# How do I show that $2$ is not a primitive root modulo $7$? [duplicate]

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?

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 $$3^2=9\equiv2\pmod7$$.]