# 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.)

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". 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 $g$.