# 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.

• I noticed that you have asked this question twice.... why? Feb 27, 2021 at 11:33
• It is here: math.stackexchange.com/questions/4041623/…. Of course my answer provides you with the "character theory". Feb 27, 2021 at 11:57

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.

• I hope $G'$ is the commutator of $G$, right? and by the way, since we already know the number of conjugacy classes beforehand, we can say the dimensions of non linear characters. My question is, is there a way to tell directly from pure character theory the type of irreducible characters? Feb 27, 2021 at 18:00
• Yes $G’$ is the commutator subgroup. What do you mean by “type”? Feb 27, 2021 at 19:36
• By type, I mean the number of representations/characters of a given dimension Mar 2, 2021 at 23:29
• Ok and this exactly what I have calculated in the last three paragraphs. So, if you deem my answer the right one please tick it as such. Thanks! Mar 3, 2021 at 6:44