# Cyclic group question

1. Sep 11, 2007

### Benzoate

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

List the cyclic subgroups of U(30)

2. Relevant equations

3. The attempt at a solution

In order to list the cyclic subgroups for U(30) , you need to lists the generators of U(30)

U(30)={1,7,11,13,17,19,23,29} . all the elements of U(30) are not generaters. in order to determine if an element is a generator of U(30) , you need to know that a^k =gcd(n,k)=

So 1 is is automatically a generator. I don't know how to determined how the rest of the elements are generators. Just to show that I'm not trying to learn how to determine the generators of U(30) and not trying to con someone into doing my homework, my textbook says the generators for U(30) are 1,7,27,19, 29

2. Sep 11, 2007

### genneth

Imagine the cyclic group on a circle -- so they're actually cyclic. Generators are a certain spacing such that you get to every number before you go back to the beginning. As you've noted, a spacing of 1 is always possible. How about 2? It would go: 1, 3, 5 ... 29, 1, 3, ... so it would miss out the even numbers. 3? 1, 4, 7 ... 28, 1, 4, ... so it also misses things out. Try all the possibilities -- it shouldn't take that long, and you should spot a pattern pretty soon. Try making a conjecture about what's happening. Try proving it.

3. Sep 11, 2007

### matt grime

There are many misunderstandings here. 1 is not a generator of U(30). It is the identity element, so it can't generate. What the generators of U(30) are is immaterial. A cyclic subgroup is something generated by a single element. So what group does 1 generate, 2, 3, etc?

4. Sep 11, 2007

### genneth

My bad -- I was assuming an additive notation, with zero as the identity.