How many conjugacy classes does a group of order $625$ have, if its center is of order $25$?
I know that the orders of conjugacy classes should divide the order of the group, so this leaves with each centralizer class order being either $5$ or $25$. But, now how do we proceed further? Will all conjugacy classes have a uniform order? How can we use the Burnside lemma here? Any hints? Thanks beforehand.
Best Answer
If $x \notin Z(G)$, then $\vert C_G(x) \vert =125$.
That's because $x \notin Z(G) \Rightarrow C_G(x) \neq G$, but $(Z(G) \subseteq C_G(x) \land x \in C_G(x) \setminus Z(G)) \Rightarrow \vert C_G(x) \vert \gt 25$.
The size of the conjugacy class containing $x$ is therefore $5$. There are $600$ non-central elements, so $G$ has $145$ conjugacy classes.
Note that this proof works for any group of size $p^4$ with center of order $p^2$, where $p$ is prime.