# Reading Burnside (and thanking Noether)

In 1905, the famous rower W. Burnside (then aged 52) proved one of the results known as Burnside’s Theorem (the other one being, usually, the striking result that finite groups of order divisible by at most two primes are solvable):

Let k be an algebraically closed field, and let
$G\subset GL_n(k)$
be a subgroup of the invertible matrices of size n over k. Let k[G] be the span of G in the matrix algebra M(n,k) of size n. Then G acts irreducibly on kn if and only if k[G]=M(n,k).

Here, recall that irreducibility (a notion apparently first introduced by Burnside himself) means that there is no proper non-zero subspace

$W\subset k^n,\quad 0\not=W\not=k^n,$

such that G leaves W invariant (globally).

This result turns out to play a role in a current research project (with O. Dinai), and since I had never looked properly at the proof(s) before, I’ve been a bit curious about it, and tried recently to understand it. There are very simple proofs known, but the shortest ones seem to be typically not very enlightening when it comes to understand why the result is true. They’re the kind of arguments you might feel you could find once you knew the result, but why would you think of proving it first? So — it was vacation time! — I had a look at Burnside’s original paper. This can be found here; if you do not have access to the Proceedings of the L.M.S, here is a fairly representative extract of the style:

As far as I’m concerned, this is barely recognizable as meaningful mathematics, and almost unreadable. I say almost, because (vacation effect) I took it as an intellectual challenge to try to reformulate Burnside’s argument in more modern terms, and I believe that I succeeded. It was a big help that the paper is only four pages long; it turns out that the one page from which the extract is taken, although I can’t explain it in any reasonable way, contains the last step of Burnside’s argument. From the fact that he needed seventeen lines to prove the “obvious” half of his theorem, there was therefore every chance that whatever is done here in one page should not be too difficult to figure out with some thought.

So here is a sketch of my reading of his proof of the non-trivial direction (that, for an irreducible action, we have k[G]=M(n,k); for full details, see this short note). We denote

$V=k^n,\quad E=M(n,k),\quad E^\prime=M(n,k)^\prime=Hom(E,k),\quad\text{the dual of } E,$

and then we define

$R=\{\phi\in E^\prime\,\mid\, \phi(g)=0\text{ for all } g\in G\}\subset E^\prime,$

which is the linear space of all linear relations satisfied by the matrices in the group. By linearity and duality, we see that the goal is to show that, if G is irreducible, then the space R is zero. The steps for this are:

• R is a subrepresentation of E’ for a natural action of G on E’ given by
$\langle g\cdot \phi,A\rangle=\langle \phi,g^{-1}A\rangle,\quad\quad g\in G,\ \phi\in E^{\prime},\ A\in E$
in duality-bracket notation (this is not the same as the usual tensor product of the tautological representation of G on V and its contragredient);
• We now attempt to analyze this representation on E’, and the next three steps do this (they are completely independent of the problem at hand); first, the representation on E’ is isomorphic to a sum of n copies of the (irreducible) contragredient of the tautological representation;
• Hence any irreducible subrepresentation in R is obtained as the image of a G-equivariant embedding
$V^\prime\rightarrow E^\prime;$
• But all such maps
$\alpha\,:\, V^\prime\rightarrow E^\prime$
are of the form
$\langle \alpha(\lambda),A\rangle=\lambda(Av)$
for some fixed vector v in V;
• And we now come back to the problem of understanding the relations R; if R were non-zero, it would contain the image of a map α of this form, for some non-zero vector v;
• But if we then specialize the definition of α with A being the identity matrix — which is in G –, we find, from the fact that the image of α is in R, that
$0=\alpha(\lambda)(1)=\lambda(v),$
for all λ, and this is a contradiction with the condition that v be non-zero… Hence we must have R=0.
• It is not obvious here where it is necessary to use the fact that k is algebraically closed, but this is hidden in an application of Schur’s Lemma. Interestingly, it seems that Schur published this result also in 1905. Since Burnside also uses it without any comment (or hint of proof), it must have been known (at least to him) before. It is also amusing to note that, in fact, there is no mention whatsoever of a base field in Burnside’s paper.

I like this proof, in part because it would make sense to try to proceed in this way, even if the result turned out to be different (say, a characterization of the relation module R instead of a proof that it is zero). Also, I may be influenced by the similarity with the study of relations between roots of polynomials that can also be done using elementary representation theory of the Galois group, as discussed in this old blog post.

But as I said, there is a part of Burnside’s paper I really don’t understand, even if I suspect it is equivalent or very similar to what I did. And I am forever thankful that Emmy Noether came along some years later to put algebra on a more reasonable track than endless talk of “successive sets of symbols with the same second suffix” (which sounds almost like one of those alliterative exercises used to detect drunkenness…)