Determine a generator of $\mathbb{Z}^*_{11}$ manually.

Question

What is the best/standard way to do this manually? Could you describe a solution in a step-by-step fashion.

Answer 1

Not step by step but unique step: find an element $\;-1\neq x\in\Bbb Z_{11}\;$ s.t. $\;a^5=-1\;$. Can you see why this works?.

For example (all the time doing arithmetic modulo $\;11\;$):

$$2^4=5\implies 2^5=-1$$

$$3^2=-2\;,\;\;3^4=(-2)^2=4\implies 3^5=1$$

So $\;2\;$ works, $\;3\;$ doesn't, etc.

Answer 2

There is no known formula for computing primitive roots modulo $p$, but you can use that an element is a multiplicative generator iff its order is $\varphi(p)=p-1.$

In your case, $\varphi(11)=10=2\cdot 5$.

This gives you the following method: For numbers $x\in (\Bbb Z/11 \Bbb Z)^\times$, check if $x^2 \not\equiv 1$ and $x^5 \not\equiv 1$. If that is the case, the multiplicative order is $10$ and therefore $x$ is a generator.