Find the number of primitive elements

Question

How can i find the number of primitive elements over the field of order q? GF(27) for example. Is there a formula that I can follow? I'm really confused on how to find them. Any help would be much appreciated!

Answer 1

By "primitive element", I think you mean an element $x$ such that every nonzero element is a power of $x$. The nonzero elements of a field of order $q$ form a group under multiplication, a group of $q-1$ elements, and there's a theorem that says it is in fact a cyclic group of order $q-1$. So all you need to know is a fact from group theory; how many generators does the cyclic group of order $n$ have? If you haven't seen this before, think about the group of integers modulo $n$, under addition; which integers are generators of this group? Try some experiments, I'm sure you'll come up with the answer.

Answer 2

For any others look for an answer: A cyclic group of order $n$ has $\phi(n)$ generators, where $\phi$ is Euler's totient function. Thus $GF(27)$ has $12$ primitive elements.