 This talk will be about the theorem due to Burnside, which states that if we have a group G, which is order the form P to the M times Q to the N for primes P and Q, then G is solvable. We're going to prove this using character theory, which is similar to Burnside's original proof. And for many years, it was an open problem to find a proof of this that didn't use character theory. There was a proof and found a few decades ago, but it's much, much harder than the proof using character theory. So the proof makes use of the center of the group ring. So suppose you call the center of the group ring CG, suppose you call this R. So this is the integral group ring rather than the complex group ring. And this is a basis as a Z module consisting of the elements where you just sum over all elements of the conjugacy class. And it's very easy to check that these span the center of the group ring. Now suppose V is an irreducible representation of G. It's an irreducible complex representation. Then we get a homomorphism from the center of the group ring to the endomorphism ring of V. And in fact, since V is irreducible, this must lie in the center of the endomorphism ring and the center of the endomorphism ring of V is just the ring of complex numbers C. So for every irreducible representation, we get a homomorphism from this ring R to the complex numbers. And we can write it down explicitly as follows. So it takes the element, if we sum over all elements of some conjugacy class, it takes this element chi of C times the size of the conjugacy class divided by the dimension of the irreducible representation V. So V has character chi and this is of course the size of the conjugacy class C. And the reason for this is that if you look at this expression here, this is just the trace of sum over C and C of C because each element of C has traced chi of C and there are this number of them. And we have to divide it by the dimension of the representation because if you've got an element of the center of a representation, then the trace is equal to A times the dimension of the representation. So to find out what A is, you have to divide by the dimension. So you notice from this, we're going to show that this number here is an algebraic integer. So chi of C times C divided by kind of one is an algebraic integer. It's pretty trivial to show that chi of C is an algebraic integer because it's just a sum of eigenvalues of C which are roots of unity. So it's much more interesting to say that you can divide it by the dimension of the representation provided you put in this factor here. And this is quite easy to show because we look at the homomorphism from the center of the group ring to the complex numbers. We notice the image is a subring of C and is finitely generated as a Z module because R is finitely generated as a Z module. So the elements are algebraic integers because if you've got a subring of the complex numbers that's finitely generated as a Z module, its elements are algebraic integers. In particular, this is the image of a certain element of R under this homomorphism. So this is also an algebraic integer. And we can give an application of this. The application is that chi of one, the dimension of the representation divides the order of the group G. And you can see this because by the orthogonality relations for characters, we know that the order of G is just sum over all conjugacy classes of the size of the conjugacy class times chi of the conjugacy class times chi bar of the element of the conjugacy class. So this is just the saying that the norm of the character chi is equal to G. Now what we can do is we can just divide both sides by chi of one. And we notice that this bit here is an algebraic integer because we just said so up there. And this bit here is also an algebraic integer because it's a sum of roots of unity. So this is an algebraic integer and as this side is rational this side must be an integer. So the dimension of any irreducible representation divides the order of the group G. By the way, this only holds for complex representations. For example, if we take G to be the group of order three then it has an irreducible representation on two dimensional real vector space because you can just rotate by third of a revolution. And the dimension of this is two which certainly doesn't divide the order of G, which is three. So this also fails for algebraically closed fields of characteristic greater than zero. The dimension of an irreducible representation need not divide the order of a group. So now we move on to Burnside's theorem. So suppose G is order P to the M, Q to the N. So we want to show that G is solvable. Well, by induction it's enough to show that if G is simple, that if G is simple this implies G is cyclic because a minimal counter-example to this must be simple and if it's cyclic that means it's solvable. Now, if G has order P to the M, Q to the Ns G has a conjugacy class of order, a power of P. At least if sort of order P to the K for some K because we can just take an element of the center of a Silov subgroup of order Q to the N and that will have the centralizer of that will have index of power of P. So the conjugacy class is order of power of P. So it's enough to show that if G is simple is any simple group with a conjugacy class with a non-trivial conjugacy class of order of power of P, order of power of any prime then G is cyclic. So this is what we're going to prove in the rest of the talk. We can now forget about G being a product of two primes because this is the only thing we're going to, this is the only thing we're going to prove. So in order to do this, what we do is we first look at the orthogonality relations. So we know that if we sum overall characters Chi we have Chi of one times Chi of C is equal to zero. So here we're taking C to be a non-trivial conjugacy class of order of power of a prime. Say let's call this prime P. Well, one term in this is if we take the trivial character Chi one of one then Chi one of C is equal to one. So this is equal to one. So we can find some character Chi not equal to the trivial character. So that first of all Chi of one is not divisible by P and secondly Chi of C is not equal to zero because if every character was either dimension divisible by P or Chi of C equals zero then all terms in this other than the first term would be divisible by P. So there's some couldn't be equal to zero. And what we're going to do is just to use these two facts get with the fact that the size of the conjugacy class C is power of prime. And we're just going to deduce from this that if G is simple, then it must be cyclic. And this is kind of rather remarkable because we don't really seem to have enough information to prove anything much about G. I mean, these two facts that Chi of one not divisible by P and Chi of C not being zero both seem to be very weak pieces of information to use. So how do we do this? Well, first of all we recall the number A which is Chi of C times the size of C divided by Chi of one is an algebraic integer. Secondly, we notice that the order of C and Chi of one are co-prime because this is a power of P and this is not divisible by P. So what we want to show is that if we emit this factor C then Chi of C divided by Chi of one is still an algebraic integer. And that's quite easy because since these two are co-prime you know that M times Chi of one plus N times order of C is equal to one for some M, N. And then if we take the number we know that Chi of C times the order of C is equal to A times Chi of one where A is this algebraic integer. And we can now just write Chi of C is equal to Chi of C times M Chi of one plus N times the order of C because that's just one which is equal to Chi of C times M times Chi of one plus N Chi of C times the order of C. And we know this is divisible by Chi of one and this is divisible by Chi of one by what we set up there. So Chi of C divided by Chi of one is an algebraic integer. So that's where we use the fact that the conjugacy classes order of power of P we can take out this factor here. And we also notice that this is not equal to zero because Chi of C is not equal to zero. And now it's fairly easy to finish off because we look at the number Chi of C over Chi of one and Chi of C is a sum of Chi of one roots of unity. So all conjugates of Chi of C over Chi of one are sum of Chi of one roots of unity divided by Chi of one. So they have absolute value less than or equal to one. And now we look at the norm of this algebraic number which is the product of all conjugates. Well, the product of all conjugates of an algebraic integer is an integer and it can't be zero because Chi of C is none zero. So is greater than or equal to one. And if you've got a product of numbers that are all less than or equal to one and the product is greater than or equal to one we see that Chi of C over Chi of one is just actually of absolute value equal to one. And now we'll show that from this fact that if G is simple then it must be set click. So we know that Chi of C is equal to Chi of one and this is a sum of Chi of one roots of unity. Well, the only way the sum of Chi of one roots of unity can be equal to Chi of one and absolute value is these roots of unity must all be the same. So Chi of C is just equal to Chi of one times Zeta for some root of unity. And now if G is simple, this implies that the conjugacy class C is in the center of the image of Chi in the ring of endomorphisms of V. So if the image of Chi in here must be non-trivial which means that C must actually be in the center of G which as Chi is simple implies Chi is set click. So this concludes the proof of Burnside's theorem that any group of order divisible by most two primes must be solvable. This is of course best possible because if you've got a group of order 60 this gives you the group A5 which is not simple and this is divisible by three different primes. In fact, another fairly easy theorem you can prove using the transfer says that if you've got a simple non-sicclet group then the smallest prime dividing it must divide it to at least the second power. And if you combine this with Burnside's theorem it immediately implies that any group of order less than 60 cannot possibly be simple because the minimum possible order is divisible by at least three primes and the smallest prime must be prepared twice. So this is obviously the smallest possible order.