Order of equivalence classes in finite fields

Question

How can i easily determine the order of $$ [3] \in \mathbb{F}_{11}^\times $$

By the way: $\mathbb{F}_{11}^\times =\mathbb{F}_{11}\setminus \{[0]\}$.

Fermat's little theorem states that the order of a group element has to be a divisor of $p-1$ if $p$ is prime. Thus it should be $1, 2, 5$ or 10.

From $$ 3² = 9 \equiv 9 \mod 11 \\ 3^3 = 27 \equiv 5 \mod 11 \\ 3^4 = 81 \equiv 4 \mod 11 \\ 3^5 = 243 \equiv 1 \mod 11 $$ we see that $ord([3])=5$. Is this correct?
and is there any faster way to "see" that without calculating it?

Answer 1

Your answer is correct.

Hint for a faster way: $5^2\equiv3\mod11$, and by Fermat's little theorem $5^{10}\equiv1\mod11.$

Answer 2

Yes, there is a faster way. Since $3^{10}\equiv1\pmod{11}$, the order of $3$ must divide $10$. Therefore, there is no way that, for instance the order is $3$ or $4$.