# Number of prime divisors of element orders from character table.

From wikipedia:

It follows, using some results of Richard Brauer from modular representation theory, that the prime divisors of the orders of the elements of each conjugacy class of a finite group can be deduced from its character table (an observation of Graham Higman).

Precisely how is this done? I would be happy with a reference. (In particular, I am looking for the location of the elements in a solvable group with which have the maximum number of prime divisors in their order. If somebody has any extra information on that particular situation that would also be appreciated.)

## 3 Answers

I think I probably wrote the quoted passage in Wikipedia. If we let $\pi$ be a prime ideal of $\mathbb{Z}[\omega]$ containing $p,$ where $\omega$ is a primitive complex $|G|$-th root of unity, then it is the case that two elements $x$ and $y$ of $G$ have conjugate $p^{\prime}$-part if and only if we have $\chi(x) \equiv \chi(y)$ (mod $\pi$) for each irreducible character $\chi$ of $G$. This is because the $\mathbb{Z}$-module spanned by the restrictions of irreducible characters to $p$-regular elements is the $\mathbb{Z}$-span of irreducible Brauer characters. The rows of the Brauer character table remain linearly independent (mod $\pi$), as Brauer showed. Furthermore, the value (mod $\pi$) of $\chi(x)$ only depends on the $p^{\prime}$-part of $x,$ since $\eta- 1 \in \pi$ whenever $\eta$ is a $p$-power root of unity. We now work inductively: for any prime $p$, we can recognise elements $g \in G$ which have $p$-power order, since $g$ has $p$-power order if and only if $\chi(g) \equiv \chi(1)$ (mod $\pi$) for all irreducible characters $\chi.$ If we choose a different prime $q,$ we can recognise $q$-elements by the same procedure. But then we can recognise the elements whose orders have the form $p^{a}q^{b}.$ Such an element $h$ must have $p^{\prime}$-part $z$ which has order $q^{b},$ so we have previously identified the possible $p^{\prime}$-parts. Furthermore, $h$ has $p^{\prime}$-part $z$ if and only if $\chi(h) \equiv \chi(z)$ (mod $\pi$) for all irreducible characters $\chi.$ And then, given a different prime $r,$ we can identify all elements whose orders have the form $p^{a}q^{b}r^{c}$ in a similar manner, etc.

It is Theorem 8.21 in Isaacs "Character Theory of Finite Groups".

As seen on Google Books: • Perhaps a few more words about what Thm. 8.21 in this reference states are helpful. The Question is fairly specific, and while the result there (and proof?) may be spot on, it would surely encourage the Reader contemplating a trip to the library stacks to know more of what is in store. – hardmath Sep 19 '14 at 21:33

A formula for the order of each conjugacy class of a finite group is given in Exercise 2.21 of Fulton and Harris Representation Theory: A first Course; namely $c(g) = {|G| \over \sum_{\chi} \overline{\chi(g)}\chi(g)}$ where the sum is over the irreducible representations, and $c(g)$ is the number of elements conjugate to $g$.

This isn't exactly what you asked for, but in any particular case you could calculate the prime divisors of the orders of the elements of each conjugacy class from this by factorisation.

• That formula gives the size of each conjugacy class, not the order of the elements in the conjugacy class. – Geoff Robinson Oct 12 '12 at 8:45