Proving Cyclic Group Generators: An Exploration

[Daveyboy]

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?

 

[Dick]

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?

 

[Symmetryholic]

An infinite cyclic group is isomorphic to Z, and you showed the two generators for it.
A finite cyclic group is isomorphic to $$Z_{n}$$, so an element "a" in $$Z_{n}$$ 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.

 

[kathrynag]

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

 

[Dick]

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.

 

[kathrynag]

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.

 

[Dick]

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?

 

[JonF]

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?

 

[kathrynag]

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

 

[Dick]

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?

 

[kathrynag]

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

 

[Dick]

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.

 

[JonF]

x is called idempotent if x*x=e

 

[vohrahul]

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! ;)