Conjugacy classes in non-abelian group of order $pq$

Question

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!

Answer 1

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 $q$).

Answer 2

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}