MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top
up vote 12 down vote accepted

Georg Frobenius, Über die Primfactoren der Gruppendeterminante, Sitzungsber. Akad. Berlin (1896) 1343-1382. The theorem is announced at the beginning, p. 1344:

Der Grad $f$ ist ein Divisor der Ordnung $h$

and proved at the very end, p. 1382:

Daher ist die Zahl $e=f$ ein Divisor der Ordnung $h$.

P.S.: As to an agreed-upon name for this theorem, I think there isn't one (or googling would reveal it). Too much competition from other results of Frobenius! So, other than in the book by Etingof et al. (who innovate by calling it "Frobenius divisibility theorem"), one finds mainly turns of phrase like "Frobenius's theorem that the degrees of the irreducible characters divide the order of the group", as in Curtis (who devotes two pages to this theorem and Burnside's second proof of it).

An interesting byproduct of this archaeology is some light shed on the words "degree" and "irreducible" in representation theory — namely, they initially qualified factors of a polynomial, the group determinant.

share|cite|improve this answer
Thank you very much for your detailed answer. I think I like the name "Frobenius divisibility theorem". About your last comment. Do you mean the determinant described in \S 4.2 of Is there a nice way to think of the factors $P_i(x)$? Thanks again – Rami Jul 25 '13 at 8:19
Yes, that's the determinant in question. I believe loc. cit. gives you the nicest way of thinking about its factors, viz.: $P_i(x) = \det(\rho_i(x)) = \det(\sum_{g\in G} x_g\rho_i(g))$ where $\rho_i$ is the $i$-th irreducible representation of $G$. For more detail see KConrad's lovely paper – Francois Ziegler Jul 25 '13 at 13:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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