The other theorem of Burnside

If a poll was held among mathematicians to identify “Burnside’s Theorem”, I guess that the result would be roughly evenly split between the irreducibility criterion for finite-dimensional representations, and the solvability of finite groups of order divisible by at most two distinct primes; far behind would come the so-called “Burnside lemma”, which I’ve heard is due to Frobenius or Cauchy, the partial results concerning Burnside’s problem, and presumably among stragglers and also-ran would be Schur’s Lemma and his work on hydrodynamics and complex function theory.

I’ve already discussed what head or tail I could make of Burnside’s proof of the irreducibility criterion, which is also incorporated in my representation theory notes (I’ve also put in there, in Section 2.7.4, a different proof based on Frobenius reciprocity which is quite cute, and presumably also well-known). I’ve just started discussing in class the $p^aq^b$ theorem. The account in the notes (Section 4.7.2) is a bit incomplete, as I’m struggling in my attempts to decide how to present in a motivated way how one comes to the integrality properties of characters which are the crucial tool (for a target audience with no prior exposure to algebraic integers). I found this easier to do on the board, and to imagine doing with hyperlinks, than within a book-like object!

I had not actually looked at the proof before the last few weeks. As far as I can see, and in striking contrast with the irreducibility criterion, Burnside’s proof is still the standard one (apart from terminological and notational changes; I do not count the purely group-theoretic arguments apparently found later by Thompson and others, which I know nothing about, save their existence). Part of the mystery of the statement is that, in fact, one proves something a bit different, and in fact weaker-looking: if $|G|=p^aq^b$, then either $G$ is abelian, or it contains a proper normal subgroup. Then by induction on the order of a group $|G|$ (divisiblie by at most two primes), one concludes easily that in fact they are solvable.

In turn, the desired normal subgroup is constructed using a more general result, which I had never heard about, but which is certainly of independent interest, and has the virtue of being a general fact about all finite groups which illustrates some of the subtle ways in which irreducible (complex) representations and conjugacy classes try to be “dual”:

If $G$ is a finite group, $g\in G$ is a non-trivial element, $\rho$ a non-trivial irreducible complex representation of $G$. If the dimension of $\rho$ is coprime to the order of the conjugacy class of $g$, then either

• The character value $\chi_{\rho}(g)$ is zero;
• Or, the element $g$ is in the kernel of the composite projective representation
$\bar{\rho}\,:\, G\rightarrow \mathrm{GL}(E)\rightarrow \mathrm{PGL}(E)$
(where $E$ is the space of $\rho$).

The second part is the main point: if, for a given group $G$, one can find a pair $(g,\rho)$ for which the result is applicable, and if the first part of the alternative can be excluded, then it follows that the kernel $N$ of $\bar{\rho}$ is a non-trivial normal subgroup. It could be that $N=G$, but that is a very special case: then the image of $\rho$ is an abelian group, and either $\rho$ is an isomorphism with such an abelian group, or the kernel of $\rho$ itself is a proper normal subgroup.)

Where having $|G|=p^aq^b$ comes in, in applying this, is in the fact that it is not so easy in general to cook up integers dividing $|G|$ which are coprime to each other (these being the size $|g^{\sharp}|$ of the conjugacy class $g^{\sharp}$ of $g$, and the dimension of the irreducible representation).

What can be done without much work is to find, for any $g\not=1$, a representation of dimension not divisible by one prime (say $p\mid |G|$) for which $\chi_{\rho}(g)\not=0$, using the orthogonality relations looked-at modulo $p$; and one can also find an element $g\not=1$ for which the conjugacy class has order coprime with another prime $q$ (looking at the partition of $G$ in conjugacy classes, modulo $q$). But there is no reason that this should ensure that $(\dim(\rho),|g^{\sharp}|)=1$, except if $|G|$ is divisible only by those two primes! In that case, we see that $\dim(\rho)$ is a power of $p$, and the conjugacy class of $g$ has size a power of $q$, and everything works.

The proof of the result on characters above is a very convincing illustration of the usefulness of algebraic integers: one first shows by basic facts about them that there is always a divisibility
$\dim(\rho)\mid \chi_{\rho}(g) |g^{\sharp}|,$
in the ring of algebraic integers. Thus if, as we assumed, the dimension and the size of the conjugacy class are coprime, we get
$\dim(\rho)\mid \chi_{\rho}(g),$
and using the fact that the character value is a sum of $\dim(\rho)$ roots of unity, the conclusion is again not too hard from the properties of algebraic integers. (It is quite obvious when $\chi_{\rho}(g)\in\mathbf{Z}$, of course, since one knows that
$|\chi_{\rho}(g)|\leq \dim\rho$, and one must see that a similar argument applies in general…)

Amusingly, one can use the divisibility relation in the “opposite” direction: if the character value is non-zero and coprime (in the ring of algebraic integers) with the dimension of the representation, then $\dim(\rho)$ divides the size of the conjugacy class. I don’t know if there are applications of this, but this can be seen in practice, e.g., for $G=\mathrm{GL}_2(\mathbf{F}_p)$ when $\rho$ is a Steinberg representation of dimension $p$ and $g$ is a semisimple (split or non-split) conjugacy class, where the character value is a root of unity, hence coprime with $p$, and indeed the conjugacy classes in question have order $p(p+1)$ or $p(p-1)$ (in the split and non-split case, respectively.)