Easy test if unitary group is cyclic


by Max.Planck
Tags: cyclic group, unitary group
Max.Planck
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#1
Feb27-13, 02:38 PM
P: 127
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.
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mfb
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#2
Feb27-13, 03:14 PM
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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...
Max.Planck
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#3
Feb27-13, 04:34 PM
P: 127
Quote Quote by mfb View Post
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
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#4
Feb27-13, 04:53 PM
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Easy test if unitary group is cyclic


This should be useful to you: http://www.math.upenn.edu/~ted/203S0...ces/multsg.pdf
Max.Planck
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#5
Feb27-13, 05:02 PM
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Quote Quote by micromass View Post
Not really, for example U(12) is not cyclic. How can I easily check without looking through all the elements for a generator.
dodo
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#6
Feb28-13, 03:14 AM
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Quote Quote by micromass View Post
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
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#7
Feb28-13, 03:24 AM
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Quote Quote by Dodo View Post
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 [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.
Max.Planck
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#8
Feb28-13, 09:43 AM
P: 127
Quote Quote by micromass View Post
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.
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
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#9
Feb28-13, 10:05 AM
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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?
Max.Planck
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#10
Feb28-13, 10:12 AM
P: 127
Quote Quote by micromass View Post
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?
No, i don't see it.
dodo
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#11
Feb28-13, 01:24 PM
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Quote Quote by micromass View Post
[tex]\mathbb{Z}_n=\mathbb{Z}_{p_1^{k_1}} \times ... \times \mathbb{Z}_{p_s^{k_s}}[/tex]
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
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#12
Feb28-13, 01:53 PM
P: 127
Quote Quote by Dodo View Post
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
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#13
Apr7-13, 03:49 PM
P: 127
Anyone?


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