Conjugacy classes in non-abelian group of order $pq$ Suppose I have a non-abelian group $G$ of order $pq$ ($\Rightarrow$ the center of $G$ is trivial) such that $p | (q-1)$. 
The Class Equation gives $|G| = \sum[G:C(x)]$, where $C(x)$ denotes the centralizer of $x$, and the sum ranges over one element $x$ from each nontrivial conjugacy class. Since the centralizer of an element $x$ of $G$ is a subgroup of $G$, each of the summands is a divisor of $|G| = pq$. Thus, we may only have conjugacy classes of order $1$, $p$, $q$, or $pq$. 
It's easy to see that, since the center of $G$ is trivial (i.e., the identity element $e$ is the only member of $Z(G)$ ), $[G:C(e)] = |G|/|C(e)| = (pq)/(pq) = 1$. Since $e$ is the only element of $G$ whose centralizer is all of $G$, it follows that we only have one conjugacy class of order $1$ in $G$. As such, we cannot have a conjugacy class of order $pq$, as then we would exceed the number of elements in the group. Thus, we only have conjugacy classes of order $p$ or $q$ remaining. 
Upon viewing Number of conjugacy classes of nonabelian group of order $pq$., the user comment states that there are $p-1$ conjugacy classes of order $q$, and $(q-1)/p$ conjugacy classes of order $p$. 
My question is, how did this user arrive at these values ? I definitely see that, taking the answer for granted for a moment, we then have $pq = 1 \cdot 1 + (p-1) \cdot q + (q-1)/p \cdot p$, so the given orders make sense. But, how to arrive at these exact orders $p-1$ and $(q-1)/p$ ? I see I haven't used the fact that $p|(q-1)$ here, so maybe that helps us out. 
In general, how can I detect the number of conjugacy classes of a certain order, if we're not given the specific group $G$ that we're working with ? 
Thanks! 
 A: Let $a$ and $b$ be elements of order $p$ and $q$ respectively.
Then
$$a^{-1}ba=b^s$$
where $s$ is a non-trivial solution of $s^p\equiv1\pmod p$.
As $\left<b\right>$ is normal in $G$, there are $q-1$ elements of
order $q$, and any element outside $\left<b\right>$ has order $p$.
So there are $pq-q$ elements of order $p$.
No element of order $p$ can commute with an element of order $q$
since otherwise their product would have order $pq$ and so the group
would be cyclic. Therefore the centraliser of an element of order
$p$ has order $p$, and each element of order $p$ has $q$ conjugates.
Likewise, the centraliser of an element of order
$q$ has order $q$, and each element of order $q$ has $p$ conjugates.
Therefore the $q-1$ elements of order $q$ fall into $(q-1)/p$
conjugacy classes (each of order $p$).
Likewise, the $pq-q$ elements of order $p$ fall into $(pq-q)/q=p-1$
conjugacy classes (each of order $q$).
A: Your nonabelian $G$ has class equation:
$$pq=1+k_pp+k_qq \tag 1$$
where $k_i$ is the number of the conjugacy classes of size $i=p,q$.  Now,  there are exactly $k_qq$ elements of order $p$ (they are the ones in the conjugacy classes of size $q$). Since each subgroup of order $p$ contributes $p-1$ elements of order $p$, and two subgroups of order $p$ intersect trivially, then $k_qq=m(p-1)$ for some positive integer $m$ such that $q\mid m$ (because $q\nmid p-1$).
Therefore, $(1)$ yields:
$$pq=1+k_pp+m'q(p-1) \tag 2$$
for some positive integer $m'(=m/q)$; but then $q\mid 1+k_pp$, namely $1+k_pp=nq$ for some positive integer $n$, which replaced in $(2)$ yields:
$$p=n+m'(p-1) \tag 3$$
In order for $m'$ to be a positive integer, it must be $n=1$ (which in turn implies $m'=1$ and hence $m=q$, whence $\bf k_q=p-1$). So, $1+k_pp=q$, whence $\bf k_p=\frac{q-1}{p}$. Therefore:
\begin{alignat}{1}
\#(G/\sim_{conj}) &= 1+k_p+k_q \\
&= 1+\frac{q-1}{p}+p-1 \\
&= p+\frac{q-1}{p} \\
\end{alignat}
