I'm trying to go in a kind of unconventional route and prove the existence of a primitive root $\pmod{p}$ (where $p$ is a prime) using group theory. Here's what I have so far:
By definition, $a$ is a primitive root $\pmod{p}$ if
$$a^{\phi(p)} \equiv 1 \pmod{p}$$
The set of congruence classes of integers $a \pmod{p}$ such that $gcd(a, p) = 1$ forms a group under multiplication. Every element of this group satisfies $a^{\phi(p)} \equiv 1$, by Fermat's Little Theorem.
How can I proceed with this proof? Thanks.
Suppose there is $K\le({\bf{Z}}/(p))^\times$ such that $K\cong C_q\times C_q$, for some prime $q$. Therefore, there are $q^2$ elements $x\in({\bf{Z}}/(p))^\times$ such that $x^q\equiv 1\pmod p$: a contradiction, because this equation has at most $q$ solutions in ${\bf{Z}}/(p)$. So, $({\bf{Z}}/(p))^\times$ does not contain any subgroup isomorphic to $C_q\times C_q$, for any prime $q$. For this characterization of the non-cyclic finite abelian groups, $({\bf{Z}}/(p))^\times$ must be cyclic.