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Primitive roots of Z_32

  1. Oct 28, 2012 #1
    What are the primitive roots of Z_32?


    However you must first check that there is a primitive root. A PR exists if
    (a) n=2,4
    (b) n=p^k

    According to the solutions, Z_32 has no primitive roots. Is this correct? 32=2^5 which fulfills one of the conditions (b) so shouldn't it have PRs?
  2. jcsd
  3. Oct 28, 2012 #2

    Well, of course it has no primitive roots as the multiplicative group [itex]\,\left(\Bbb Z/2^5\Bbb Z\right)^*\cong C_2\times C_8\,[/itex] is not cyclic...

  4. Oct 28, 2012 #3
    This is a stupid question but how do I determine it's cyclic? I don't remember covering this in our notes. I was looking at more examples, Z_97,98 and 99 and was wondering what approach I would need to take then.

    98=2 x 7^2=2p^k
    99=3^2 x 11=/=2p^k or p^k

    97 and 98 have PRs but 99 does not.
  5. Oct 28, 2012 #4

    It's not a stupid question at all, and it is not trivial. I could tell you the different cases depending on m in Z_m but you can either

    find it in almost any decent group theory book or even google it by "group of units modulo m".

    About your results: they are correct as (Z_m)* is cyclic iff [itex]\,m = 2,4,p^k, 2p^k\,[/itex] , with p a prime number and k a natural one.

  6. Oct 28, 2012 #5


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    (b) and (c) are only for odd p.
  7. Oct 28, 2012 #6
    ugh I forgot about that. Every prime except 2 in an odd number so that left my mind. Thanks to both of you
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