Primitive Root

Finally i understand primitive root, so happy so share here.

1st - How many primitive root of mod n ?

let n = 11

total primitive root is:-

phi(n-1) = phi(10)

...How to find phi(n-1) in case it is too big ???????

Got method 10 = 2 x 5 ( 2 also prime, 5 also prime),
let p = 2, q = 5,
so formula is phi(a) = a x ( 1 - 1/p) x (1 - 1/q)
phi(10) = 10 x (1 - 1/2) x (1 - 1/5)
= 10 x (1/2) x(4/5)
=4

**** teacher method is
phi(10) = phi(5) x phi(2)
= ( how many prime number less than 5) x
( how many prime number less than 2)
= 4 x 1
= 4

2nd - What is primitive root of mod n?
let n = 11
primitive root of n where all the integer numbers < 10 that
your power of mod n there is no repeatation
we start to test with 2
2^k, where 0 < k < 10
a , b , c , d, e, f , g , h , i
2^1, 2^2,2^3,2^4,2^5,2^6,2^7,2^8,2^9,
2 , 4 , 8, 5 ,10 , 9, 7, 3, 6 ------------>( mod 11)

.... see, there is no number repeat,
so we conclude 2 is primitive root of mod 11

3rd - Find all primitive root of mod n
continue..... let n = 11
once we have found one primitive root, we can find the rest of primitive root already
how?
let's look at the above calculation,
b is power 2, is gcd(2,10)= 1? no, so 4 is not primitive root
c is power 3, is gcd(3,10)= 1? yes, so 8 is not primitive root
d is power 4, is gcd(4,10)= 1? no, so 5 is not primitive root
e is power 5, is gcd(5,10)= 1? no, so 10 is not primitive root
f is power 6, is gcd(6,10)= 1? no, so 9 is not primitive root
g is power 7, is gcd(7,10)= 1? yes, so 7 is not primitive root
h is power 8, is gcd(8,10)= 1? no, so 3 is not primitive root
i is power 9, is gcd(9,10)= 1? yes, so 6 is not primitive root

so, we conclude that all primitive root for mod 11 are:-
2,8,7,6 ( total 4 as proved in part 1)