Let \(\alpha(G)\) represent the maximal size of any product-free subset of a finite abelian group \(G\). It is well known that \(\alpha(G) = \frac{|G|}{3}\left(1 + \frac{1}{p}\right)\) if \(|G|\) is divisible by a prime \(p \equiv 2 \pmod{3}\) and \(p\) is the smallest such prime, \(\alpha(G) = \frac{|G|}{3}\) if \(|G|\) is not divisible by a prime \(p \equiv 2 \pmod{3}\) but \(3\) divides \(|G|\), and \(\alpha(G) = \frac{|G|}{3}\left(1 – \frac{1}{m}\right)\) if \(|G|\) is divisible only by primes \(p \equiv 1 \pmod{3}\) and \(m\) is the exponent of \(|G|\). In this paper, we use only basic group theory and number theory to derive exact expressions for the number of different maximal product-free subsets of \(G\) in the first two cases. The formulas are given in terms of the sizes of the subgroups of \(G\).