# Characters of nonabelian group of order 57

I have the following problem:

Let G be the nonabelian group of order 57.

(a.) How many 1-dimensional characters does G have?

(b.) What are the dimensions (aka degrees) of the other irreducible characters of G?

For part (a) I have already found that $$G$$ has 3 irreducible representations of degree 1 by analyzing the Sylow subgroups of $$G$$, realizing that the 19-Sylow is normal and the 3-Sylow's are not. Hence $$G$$ has three 1-dimensional characters.

However for part (b) I am not so sure what to do. I know two things that might be useful:

1. The number of irreducible representations of $$G$$ must equal the number of conjugacy classes of $$G$$.
2. The sum of the squares of the degrees of the irreducible representations of $$G$$ must equal the order of $$G$$.

So in particular the sum of the squares of the degrees of the remaining representations must equal $$57-3 =54$$ and none of those may be degree one. However this alone is not enough to give the answer as there are multiple ways of expressing 54 as a sum of squares, for example $$54= 36+9+9 = 25+16+9+4$$.

Any help would be appreciated!

• It is a well-known theorem that the dimension of an irreducible representation of a finite group must divide the order of the group. Does this help? – Sameer Kailasa Mar 24 at 19:40
• Oh shoot that's huge. So all the remaining characters have to be degree 3, which works since 54= (9)(6), right? – Edgar Jaramillo Rodriguez Mar 24 at 19:52

A theorem of Ito asserts that if $$A$$ is an abelian normal subgroup of $$G$$, then the degree of an irreducible character divides $$|G:A|$$. Since the Sylow $$19$$-subgroup is normal, it follows that the non-linear irreducible characters all have degree $$3$$. Hence there are $$6$$ of them, since $$57$$ equals the sum of squares of all degrees.
If you do not want to use Ito, $$\chi(1)^2 \leq |G|$$ and $$\chi(1)$$ divides $$|G|$$, for all $$\chi \in Irr(G)$$.