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

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

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

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?