Given one primitive root, how do you find all the others?

Question

For example: if $5$ is a primitive root of $p = 23$.

Since $p$ is a prime there are $\phi(p - 1)$ primitive roots. Is this correct?

If so, $\phi(p - 1) = \phi(22) = \phi(2) \phi(11) = 10$. So $23$ should have $10$ primitive roots?

And, to find all the other primitive roots we need powers of $5$, say $k$, sucht that $gcd(k, p - 1) = d> 1$. Again, please let me know If this true or not

So, the possible powers of $5$ are: $1, 2, 11, 22$. But this only gives four other primitive roots. So I don't think I'm on right path.

Answer 1

If $d=ord_ma,$ we know from here $ord_m(a^n)=\frac d{(n,d)}$ where $m,d$ are positive integers and $a,n$ are integers

If $a$ is a primitive root $\pmod m, ord_ma=\phi(m)$

(Remember all integers may not have a primitive root )

So, $ord_m(a^n)=\frac {\phi(m)}{(n,\phi(m))}$

So, $a^n$ is a primitive root $\pmod m,$ if $ord_m(a^n)=\phi(m)$ which needs $(n,\phi(m))=1$

If $m$ is prime, $\phi(m)=m-1$

Answer 2

You simply need $\,k\,$ such that $$\gcd(k, p-1) = 1$$

Recall that there are $\varphi (n)$ primitive $n$-th roots of unity. And for any prime $\,p,\;$ we know there are $\varphi(p)=p−1\;$ such roots.

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For a more detailed explanation of why $g^k$ is a primitive root modulo n if $\gcd(k, \varphi(n)) = 1$, see particularly the answer by Arturo Magidin.

Answer 3

Where you are going wrong is the part where you said that $\gcd(k,p−1) = d > 1$. Infact, what you want is $\gcd(k,p-1) = 1$. After you've found the first primitive root $= 5$ , the powers of $5$ will be the elements in $\phi(\phi(23)) = \phi(22) = \{1,3,5,7,9,13,15,17,19,21\}$.

This will give the required 10 primitive roots.

Answer 4

The possible powers of 5 are all the $k$'s that $gcd(k,p−1)=1$, so $k$ is in the set {$1, 3, 5, 7, 9, 13, 15, 17, 19, 21$} and $5^k$ is in the set {$5, 10, 20, 17, 11, 21, 19, 15, 7, 14$} which is exactly of length 10.