Let $G$ be a group with $|G :Z(G)|=n$ finite. Let $A\subseteq G$. Show that the number of conjugates of $A$ is finite and divides $n.$ > Let $G$ be a group with $|G :Z(G)|=n$ nite. Let $A\subseteq G$. Show that the number of conjugates of $A$ is nite and divides $n.$ I know that $G$ acts on the set of nonempty subsets of $G$ by conjugation. I thought the orbit stabilizer theorem would be useful for an answer, but I can't use it well. How do I do it?

By conjugation, $G$ acts transitively on the set of conjugates of $A$. For $g\in Z(G)$, the action is trivial, hence the action factors over $G/Z(G)$. Then orbit length divides $| G/Z(G)|$.