# Prove any cyclic group with more than one two elements has at least two different gen

1. Jan 25, 2009

### Daveyboy

1. The problem statement, all variables and given/known data

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

2. Relevant 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.

3. 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?

2. Jan 25, 2009

### Dick

Re: Prove any cyclic group with more than one two elements has at least two different

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. Jan 25, 2009

### Symmetryholic

Re: Prove any cyclic group with more than one two elements has at least two different

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.

Last edited: Jan 25, 2009
4. Nov 6, 2010

### kathrynag

Re: Prove any cyclic group with more than one two elements has at least two different

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. Nov 6, 2010

### Dick

Re: Prove any cyclic group with more than one two elements has at least two different

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. Nov 7, 2010

### kathrynag

Re: Prove any cyclic group with more than one two elements has at least two different

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. Nov 7, 2010

### Dick

Re: Prove any cyclic group with more than one two elements has at least two different

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. Nov 7, 2010

### JonF

Re: Prove any cyclic group with more than one two elements has at least two different

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?

Last edited: Nov 7, 2010
9. Nov 7, 2010

### kathrynag

Re: Prove any cyclic group with more than one two elements has at least two different

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

10. Nov 7, 2010

### Dick

Re: Prove any cyclic group with more than one two elements has at least two different

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. Nov 7, 2010

### kathrynag

Re: Prove any cyclic group with more than one two elements has at least two different

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

12. Nov 7, 2010

### Dick

Re: Prove any cyclic group with more than one two elements has at least two different

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

13. Nov 7, 2010

### JonF

Re: Prove any cyclic group with more than one two elements has at least two different

x is called idempotent if x*x=e

14. May 13, 2012

### vohrahul

Re: Prove any cyclic group with more than one two elements has at least two different

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! ;)