Let $G$ be a non-abelian group of order $p^3$. How many are its
conjugacy classes?
The conjugacy classes are the orbits of $G$ under conjugation of
$G$ by itself. Since $G$ is non-abelian, its center has order $p$.
So the class equation yields
$p^3 = p + \sum_{[x]} (G: G_x)$, where $G_x$ is the centralizer of $x$
and the sum is taken over disjoint orbits $[x]$. We can also see that
$(G:G_x)$ can only be $p$ or $p^2$. So we will have $p$ orbits of length
$1$ and then orbits of length $p$ and $p^2$. Any hints on determining the
number of the latter?
Best Answer
We know that $Z(G) = p$. Suppose $[G:G_x] = p^2$ for some $x \notin Z(G)$. Then $G_x$ has $p$ elements, and since $Z(G) \subseteq G_x$, we have $Z(G) = G_x$. Now since $x$ is in $G_x$, it is also in $Z(G)$. From this it follows that $G_x = G$, and thus $Z(G) = G$, a contradiction. Thus $[G:G_x] = p$ for all $x \notin Z(G)$, showing that $G$ has $p^2 + p - 1$ conjugacy classes.