How do I show that $2$ is not a primitive root modulo $7$?

Question

How do I show that $2$ is not a primitive root modulo $7$?

From discrete math. How many times would I need to do the form?

Answer 1

A primitive root modulo $7$ would have order $6$, but $2^3=8\equiv 1\pmod 7$,

so $2 $ is not a primitive root modulo $7$.

[$3$ is a primitive root modulo $7$, and $3^2=9\equiv2\pmod7$.]