Understanding Primitive roots I am trying to find a single primitive root modulo $11$. The definition in my textbook says "Let  $a$  and  $n$  be relatively prime integers with  ($a \neq 0$)  and  $n$  positive. Then the least positive integer  $x$  such that  $a^x\equiv1\pmod{\! n}$  is called the  order of $a$ modulo $n$ and is denoted by $\text{ord}_{n}a$".
So what I don't understand is how I can find a single primitive root modulo $11$ if I am not also given $a$.
Or is it that maybe I understand things after all since $2$ is a primitive root modulo $11$ since $2^{10} \equiv \phi(11)\equiv 10\pmod{\! 11}$ and $2$ is a generator for the group $\mathbb{Z}/11\mathbb{Z}$?
In any case, I am confused since I need to find a second primitive root modulo $11$ and I'm not sure how to do that other than by guessing and checking $a^{10}$ for $a \in \{3,4,5,6,7,8,9,10\}$. Any help would be appreciated. 

Using only simple language:
$a$ is a primitive root mod $p$ iff $\text{ord}_p (a)=p-1$. Also,  
$$\text{ord}_p(a)=k\iff (a^k\equiv 1\!\!\pmod{\! p}\ \text{ and }\ q\mid k\,\Rightarrow\, a^{k/q}\not\equiv 1\!\!\pmod{\! p})$$
$a\not\equiv 0$ is a primitive root mod $p$ iff 
$$q\mid p-1\,\Rightarrow\, a^{(p-1)/q}\not\equiv 1\!\!\pmod{\! p}$$  
$a\not\equiv 0$ is a primitive root mod $11$ iff 
$$a^5\not\equiv 1,\, a^2\not\equiv 1\pmod{\! 11}$$
None of $1,3,4,5,9,10$ is a primitive root mod $11$, because  
$$1^2\equiv 3^5\equiv 4^5\equiv 5^5\equiv (-2)^5\equiv (-1)^2\equiv 1\pmod{\! 11}$$
$\{2,6,7,8\}$ is the set of primitive roots mod $11$, because $$2^{5}\not\equiv 1,\, 2^2\not\equiv 1,\,(-5)^5\not\equiv 1,\, (-5)^2\not\equiv 1,\, (-4)^5\not\equiv 1,$$  
$$(-4)^2\not\equiv 1,\, (-3)^5\not\equiv 1,\, (-3)^2\not\equiv 1\pmod{\! 11}$$
