Easy test if unitary group is cyclic

In summary, the conversation discusses the unitary group U(n) and its definition as well as the concept of cyclic groups and their generators. The operation of addition mod n and its relation to prime numbers is also mentioned. The conversation explores the use of the Chinese Remainder theorem to deduce information about unitary groups, but no concrete conclusions are reached.
  • #1
Max.Planck
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Is there an easy way to see if a unitary group is cyclic? The unitary group U(n) is defined as follows [itex]U(n)=\{i\in\mathbb{N}:gcd(i,n)=1\}[/itex]. Cyclic means that there exits a element of the group that generates the entire group.
 
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  • #2
That does not look like the usual "unitary group".
Those groups needs some operation.
- Addition mod n (and restrict i to 0...n-1)? In that case, prime numbers could be interesting.
- Addition as in the natural numbers? 1, n and n+1 might be interesting to consider...
 
  • #3
mfb said:
That does not look like the usual "unitary group".
Those groups needs some operation.
- Addition mod n (and restrict i to 0...n-1)? In that case, prime numbers could be interesting.
- Addition as in the natural numbers? 1, n and n+1 might be interesting to consider...
The operation is multiplication mod n, sorry forgot to mention.
 
  • #6
micromass said:
I think the problem is that, if G is to be a multiplicative subgroup of a field K, the operation cannot be the ordinary multiplication modulo n (unless n is prime). For example, U(12) or Z/12Z can be subsets of Q, but not subgroups; they are groups in their own right, by virtue of a different operation than in Q.
 
  • #7
Dodo said:
I think the problem is that, if G is to be a multiplicative subgroup of a field K, the operation cannot be the ordinary multiplication modulo n (unless n is prime). For example, U(12) or Z/12Z can be subsets of Q, but not subgroups; they are groups in their own right, by virtue of a different operation than in Q.

I'm not talking about [itex]\mathbb{Q}[/itex]. I'm talking of the field [itex]\mathbb{Z}_p[/itex] and the subgroup U(p). This answers the OP his question in the case that n is prime.
Now he should think about the nonprime cases.
 
  • #8
micromass said:
I'm not talking about [itex]\mathbb{Q}[/itex]. I'm talking of the field [itex]\mathbb{Z}_p[/itex] and the subgroup U(p). This answers the OP his question in the case that n is prime.
Now he should think about the nonprime cases.

In case it is prime it is cyclic then. But when it is nonprime I only see looking through the elements for a generator as the only solution.
 
  • #9
Something else that might be worth to look at would be the Chinese Remainder theorem. Se http://en.wikipedia.org/wiki/Chinese_remainder_theorem

If [itex]n=p_1^{k_1}...p_s^{k_s}[/itex], this says that there is an isomorphism of rings

[tex]\mathbb{Z}_n=\mathbb{Z}_{p_1^{k_1}} \times ... \times \mathbb{Z}_{p_s^{k_s}}[/tex]

Can you deduce anything about the unitary groups?
 
  • #10
micromass said:
Something else that might be worth to look at would be the Chinese Remainder theorem. Se http://en.wikipedia.org/wiki/Chinese_remainder_theorem

If [itex]n=p_1^{k_1}...p_s^{k_s}[/itex], this says that there is an isomorphism of rings

[tex]\mathbb{Z}_n=\mathbb{Z}_{p_1^{k_1}} \times ... \times \mathbb{Z}_{p_s^{k_s}}[/tex]

Can you deduce anything about the unitary groups?

No, i don't see it.
 
  • #11
micromass said:
[tex]\mathbb{Z}_n=\mathbb{Z}_{p_1^{k_1}} \times ... \times \mathbb{Z}_{p_s^{k_s}}[/tex]
Hmm, that was clever. It should tell you (at least) that, for moduli which are a power of a squarefree number, a generator exists and is the product of the generators under the composing primes. I don't know if more can be read from this, though.

Edit: Ehem, no, erase what I just said. You can generate numbers as product of powers of the generators, but you need differents powers to generate them all. So I'm shutting up for the moment. :)
 
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  • #12
Dodo said:
Hmm, that was clever. It should tell you (at least) that, for moduli which are a power of a squarefree number, a generator exists and is the product of the generators under the composing primes. I don't know if more can be read from this, though.

Edit: Ehem, no, erase what I just said. You can generate numbers as product of powers of the generators, but you need differents powers to generate them all. So I'm shutting up for the moment. :)

U(15) is not cyclic, but it is a power of a squarefree number right?
 
  • #13
Anyone?
 

1. What is a unitary group?

A unitary group is a mathematical group consisting of all invertible unitary matrices. These matrices have special properties, such as being square and having complex entries, and are commonly used in linear algebra and quantum mechanics.

2. How do you test if a unitary group is cyclic?

To test if a unitary group is cyclic, you can check if the group has a generator, which is an element that can generate all other elements in the group through repeated multiplication. If the group has a generator, it is cyclic.

3. What is the significance of a unitary group being cyclic?

A unitary group being cyclic means that it has a simple structure and can be easily studied and understood. Additionally, cyclic groups have many unique properties that make them useful in various mathematical applications.

4. Can a unitary group be cyclic if it is not finite?

No, a unitary group cannot be cyclic if it is not finite. Cyclic groups are always finite, meaning they have a finite number of elements. Unitary groups, on the other hand, can be infinite and therefore cannot be cyclic.

5. How does testing for a cyclic unitary group differ from testing for a cyclic group in general?

The main difference in testing for a cyclic unitary group versus a cyclic group in general is the use of unitary matrices. In a general cyclic group, any element can be a generator, while in a unitary group, only specific elements (unitary matrices) can be generators.

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