Find the number of elements of orders $p$ and $q$ in a nonabelian group $G$ of order $pq$, with $p<q$ (both prime).

Approach: Let the number of elements of order $t$ with $n(t)$.

Let $n(p) = m$ and $n(q)=pq-1-m$. By Sylow's theorem, the number of subgroups of order $p$ is either $1$ or $p$. It cannot be $1$, so it must be $p$. ($q$ must be of the form $pk+1$ in order for $G$ to be nonabelian ).

Now, each of the sylow-p subgroups being of order $p$, it has $\phi(p)=p-1$ generators in it. There are $p$ of them. So, $n(p)=p(p-1)=m$, and $n(q)=pq-1-p(p-1)$.

Context: cryptography

-Choose two distinct primes p and q of approximately equal size so that their product n = pq is of the required length.
-Compute φ(n) = (p-1)(q-1).
-Choose a public exponent e, 1 < e < φ(n), which is coprime to φ(n), that is, gcd(e, φ(n))=1.
-Compute a private exponent d that satisfies the congruence ed ≡ 1 (mod φ(n)).
-Make the public key (n, e) available to others. Keep the private values d, p, q, and φ(n) secret.

**What does the trible equal sign indicate in this context?** [According to Wikipedia, it can mean different things depending on the context](https://en.…