Number of conjugacy classes of a group of order $5^4$ whose center is $25$ order

Question

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.

Answer 1

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.

Answer 2

The problem is a standard application of the fact that the center of the group is a factor of the centralizer of every element of the group, as the center is the normal subgroup of the centralizers. Thus, the only possible order for centralizer of each element is $125$, as otherwise the group should have been abelian. Hence, the indices of centralizers are all $5$, whence the class equation is $1+1\ldots (25\ \text{times})+1+5+5+\ldots (120\ \text{times})+5$, thus giving us a total of $145$ conjugacy classes.