1. ## Primitive roots

If g is a primitive root mod 13 then I can show that g^11 is also a primitive root.

My question is how can I deduce that the product of all the primitive roots mod 13 are congruent to 1 (mod 13) from this result?

Is there a corresponding result for the product of all the primitive roots mod 169?

2. find the condition for k such that $\displaystyle g^k$ is also a primitive root mod 13.

3. I'm aware of the condition for k to such that g^k is a primitive root mod 13. In fact, I have shown that g^11 is a primitive root mod 13.

But my question is how can I use this information to deduce that the product of all the primitive roots mod 13 is congruent to 1 mod 13.

I was then curious if a similar result holds for the product of all primitive roots mod 169.

4. Originally Posted by Cairo
I'm aware of the condition for k to such that g^k is a primitive root mod 13. In fact, I have shown that g^11 is a primitive root mod 13.

But my question is how can I use this information to deduce that the product of all the primitive roots mod 13 is congruent to 1 mod 13.

I was then curious if a similar result holds for the product of all primitive roots mod 169.
Given that $\displaystyle g$ is a primitive root of $\displaystyle 13$, all the primitive roots are given by $\displaystyle g^k$, where $\displaystyle (k,12)=1$; so the primitive roots of $\displaystyle 13$ are $\displaystyle g^1, g^5, g^7,$ and $\displaystyle g^{11}$.

Then the product of all the primitive roots of $\displaystyle 13$ is congruent to $\displaystyle g^{1+5+7+11}=g^{24}$ modulo $\displaystyle 13$. By Fermat's Theorem, $\displaystyle g^{24}=(g^{12})^2\equiv1\pmod {13}$.

The general result is the following: Let $\displaystyle m$ be a positive integer which has a primitive root. Then the product of the primitive roots of $\displaystyle m$ is congruent to $\displaystyle 1$ modulo $\displaystyle m$; the exceptions are $\displaystyle m=3, 4,$ and $\displaystyle 6$.

By the way, the integers $\displaystyle m$ that have primitive roots are $\displaystyle 1, 2, 4, p^k,$ and $\displaystyle 2p^k$, where $\displaystyle p$ is an odd prime.

5. Thank you both for your help with this.

I managed to prove the first result in the end. I never thought to use Fermat's Theroem until It struck me at the last minute.

I would like some clarification on your second point melese if that is possible?

If g is a primitive root mod 169, then so is g+13. So how are the rest of the primitive roots obtained and how does their product become congruent to 1 mod 169?