Question

Show that 2 is a primitive root of 19.

Short answer

2 is a primitive root of 19.

Step-by-step solution

Step: 1

A Number a is a primitive root modulo n if its powers span a reduced set of residues mod n.

If the order of a is $\varnothing \left(n\right)$.

For n=19 we have $\varnothing \left(19\right)=18$. So the possible orders are 1,2,3,6,9,18.

Step: 2

Note that if the order is not 18, then it must divide either 9 or 6. But

$\begin{array}{r}{2}^6=7(mod19)\\ 2^9\equiv -1(mod19)\end{array}$

We conclude that the order of 2 modulo 19 must be $\varnothing \left(19\right)=18$and therefore 2 is a primitive root.