# Easy test if unitary group is cyclic

by Max.Planck
Tags: cyclic group, unitary group
 P: 127 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.
 Mentor P: 10,523 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...
P: 127
 Quote by 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...
The operation is multiplication mod n, sorry forgot to mention.

Mentor
P: 15,941

## Easy test if unitary group is cyclic

This should be useful to you: http://www.math.upenn.edu/~ted/203S0...ces/multsg.pdf
P: 127
 Quote by micromass This should be useful to you: http://www.math.upenn.edu/~ted/203S0...ces/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.
P: 688
 Quote by micromass This should be useful to you: http://www.math.upenn.edu/~ted/203S0...ces/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.
Mentor
P: 15,941
 Quote by 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.
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.
P: 127
 Quote by 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.
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.
 Mentor P: 15,941 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?
P: 127
 Quote by 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?
No, i don't see it.
P: 688
 Quote by micromass $$\mathbb{Z}_n=\mathbb{Z}_{p_1^{k_1}} \times ... \times \mathbb{Z}_{p_s^{k_s}}$$
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. :)
P: 127
 Quote by 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. :)
U(15) is not cyclic, but it is a power of a squarefree number right?
 P: 127 Anyone?

 Related Discussions Linear & Abstract Algebra 6 Calculus & Beyond Homework 3 Differential Geometry 7 Calculus & Beyond Homework 3 Differential Geometry 0