Amount of generators for cyclic group and Euler's function I have this question, since I'm confused about this concept. Let's say I have a cyclic group, for example $\mathbb{Z}_7^{\times} = \left\{1, 2, 3, 4, 5, 6\right\}$. What is the relationship between the amount of generators for this group and $\phi(7)$, with $\phi$ Euler's function? 
I found the generators as $\left\{3, 4, 5, 6 \right\}$. I know that if $G$ is a cyclic group of order $n$, and $g$ is a generator, then $g^s$ will also be a generator if and only if $gcd(s,n) = 1$. But then I read somewhere that if $n$ is prime (which in our case it is), then $\phi(7) = 6$ and the amount of generators should be $6$? This cannot be true for the group $\mathbb{Z}_7^{\times}$, since $1$ cannot possibly be a generator for multiplication.
For example, if I have the cyclic group $\mathbb{Z}_{13}^{\times}$, what is the best method to find a generator, and to find how many there are? 

As for 

best method to find a generator, that is a far more advanced problem

There is fine way to find generators:
For example lets take Z×22.
1. Z×22 = {1,3,5,7,9,13,15,17,19,21}
2. φ(22) = 10 = 5*2
3. Find the first generator(7) and remember the order it generates the group in: 
{1:7, 2:5, 3:13, 5:13, 5:21, 6:15, 7:17, 8:19, 9:19, 10:1} 
4. Take all elements which indexes have gcd(5, index) = gcd(2, index) = 1. 
5. They are: {7,13,17,19} - that will be your generators.

