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 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 , then either is abelian, or it contains a proper normal subgroup. Then by induction on the order of a group (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 is a finite group, is a non-trivial element, a non-trivial irreducible complex representation of . If the dimension of is coprime to the order of the conjugacy class of , then either

- The character value is zero;
- Or, the element is in the kernel of the composite projective representation

(where is the space of ).

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

Where having comes in, in applying this, is in the fact that it is not so easy in general to cook up integers dividing which are coprime to each other (these being the size of the conjugacy class of , and the dimension of the irreducible representation).

What *can* be done without much work is to find, for any , a representation of dimension not divisible by one prime (say ) for which , using the orthogonality relations looked-at modulo ; and one can also find an element for which the conjugacy class has order coprime *with another prime* (looking at the partition of in conjugacy classes, modulo ). But there is no reason that this should ensure that , *except* if is divisible only by those two primes! In that case, we see that is a power of , and the conjugacy class of has size a power of , 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

in the ring of algebraic integers. Thus if, as we assumed, the dimension and the size of the conjugacy class are coprime, we get

and using the fact that the character value is a sum of roots of unity, the conclusion is again not too hard from the properties of algebraic integers. (It is quite obvious when , of course, since one knows that

, 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 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 when is a Steinberg representation of dimension and is a semisimple (split or non-split) conjugacy class, where the character value is a root of unity, hence coprime with , and indeed the conjugacy classes in question have order or (in the split and non-split case, respectively.)

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