Why is the multiplicative group of a finite field cyclic? [duplicate]

In a field, $X^d-1$ can have at most $d$ roots, so there can always be at most $d$ elements whose order divides $d$.

The only commutative finite groups with this property are the cyclic groups.

If $G$ is any other commutative finite group then it has a subgroup of the form $(\Bbb Z/p \Bbb Z)^2$ for some integer $p$, which means it has $p^2$ elements of order dividing $p$.

Let $G$ be the multiplicative group of a finite field, $n$ its order. Let $d$ be a divisor of $n$, $\psi(d)$ the number of elements order $d$ in $G$. Suppose there exists an element $a$ of $G$ whose order is $d$. Let $H$ be the subgroup of $G$ generated by $a$. Then every element of $H$ satisfies the equation $x^d = 1$. Since the number of the solutions of $x^d = 1$ is less than or equal to $d$ and the order of $H$ is $d$, $H = \{x \in G\mid x^d = 1\}$. Therefore $\psi(d) = 0$ or $\phi(d)$, where $\phi(d)$ is the Euler's function, i.e. the number of elements of order $d$ of a cyclic group of order $d$. Since $\sum_{d\mid n} \psi(d) = n = \sum_{d\mid n} \phi(d)$, $\psi(d) = \phi(d)$ for all $d\mid n$. In particular $\psi(n) = \phi(n)$, which means there exists an element of order $n$ in $G$. This completes the proof.

Let $G$ be the multiplicative group of a finite field, $n$ its order. Let $x$ be an element of $G$, $d$ its order. Let $H$ be the subgroup of $G$ generated by $x$. If $G = H$, we are done. Suppose otherwise. There exists $y \in G - H$. Let $m$ be the order of $y$, $l$ = lcm$(d, m)$. Suppose $d = l$. Then $m\mid d$. Hence $y^d = 1$. This contradicts the fact that the number of solutions of the equation $X^d = 1$ is less than or equal to $d$. Hence $d \lt l$. By the answer to this question, there exists an element $z$ of order $l$ in $G$. We can repeat this process until we find a generator of $G$. This completes the proof.