Find order of element Hi i have a very basic problem with group theory.
1)I have x element, and i know that $|x^2|=5$ and i need to find order of x. 
I see it like that : $(x^2,x^4,x^6,x^8,x^{10})$ so order of x is 10 because : $(x,x^2,x^3,x^4,x^5,x^6,x^7,x^8,x^9,x^{10})$ but in task i got it can be also 5, i don't see it so if you could show me specific group i would be more than glad
2) I have $Z_{17}^{*}$ and i know it's order is 16. How can i tell how elements that have 16 order are in it? (Other than manually counting for each) 
 A: A really simple example is inside the group $\mathbb Z/5\mathbb Z$.
$2$ has order $5$ since it generates: $2, 4, 1, 3, 0$.
but $2^2 = 4$ and it also has order $5$ since it generates: $4, 3, 2, 1, 0$.
A: For the first part, consider the additive group $\mathbb{Z}/5\mathbb{Z}$ and pick appropriate elements (try 1 and 2).
For the second, I'm not sure there's an elementary way to do it other than manually counting, but for a prime $p$, I believe the number of primitive roots (elements that generate the cyclic group) is $\phi(\phi(p))$, where $\phi$ denotes Euler's totient function, but I'm not sure of a basic proof of this off the top of my head.
A: The key result for both questions is

If $ord(x)=n$, then 
  $
ord(x^k) = \dfrac{n}{(n,k)}
$



*

*$ord(x^2)=5$ implies $n=5(n,2)$ and so $n=5$ or $n=10$. Both cases are possible.

*If you know that $\mathbb Z_{17}^{\times}$ is cyclic of order $16$, then there are as many generators as $k$ such that $0\le k \le 15$, $(16,k)=1$ and so there are $\phi(16)=8$ generators.
