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. 

One cannot in general find primitive roots without trying, but it usually does not take many trials. (If your prime $p$ is so large that factoring $p-1$ is a problem, then just testing whether a given number is a primitive root may be a stumbling block, but that is a different matter.)
In the given case, you can just write down the powers of $2$ modulo $11$, to find the sequence $1,2,4,8,5,10,9,7,3,6,1$, which returns to its starting value only after seeing all nonzero classes modulo$~11$, so indeed the class $[2]$ of $2$ is a primitive element for this field; the first trial succeeds. Once you've got this, you know that $i\mapsto[2^i]$ (the brackets meaning the class modulo$~11$ induces a group isomorphism from the additive group $\Bbb Z/10\Bbb Z$ to the multiplicative group $(\Bbb Z/11\Bbb Z)^\times$; the other generators of the latter group correspond to the generators of $\Bbb Z/10\Bbb Z$, which are the classes relatively prime to$~10$: for $3$ one gets $[2^3]=[8]$, for $7$ one gets $[2^7]=[7]$, and for $9$ one gets $[2^9]=[6]$.
