How to tell if U(n) is cyclic?

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In summary, the group U(n) is defined as all numbers less than n that are relatively prime to n. It is cyclic for some n but not for all. To determine if U(n) will be cyclic, we can use the Chinese remainder theorem and consider the prime factorization of n. If n can be written as 2, 4, p^a, or 2p^a where p is an odd prime, then U(n) is cyclic. Otherwise, it is not cyclic. This group is also known as the group of units mod n, or the group of units in the ring Z/nZ. The structure of U(n) is fully known and it is nontrivial to prove. When U(n
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In my book U(n) is defined as all numbers less than n that are relatively prime to n. U(n) is cyclic for some n but not for all. I was wondering if there is theory behind how to tell if U(n) will be cyclic or, even better, what elements of U(n) generate U(n). Also, the formal name of this group would help a lot too, if anyone knows what it is...
 
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This is the group of units mod n (i.e. the group of units in the ring Z/nZ). Its isomorphism type is fully known. First note that if ##n=p_1^{a_1} \cdots p_r^{a_r}## is the prime factorization of n then $$U(n) = U(p_1^{a_1}) \times \cdots \times U(p_r^{a_r})$$ by the Chinese remainder theorem. Thus it suffices to determine what ##U(p^a)## looks like for a prime p. Here there are two cases to consider.
1) If p is odd then ##U(p^a)## is cyclic (of order ##\varphi(p^a)##).
2) If p=2 then ##U(2)## is trivial (hence cyclic) but for a>1 ##U(2^a) = \mathbb Z / 2\mathbb Z \times \mathbb Z/2^{a-2} \mathbb Z## (which is noncyclic if a>2).
In particular, it follows that U(n) is cyclic iff n=2, 4, p^a or 2p^a, where p is an odd prime.

Depending on how much group theory you know, these results might not be so easy to prove. They follow most easily from the structure theorem for finitely generated abelian groups, which is a nontrivial theorem.

When ##U(n)## is cyclic, a generator is called a primitive root mod n. In general finding a primitive root explicitly isn't an easy task.
 
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Ahh I'm not familiar with a lot of this notation (I'm still in an introductory Abstract Algebra class), and I'm also not familiar with the Chinese remainder theorem (although I'm searching it up right now). However, you are saying that if I can write n as 2,4, or p^a or 2p^a where a is any natural number?
Also, thanks for providing the group name...it will help me alot.
 

1. How do you determine if a group is cyclic?

In order for a group U(n) to be cyclic, there must exist an element g in U(n) such that every element in U(n) can be expressed as a power of g. This means that g generates all other elements in U(n) and is called a generator of the group.

2. Can a finite group be cyclic?

Yes, a finite group can be cyclic. In fact, all finite cyclic groups are isomorphic to the group of integers modulo n, denoted as Z/nZ, where n is a positive integer.

3. What is the order of a cyclic group?

The order of a cyclic group is the number of elements in the group. For a finite cyclic group, the order is equal to the number of integers that are relatively prime to n, where n is the modulus of the group.

4. How can you prove that a group U(n) is cyclic?

To prove that a group U(n) is cyclic, you must show that there exists an element g in U(n) that generates all other elements in the group and that every element in U(n) can be expressed as a power of g. Additionally, you can show that the order of the group is equal to the number of elements in U(n).

5. Are all abelian groups cyclic?

No, not all abelian groups are cyclic. While all cyclic groups are abelian, there are non-cyclic abelian groups, such as the Klein four-group, which have more than one generator and do not follow the definition of a cyclic group.

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