Easy test if unitary group is cyclic

[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.

 

[mfb]

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...

 

[Max.Planck]

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.

 

[micromass]

This should be useful to you: http://www.math.upenn.edu/~ted/203S06/References/multsg.pdf

 

[Max.Planck]

This should be useful to you: http://www.math.upenn.edu/~ted/203S06/References/multsg.pdf

Not really, for example U(12) is not cyclic. How can I easily check without looking through all the elements for a generator.

 

[dodo]

This should be useful to you: http://www.math.upenn.edu/~ted/203S06/References/multsg.pdf

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.

 

[micromass]

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 \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.

 

[Max.Planck]

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.

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.

 

[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?

 

[Max.Planck]

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?

No, i don't see it.

 

[dodo]

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

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. :)

 

[Max.Planck]

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?

 

[Max.Planck]

Anyone?