Definition 1   Let $N>2$ be a natural number. We say that $n$ is the least universal exponent for $\Ints/N\Ints$ if $a^n\equiv 1\mod N$ for all $a\in (\Ints/N\Ints)^\times$ and $n$ is the least positive integer with this property.

Lemma 1   Let $p$ be a prime and let $n$ be the least universal exponent of $\Ints/p\Ints$ . Then there is a number $a\in (\Ints/p\Ints)^\times$ whose order is precisely $n$ .

Proof. In this proof we will use the properties of the multiplicative order of an integer. We shall prove that (a) the least universal exponent is the least common multiple of all the orders of all the elements of $(\Ints/N\Ints)^\times$ , and (b) there is an element whose order is exactly the least common multiple. In order to show (a) and (b), it suffices to show that if $a$ and $b$ have orders $\ord(a)=h$ and $\ord(b)=k$ then there is an element $c$ whose order is $\ord(c)=\lcm(h,k)$ . This is clear: if $q$ is a prime and $q^i$ divides $h=\ord(a)$ then, there is an element whose order is $p^i$ (namely $a^{h/q^i}$ has such order); and if $h'=\ord(c)$ and $k'=\ord(d)$ are relatively prime then $\ord(cd)=h'k'$ . Finally, suppose that $\lcm(h,k)=q_1^{i_1}\cdots q_t^{i_t}$ , for some distinct primes $q_1,\ldots, q_t$ and positive integers $i_1,\ldots,i_t$ . Then, each $q_j^{i_j}$ divides either $h$ or $k$ and there is an element $c_j$ of order exactly $q_j^{i_j}$ . Thus, the element $c=c_1\cdots c_t$ has exact order $\lcm(h,k)$ .

Theorem 1   Every prime $p$ has a primitive root.

Proof. If $p=2$ then $g=1$ is a primitive root. Let us assume that $p>2$ is prime and let $n$ be the least universal exponent for $p$ , i.e. $n$ is the smallest positive integer such that $x^n\equiv 1 \mod p$ , for all non-zero $x\in \Ints/p\Ints$ . Notice that, in particular by the Lemma, there is some element $g\in \Ints/p\Ints$ such that $g^n\equiv 1$ but $g^m\neq 1 \mod p$ for any $m<n$ , i.e. the multiplicative order of $g$ is precisely $n$ . Also, notice that by Fermat's little theorem, $n\leq p-1$ .

Now, the polynomial $f(x)=x^n-1$ has at most $n$ roots over the field $\Ints/p\Ints$ (see this entry), and $f(x)\equiv 0 \mod p$ for all non-zero $x \mod p$ . Thus $n\geq p-1$ . Hence, $n=p-1$ and $g$ is of exact order $p-1$ , therefore $g$ is a primitive root.