# Multiplication of primitive roots

1. May 5, 2007

### trisacloon

Hi

I noticed that multiplication of all primitive roots modulo p ,p>3,
congruent to 1 modulo p...

I have tried some examples (13,17,19...) but i couldn't prove the general case
(let g1....gk be primitive roots modulo p,p>3 ==> g1*g2*...*gk=1(p))

I need help to prove or disprove it...

2. May 5, 2007

### Hurkyl

Staff Emeritus
Try multiplying them in a clever order.

3. May 6, 2007

### trisacloon

thanks for the replay

4. May 6, 2007

### robert Ihnot

I would suggest looking at detail, leave no stone unturned!

5. May 7, 2007

### trisacloon

I will rephrase my question...
How can i explain that if g is a primitive root then g^-1 is primitive root?

6. May 7, 2007

### Hurkyl

Staff Emeritus
I would have expected it to be straightforward, although I haven't worked it out explicitly.

Where, when proving the inverse of a primitive root is primitive, do you get stuck?

Last edited: May 7, 2007
7. May 7, 2007

### matt grime

It's a group. The order x and x^-1 are trivially shown to be the same.

8. May 7, 2007

### robert Ihnot

If 1/g^k ==1, Mod M, then of course, g^k==1 Mod M.