Example 2 Find a primitive root modulo 41.
Solve $41-1=40=2^{3} \cdot 5$, divisors $d=1,2,4,8,5,10,20,40$. Now we will find the orders of $a=2,3, \cdots$.
$$\begin{array}{c}
2^{2} \equiv 4(\bmod 41), 2^{4} \equiv 16(\bmod 41), 2^{5} \equiv-9(\bmod 41) \\
2^{10} \equiv-1(\bmod 41), \quad 2^{20} \equiv 1(\bmod 41)
\end{array}$$

So $\delta_{41}(2) \mid 20$. Since $d \mid 20, d<20$ when $2^{d} \not \equiv 1(\bmod 41)$, we have $\delta_{41}(2)=20$. Therefore, 2 is not a primitive root.
$$3^{2} \equiv 9(\bmod 41), \quad 3^{4} \equiv-1(\bmod 81), \quad 3^{8} \equiv 1(\bmod 41)$$

Similarly, $\delta_{41}(3)=8$. So 3 is also not a primitive root.
Notice that $\left[\delta_{41}(2), \delta_{41}(3)\right]=[20,8]=40$. Therefore, we do not need to calculate the orders of $4,5, \cdots$ one by one, but can use property XI in $\S 1$ to find the primitive root. Notice that
$$\delta_{41}(2)=4 \cdot 5, \quad \delta_{41}(3)=1 \cdot 8$$

By property XI, $c=2^{4} \cdot 3=48$ is a primitive root. Therefore, 7 is also a primitive root modulo 41. This method of finding primitive roots is essentially the proof 1 of Theorem 2, so proof 1 can be a method to find primitive roots in certain cases.

To find the primitive roots of $41^{\alpha}, 2 \cdot 41^{\alpha}$, we need to calculate $7^{40}$ modulo $41^{2}=1681$.
$$\begin{array}{l}
7^{2} \equiv 41+8\left(\bmod 41^{2}\right) \\
7^{4} \equiv(41+8)^{2} \equiv 18(41-1)\left(\bmod 41^{2}\right) \\
7^{5} \equiv 3 \cdot(41+1)(41-1) \equiv-3\left(\bmod 41^{2}\right) \\
7^{10} \equiv 9\left(\bmod 41^{2}\right) \\
7^{20} \equiv 81(\bmod 41)^{2} \\
7^{40} \equiv(2 \cdot 41-1)^{2} \equiv 1+(-4) \cdot 41\left(\bmod 41^{2}\right)
\end{array}$$

Since $41 \nmid-4$ and 7 is odd, it follows that 7 is a primitive root for all moduli $41^{\alpha}, 2 \cdot 41^{\alpha}$.