also works for a general cyclic group of order n, {e,a,a^2,...,a^(n-1)} generated by a. Does a^(n-1) (which is a^(-1)) also generate? What's (a^(n-1))^2?
 
  • #11


Yes a^(n-1) generates. (a^(n-1))^2=a^(2(n-1))=a^(2n)a^-1
 
  • #12


kathrynag said:
Yes a^(n-1) generates. (a^(n-1))^2=a^(2(n-1))=a^(2n)a^-1

Yes, a^(n-1) generates. But your calculation above is quite wrong.
 
  • #13


x is called idempotent if x*x=e
 
  • #14


Consider a group G = {1,3,5,7,11,13,17} under the multiplication modulo 18 .

Now this group is CYCLIC and have two generators : 5 and 11..
5^1 = 5
5^2 = 7
5^3 = 17
5^4 = 13
5^5 = 11
5^6 = 1

thus giving it order of 6 which is a divisor of order of G. 6(1) = 6 {hence proving lagrange's theorem also }

Similarly,

11^1 = 11
11^2 = 13
11^3 = 17
11^4 = 7
11^5 = 5
11^6 = 1

Again has order 6, a divisor of order of G.

Check and verify!

I know the post is Q-U-I-T-E old.. But I couldn't find any simple answers here.. So posted one! ;)
 

Related to Proving Cyclic Group Generators: An Exploration

1. What is a cyclic group?

A cyclic group is a group of elements that can be generated by repeatedly applying a single element, known as a generator, to itself. This process creates a cycle of elements within the group.

2. What is the importance of proving cyclic group generators?

Proving cyclic group generators is important because it allows us to understand the structure and properties of cyclic groups. It also helps us to identify which elements can generate the entire group and which elements cannot.

3. How do you prove that an element is a generator of a cyclic group?

To prove that an element is a generator of a cyclic group, we need to show that the element, when repeatedly applied to itself, can generate all the elements in the group.

4. Can all elements in a cyclic group be generators?

No, not all elements in a cyclic group can be generators. Only certain elements, known as primitive elements, have the ability to generate the entire group through repeated application.

5. What is the significance of exploring cyclic group generators?

Exploring cyclic group generators allows us to better understand the structure and properties of cyclic groups. This knowledge can then be applied to various fields of mathematics, such as cryptography and number theory.

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