# Cyclic groups

1. Sep 5, 2007

### Benzoate

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

Find all generators of Z(6), Z(8) , and Z(20)

2. Relevant equations

3. The attempt at a solution

I should probably list the elements of Z(6), Z(8) and Z(20) first.

Z(6)={0,1,2,3,4,5}
Z{8}={0,1,2,3,4,5,6,7}
Z(20)={0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19}

perhaps taking the divisors of Z(sub 6) , Z(sub 8) and Z(sub 20)

For 6 , the divisors are 1,2,3,and 6
For 8 , the divisors are 1,2,4,and 8
For 20 , the divisors are 1,2,4,5, and 20
any element in <..> represent the generater
Z(6), its generators are <a^1> , <a^2> , <a^3> and <a^6>
<a^1> ={0,1,2,3,4,5}
<a^2>={0,2,4}
<a^3>={0,3}
and <a^6>={0}
<a^1> has order 6, <a^2> has order 3 ,<a^3> has order 2 and <a^6> has order 1.

the back of my book says the generators for generator 6 are 1 and 5. I don't understand why generator 6 is 1 and 5.

2. Sep 5, 2007

### HallsofIvy

Do you understand what a "generator" is? Divisors of 6, except for 1 itself, can't be generators.
Your "(6), its generators are <a^1> , <a^2> , <a^3> and <a^6>
<a^1> ={0,1,2,3,4,5}
<a^2>={0,2,4}
<a^3>={0,3}
and <a^6>={0}
<a^1> has order 6, <a^2> has order 3 ,<a^3> has order 2 and <a^6> has order 1. "
pretty much shows that, of those, only 1 is a generator!
For Z6, look at multiples:
1, 2, 3, 4, 5, 0: covers the whole group.
2, 4, 0: does not
3, 0: does not
4, 2*4= 8= 2 (mod 6), 3*4= 12= 0 (mod 6): does not
5, 2*5= 10= 4 (mod 6), 3*5= 15= 3 (mod 6), 4*5= 20= 2 (mod 6), 5*5= 25= 1 (mod 6), 6*6= 36= 0 (mod 6) covers the whole group.
1 and 5 are the generators.

3. Sep 5, 2007

### Benzoate

what do you mean something covers the whole group? I think I understand why <a^2> and <a^3> would cover Z(6 ) because all the elements that are in Z(6) are not in <a^2> or <a^3>.
The last part of your answer threw me completely off the wall. Where did you get 4,5, and 6 from?

I don't understand why you to

4. Sep 5, 2007

### ChaoticOrder

Um
1+1 =2
2+1 =3
3+1 =4
4+1 =5
5+1 =6=0
0+1 =1
So 1 generates Z6

0+5=5
5+5=10=4
4+5=9=3
3+5=8=2
2+5=7=1
1+5=6=0

So does 5!

Wait... aren't 5 and 6 relatively prime? Hmmmm...what to do about those other groups...?

5. Sep 5, 2007

### HallsofIvy

By "cover the group" I mean that every member of the group is some multiple of the base number. That's what a generator does: multiples of that single element produce all elements of the group.
[/quote] I think I understand why <a^2> and <a^3> would cover Z(6 ) because all the elements that are in Z(6) are not in <a^2> or <a^3>.

6. Sep 5, 2007

### matt grime

The identity is never a generator and 1 is not the identity in Z(6) (which is horrendously bad notation, by the way).

7. Sep 5, 2007

### Kummer

Theorem: Given $$\mathbb{Z}_{n}$$ for $$n\geq 1$$ the generators of this group are $$\{k | \gcd(k,n) = 1 \mbox{ and }1\leq k \leq n \}$$.

8. Sep 5, 2007

### Benzoate

I think I understand why <a^2> and <a^3> would cover Z(6 ) because all the elements that are in Z(6) are not in <a^2> or <a^3>.
I clearly understand now how the concept of generators. My only question is , isn't their an easier way to determine the generators of Z(20) without finding the remainders of 20

9. Sep 5, 2007

### Dick

What Kummer is saying is that a generator is relatively prime to 20. It has no common divisor (except 1) with 20. Do you see why x having a common divisor with 20 would really screw up it's chances of being a generator? To be concrete, suppose it is divisible by 5. Then it's equal to 5*k for some k. That means x*4=20*k. What's 20*k mod 20?

Last edited: Sep 5, 2007
10. Sep 6, 2007

### matt grime

I don't think you do undestand what is going on. First, what the heck is a? You've never said what a is. Second, you should stick to additive notation for Z/6Z.

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