Are there primitive roots in Z_32?

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Discussion Overview

The discussion revolves around the existence of primitive roots in the group of units modulo 32, denoted as Z_32. Participants explore the conditions under which primitive roots exist and analyze specific cases, including comparisons with other integers like 97, 98, and 99.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that Z_32 has no primitive roots based on the structure of its multiplicative group, which is not cyclic.
  • Others question how to determine if a group is cyclic and express uncertainty about the conditions for the existence of primitive roots.
  • One participant lists the conditions for the existence of primitive roots, noting that Z_32 fulfills one of the conditions but still concludes it has no primitive roots.
  • There is a clarification that conditions (b) and (c) for the existence of primitive roots apply only for odd primes, which some participants initially overlooked.
  • Comparative examples of Z_97, Z_98, and Z_99 are discussed, with participants noting that 97 and 98 have primitive roots while 99 does not.

Areas of Agreement / Disagreement

Participants generally agree that Z_32 does not have primitive roots, but there is some confusion regarding the conditions that determine the existence of primitive roots, particularly concerning the role of the prime number 2.

Contextual Notes

Some participants express uncertainty about the cyclic nature of groups and the specific conditions for primitive roots, indicating a need for further clarification on these mathematical concepts.

Applejacks
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What are the primitive roots of Z_32?

[itex]\varphi[/itex]([itex]\varphi(32)[/itex])=8

However you must first check that there is a primitive root. A PR exists if
(a) n=2,4
(b) n=p^k
(c)n=2p^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?
 
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Applejacks said:
What are the primitive roots of Z_32?

[itex]\varphi[/itex]([itex]\varphi(32)[/itex])=8

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


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...

DonAntonio
 
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.

97=prime=p
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.
 
Applejacks said:
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.

97=prime=p
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.


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.

DonAntonio
 
Applejacks said:
What are the primitive roots of Z_32?
A PR exists if
(a) n=2,4
(b) n=p^k
(c)n=2p^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?
(b) and (c) are only for odd p.
 
haruspex said:
(b) and (c) are only for odd p.

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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