# 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!

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 $$\left$$ is normal in $$G$$, there are $$q-1$$ elements of order $$q$$, and any element outside $$\left$$ 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 $$q$$).