I'm trying to find the conjugacy classes of non-abelian group of order $pq$ $(p>q)$. It is known that this group is uniquely expressed as
$$\begin{align*} G&=F_{p,q}\\ &=\langle a,b:a^p=b^q=1,b^{-1}ab=a^u\rangle, \end{align*} $$
where $u$ has order $q$ modulo $p$ in $\mathbb{Z}^*_p$.
Here, the choice of $u$ doesn't matter. I have a question in finding the conjugacy class of $b^n$ specifically. I was convinced that it has the conjugacy class of size $p$. Then the book I'm reading states the following argument, which I'm not sure of.
On the other hand, as $G/\langle a\rangle$ is abelian, every conjugate of $b^n$ has the form $a^mb^n$ for some $m$. Hence $(b^n)^G=\{a^mb^n:0\leq m\leq p-1\}$.
Here $(b^n)^G$ is just the conjugacy class of $b^n$. Could anyone give some detailed explanation of this? Thanks!
$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$Take $0 < n < q$. Note first of all that $b^{n}$ is not in the centre of $G$, as it does not commute with $a$. Hence its centraliser is $\Span{b^{n}}$, of order $q$, so that its conjugacy class has indeed order $p$, as you correctly stated.
Now consider, as suggested, $K = \Span{a}$, and the factor group $G / K$, which has prime order $q$, and is thus abelian. If you take any $x \in G$, you have then in $G / K$ $$ b^{n} K = (x K)^{-1} (b^{n} K) (x K) = x^{-1} b^{n} x K. $$ This shows that $$ x^{-1} b^{n} x = b^{n} a^{i} $$ for some $i$. There are $p$ possibilities for $a^{i}$. Since you already know that $b^{n}$ has $p$ conjugates, they must be exactly the $b^{n} a^{i}$, for $0 \le i < p$.