Mathematical Reflections 2017, Issue 5 - Problem U425

Problem:
Let $p$ be a prime number and let $G$ be a group of order $p^3$. Define $\Gamma(G)$ as the graph whose vertices represent the noncentral conjugacy class sizes of $G$ and two vertices are joined if and only if the two associated conjugacy class sizes are not coprime. Determine the structure of $\Gamma(G)$.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
We have two cases.

(i) $G$ is abelian. In such a case, all the conjugacy classes are central, so $\Gamma(G)$ has no vertex, i.e. $\Gamma(G)$ is the null graph.

(ii) $G$ is nonabelian. Since $Z(G)$ is a subgroup of $G$, then $|Z(G)| \ | \ |G|$. Since $G$ is nonabelian, then $|Z(G)| \neq |G|$ and since $G$ is a $p$-group, then $|Z(G)| \neq 1$. It follows that $|Z(G)| \in \{p,p^2\}$. If $|Z(G)|=p^2$, then $|G/Z(G)|=p$, so $G/Z(G)$ is cyclic and this implies that $G$ is abelian, contradiction. So, it must be $|Z(G)|=p$. Now, take a noncentral conjuacy class $\textrm{Cl}(a)$. Clearly, $|\textrm{Cl}(a)| \neq 1$. By the Orbit-Stabilizer Theorem, we have that $$|\textrm{Cl}(a)|=[G:C_G(a)],$$ where $C_G(a)$ is the centralizer of $a$ in $G$. It follows that $|\textrm{Cl}(a)| \ | \ |G|$. Since $Z(G)=\bigcap_{a \in G} C_G(a)$ and $a \in C_G(a)$, but $a \notin Z(G)$, then $Z(G) \subset C_G(a)$, which gives $|Z(G)| < |C_G(a)|$. Since $C_G(a)$ is a subgroup of $G$, then $|C_G(a)| \ | \ |G|$, which implies that $|C_G(a)|=p^2$. So, $|\textrm{Cl}(a)|=p$, i.e. each noncentral conjugacy class has size $p$, so the graph is a complete graph. In order to determine the number of its vertices, we use the Conjugacy Class Equation $$|G|=|Z(G)|+\sum_{i}[G:C_G(x_i)]$$ and we get $$p^3=p+kp,$$ i.e. $k=p^2-1$. We conclude that $\Gamma(G)=K_{p^2-1}$.