
#1
Feb2713, 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.




#2
Feb2713, 03:14 PM

Mentor
P: 10,813

That does not look like the usual "unitary group".
Those groups needs some operation.  Addition mod n (and restrict i to 0...n1)? 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
Feb2713, 04:34 PM

P: 127





#4
Feb2713, 04:53 PM

Mentor
P: 16,593

Easy test if unitary group is cyclic
This should be useful to you: http://www.math.upenn.edu/~ted/203S0...ces/multsg.pdf




#5
Feb2713, 05:02 PM

P: 127





#6
Feb2813, 03:14 AM

P: 688





#7
Feb2813, 03:24 AM

Mentor
P: 16,593

Now he should think about the nonprime cases. 



#8
Feb2813, 09:43 AM

P: 127





#9
Feb2813, 10:05 AM

Mentor
P: 16,593

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? 



#10
Feb2813, 10:12 AM

P: 127





#11
Feb2813, 01:24 PM

P: 688

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. :) 



#12
Feb2813, 01:53 PM

P: 127





#13
Apr713, 03:49 PM

P: 127

Anyone?



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