Sum of squares of the dimension of irreducible characters and conjugacy classes While working through a prior problem on classifying a group of order $625$, I stumbled upon (by virtue of a mistaken answer by a user), the following problem. Since the dimension of irreducible representations divide the order of a group and have their sum of squares equal to the order of the group, how can we determine the number of conjugacy classes of the $625$ order group with center of order $25$?
Specifically, the center of the group would correspond to $25$ trivial representations of dimension $1$. Now, how to show that there are $120$ other irreducible representations using only representation/ character properties, as we must have $625=1^2A+5^2B+25^2C+125^2D+625^2E$, ($A,B,C,D,E$ are number of irreducible representations of degrees $1, 5, 25, 125, 625$ respectively) which does not give us anything special? Any hints? Thanks beforehand.
 A: It is not the center, but the index $|G:G'|$ that corresponds to the number of linear characters. 
If $|G|=625$ and $Z(G)=25$, then for every $g \in G-Z(G)$ we must have $|Cl_G(g)|=5$. How can you see this? Well, if $g$ is non-central, then $Z(G) \subset C_G(g) \subset G$, where the inclusions are all strict. So the only way this can happen is when $|Cl_G(g)|=|G:C_G(g)|=|C_G(g):Z(G)|=5$. Hence the class equation amounts to $625=25+5a$, whence $a=120$ and $k(G)=25+120=145$ is the number of conjugacy classes.
Note that $|G/Z(G)|=25$, so $G/Z(G)$ is abelian, hence $G' \subseteq Z(G)$. $G'$ cannot be trivial ($G$ is not abelian), so $|G:G'|=25$ or $=125$.
Now, if $|G|=625=5^4$, $\chi \in Irr(G)$, non-linear, then $\chi(1) \mid 625$ and $1 \lt \chi(1)^2 \lt 625$, hence $1 \lt \chi(1) \lt 25$. So $\chi(1)=5$.
If $G'=Z(G)$, then $|G:G'|=25$ and we would have $625=25+25a$, so $a=24$ and thus $k(G)=24+25=49$. This rules out $|G:G'|=25$.
So $|G:G'|=125$ and a similar calculation shows that the number of non-linear irreducible characters is $20$ and indeed $k(G)=20+125=145$ as calculated above.
