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Easy test if unitary group is cyclic

  1. Feb 27, 2013 #1
    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.
  2. jcsd
  3. Feb 27, 2013 #2


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    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...
  4. Feb 27, 2013 #3
    The operation is multiplication mod n, sorry forgot to mention.
  5. Feb 27, 2013 #4
  6. Feb 27, 2013 #5
  7. Feb 28, 2013 #6
    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.
  8. Feb 28, 2013 #7
    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.
  9. Feb 28, 2013 #8
    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.
  10. Feb 28, 2013 #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?
  11. Feb 28, 2013 #10
    No, i don't see it.
  12. Feb 28, 2013 #11
    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
  13. Feb 28, 2013 #12
    U(15) is not cyclic, but it is a power of a squarefree number right?
  14. Apr 7, 2013 #13
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