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The problem asks to show that a nonabelian group of order $pq$ has $p+\frac{q-1}{p}$ conjugacy classes.

I have shown:

a. $p$ divides $q-1$,

b. $|Z(G)| = 1$,

Now I'm using the class equation to write $$pq = 1 + \sum_{i=1}^{k} |G:C_{G}(x_{i})|$$ where $C_{G}(x_{i})$ is the centralizer of the distinct representatives $x_{i}$ of noncentral congjugacy classes. Since $|C_{G}(x_{i})|$ must divide $|G|$ and cannot be $1$ or $pq$ they are of order either $p$ or $q$. Since there is only one Sylow $q$-subgroup, there is only one conjugacy class of order $p$ the rest of them must be order $q$. So we can write $$pq = 1 + p + nq$$ where $n$ is the number of distinct conjugacy classes of order $q$.

From here I would think that the number of conjugacy classes is $n+2$ (since there is also one conjugacy class of order $q$ and one conjugacy class of order $1$. However, solving the final equation for $n$ and adding $2$ to it does not give me what we are supposed to be getting.

What am I missing here?

Edit: I think the issue is that $\mathcal{O}_{x}$ and $\mathcal{O}_{y}$ being distinct conjugacy classes does not imply that $C_{G}(x)$ is distinct from $C_{G}(y)$.

Tuo
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    You wrote first that there is only one conjugacy class of order $q$ (which is false) and then you let $n$ be the number of conjugacy classes of order $q$. In fact there is one class of order $1$, $p-1$ of order $q$ and $(q-1)/p$ of order $p$. Just count up all of the elements. – Derek Holt Jul 25 '15 at 19:21
  • @ Derek Holt Where are you getting these numbers ? – Tuo Jul 25 '15 at 19:39
  • Never mind, I figured it out. – Tuo Jul 25 '15 at 20:25

2 Answers2

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Without loss of generality, we can assume that $p<q$.

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 $\color{blue}{k_qq=m(p-1)}$ for some positive integer $m$ such that $q\mid m$ (because $p<q \Rightarrow q\nmid p-1$). 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 $\color\red{1+k_pp=nq}$ 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 $\color{blue}{m=q}$, and finally $k_q=p-1$. And from $\color\red{n=1}$ follows also $1+k_pp=q$, hence $k_p=\frac{q-1}{p}$. Therefore: \begin{alignat}{1} \left|G/\sim_{conj}\right| &= 1+k_p+k_q \\ &= 1+\frac{q-1}{p}+p-1 \\ &= p+\frac{q-1}{p} \\ \end{alignat}

Shuning Zhang
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Kan't
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    We actually don't need the fact that elements of order $p$ are the ones in the conjugacy classes of size $q$. Here's my argument. Suppose we have already argued that $|Z(G)|=1$ and $q\equiv1\pmod p$. By taking mod $p$ on the class equation, we have $0\equiv1+k_q\pmod p$. Then $k_q=np-1$. If $n>1$, then $G$ would have at least $2pq-q>pq$ elements, a contradiction. So $n=1$. Looking at the class equation again, we would have $k_p=\frac{q-1}{p}$. – 19021605 May 31 '25 at 07:35
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Alternatively to the counting argument in the other answer, you can apply Burnside's orbits counting lemma. In fact, there are:

  • $q-1$ noncentral (i.e. nontrivial) elements of order $q$, whose centralizers must then have order $q$ either;
  • $pq-(q-1)-1=pq-q$ noncentral (i.e. nontrivial) elements of order $p$, whose centralizers must then have order $p$ either.

Then, the lemma yields: \begin{alignat}{1} |G/\sim_{conj}| &= \frac{1}{|G|}\sum_{g\in G}|C_G(g)| \\ &= \frac{1}{pq}[pq+(q-1)q+(pq-q)p] \\ &= 1+\frac{q-1}{p}+p-1 \\ &= p+\frac{q-1}{p} \\ \end{alignat}

Kan't
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