A family of difference sets (paper by A. L. Whiteman)

Question

I was reading A. L. Whiteman - A family of difference sets. On page 109, the author generalizes the reduced residue system modulo $v$ where $v=pq$, $p$ and $q$ are primes. At the last paragraph, can someone explain why if $p-1=df$ and $q-1=ef'$ and $(f,f')=1$, if $ff'$ is odd, then $-1 \equiv g^{(d/2)} \pmod v$, and if $ff'$ is even, there does not exist any $s$, $s=0,\dotsc, d-1$, where $d$ is $\operatorname{lcm}(p-1,q-1)$ such that $-1 \equiv g^{s}\pmod v$?

Answer 1

The multiplicative group of $\mathbb{Z}_{pq} \cong \mathbb{Z}_{p} \times \mathbb{Z}_{q}$ is not cyclic, but is generated by $x =(a,1)$ and $y = (1,b)$ where $a$ is a primitive root mod $p$ and $b$ is primitive mod $q$. Whiteman's Lemma 1 shows that $xy$ has order the least common multiple of $p-1$ and $q-1$. Clearly $\langle x, xy\rangle$ is the whole multiplicative group.

In the portion of the paper to which the question refers, Whiteman wants to decide whether $-1 \in \langle xy\rangle$. Equivalently, whether there exists an integer $m$ such that $a^m \equiv -1 \bmod p$ and $b^m \equiv -1 \bmod q$.

Whiteman's choice of notation is standard for working with cyclotomic difference sets, but slightly hard to follow otherwise. Let $e = \gcd(p-1, q-1)$, and suppose that both $\frac{p-1}{e}$ and $\frac{q-1}{e}$ are odd. Taking $m = \frac{(p-1)(q-1)}{2e}$, it's easy to see that $a^m = (a^{(q-1)/e})^{(p-1)/2} \equiv -1 \bmod p$ and similarly for $b$. So $(xy)^m \equiv -1 \bmod pq$.

On the other hand, if $\frac{p-1}{e}$ is even and $\frac{q-1}{e}$ is odd, the conditions on $a^{m} \equiv -1 \bmod p$ and $b^{m} \equiv -1 \bmod q$ are inconsistent and $-1 \notin \langle xy \rangle$. Whiteman goes on to express $-1$ as a product of generators of the multiplicative group.