# Number of conjugacy classes of nonabelian group of order $pq$.

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 $C_{G}(y)$.

• 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 ? – TuoTuo Jul 25 '15 at 19:39
• Never mind, I figured it out. – TuoTuo Jul 25 '15 at 20:25