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!

  • 1
    $\begingroup$ 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? $\endgroup$ Mar 24, 2019 at 19:40
  • $\begingroup$ Oh shoot that's huge. So all the remaining characters have to be degree 3, which works since 54= (9)(6), right? $\endgroup$ Mar 24, 2019 at 19:52

1 Answer 1


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 $\chi \in Irr(G)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.