Using Fermat's Little Theorem to find the smallest k for which $a ^k $≡ 1(mod 11) for a = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. We know that $a^{11-1} ≡1\bmod{11}$ for all positive integers $a$ less than $11$, by Fermat’s Little
Theorem.
But $(11 − 1)$= $10$ is not the smallest k for which $a ^k$ ≡ 1(mod 11) for all such a.How do we find the smallest k for which $a ^k $≡ $1(mod 11)$ for a = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
$[$Examples: (i) $3^5$ ≡ 1(mod 11). Here k = 5 for a = 3.
(ii)$10^2$ ≡ $1(mod 11)$. Here k = 2 for a =10.$]$
How are the values of k related to $(11 − 1)= 10?$

The multiplicative group $\mathbf F^\times_{11}$ is cyclic, so you can determine a generator of this group (i.e.  a primitive root of unity in $\mathbf F_{11}$. It happens that (as is often the case) $2$ is such a generator:$\DeclareMathOperator{\ord}{ord}$
\begin{array}{|r|cccccccccc|}
\hline
n=& 1&2&3&4&5&6&7&8&9&10 \\\hline
2^n& 2 & 4 & 8 & 5 & 10 & 9 & 7 & 3 & 6  & 1 \\\hline
\end{array}
For the orders of the other elements, write each of them as $2^k$ and use that the order of such an element is
$$\ord(2^k)=\frac{\ord(2)}{\gcd(k,\ord 2)}.$$
