Exercise 1587Number theory › CongruencesOlympiadAdded 3 Nov 2026

Primitive roots

#primitive root#generator#cyclic group

Statement

Show that $2$ is a primitive root mod $11$ .

Hint

The order of $2$ mod $11$ divides $\varphi(11)=10$ . Check $2^1,2^2,2^5\not\equiv1$ .

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Order divisor. $\operatorname{ord}_{11}(2)\mid10$ .

Check. $2^1=2\neq1$ , $2^2=4\neq1$ , $2^5=32\equiv10\equiv-1\neq1$ .

Conclusion. The order is not $1,2,5$ , so it must be $10$ . $2$ is a primitive root mod $11$ .

Corollary. $\{2^k:0\le k<10\}=\mathbb{F}_{11}^*$ .