Primitive roots of Z_32

1. Oct 28, 2012

Applejacks

What are the primitive roots of Z_32?

$\varphi$($\varphi(32)$)=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?

2. Oct 28, 2012

DonAntonio

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

DonAntonio

3. Oct 28, 2012

Applejacks

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.

4. Oct 28, 2012

DonAntonio

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 $\,m = 2,4,p^k, 2p^k\,$ , with p a prime number and k a natural one.

DonAntonio

5. Oct 28, 2012

haruspex

(b) and (c) are only for odd p.

6. Oct 28, 2012

Applejacks

ugh I forgot about that. Every prime except 2 in an odd number so that left my mind. Thanks to both of you