# Easy test if unitary group is cyclic

1. Feb 27, 2013

### Max.Planck

Is there an easy way to see if a unitary group is cyclic? The unitary group U(n) is defined as follows $U(n)=\{i\in\mathbb{N}:gcd(i,n)=1\}$. Cyclic means that there exits a element of the group that generates the entire group.

2. Feb 27, 2013

### Staff: Mentor

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. Feb 27, 2013

### Max.Planck

The operation is multiplication mod n, sorry forgot to mention.

4. Feb 27, 2013

### micromass

Staff Emeritus
5. Feb 27, 2013

### Max.Planck

6. Feb 28, 2013

### dodo

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. Feb 28, 2013

### micromass

Staff Emeritus
I'm not talking about $\mathbb{Q}$. I'm talking of the field $\mathbb{Z}_p$ 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. Feb 28, 2013

### Max.Planck

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. Feb 28, 2013

### micromass

Staff Emeritus
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 $n=p_1^{k_1}...p_s^{k_s}$, this says that there is an isomorphism of rings

$$\mathbb{Z}_n=\mathbb{Z}_{p_1^{k_1}} \times ... \times \mathbb{Z}_{p_s^{k_s}}$$

Can you deduce anything about the unitary groups?

10. Feb 28, 2013

### Max.Planck

No, i don't see it.

11. Feb 28, 2013

### dodo

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. :)

Last edited: Feb 28, 2013
12. Feb 28, 2013

### Max.Planck

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

13. Apr 7, 2013

Anyone?