 This lecture is part of an online course on mathematical group theory and will be about the transfer homomorphism between groups. So as motivation for this, in previous lectures we've classified the groups of order up to 24 and we're now going to look at the groups of order 30. So group of order 30 has ordered 2 times 3 times 5 and this is the first example of an order that's a product of three different primes. And in order to do this we're going to show that if she has ordered 30 then she has a normal subgroup of index 2. And the way we will do this is we will construct a transfer homomorphism from G to it's a subgroup of order 2 of it and this map will turn out to be on 2. So here this is a seal of 2 subgroup of G and this map will be on 2 and so its kernel would be a normal subgroup of order 15. This makes G easier to classify because the only subgroup of order 15 must be cyclic of order 15. So we get an exact sequence like this and this is split because G has an element of order 2. So G is isomorphic to a semi-direct product Z modulo 15Z with Z modulo 2Z and these are just classified by the ways the group of order 2 can act on the group of order 15 and there are four ways you can do this and this gives us four groups for G none of which are terribly interesting. It can be cyclic or it can be a product of a group of order 2 by sorry group of order 3 by a cyclic group of order by the dihedral group of order 10 or it can be a product of a group of order 5 by the dihedral group of order 6 or it can be the dihedral group of order 30. So there's nothing new there are just products of groups we already know about. So what we've got to do is to construct this magical transfer homomorphism. I should say this is transfer and the way it works is as follows. Suppose H is a subgroup of a group G then the transfer homomorphism will be a homomorphism from the abelianization of G to the abelianization of H. So what is the abelianization of G? Well this is the biggest abelian quotient of G. So informally you can think of it as being what happens if you take a group and force it to be abelian. So what we do is we take the set of all elements ab 8 minus 1b to the minus 1 in G and these generate a subgroup called the commutator subgroup of G sometimes denoted like this because ab sometimes used to mean ab 8 minus 1b to the minus 1 so and G modulo the commutator subgroup is abelian because you've kind of forced all the elements to commute by just quotienting out the bits that stop everything commuting. If you have any homomorphism from G to an abelian group then you can easily see that this actually factorizes through the abelian quotient of G. So this is a sort of universal way of making G into an abelian group. And what you notice is we seem to have made a mistake here because this arrow goes in the wrong direction. It seems to be going in the wrong way. So if we've got a map from H to G then it's pretty easy to see we've got a map from the we get an induced map from the abelianization of H to the abelianization of G so it looks as if it made a mistake there but in fact the transfer really does go the wrong way. The way of defining the transfer we're going to give looks a bit mysterious. The most natural way of defining the transfer is to use the fact that the abelianization of G is in fact some sort of Tate homology group of G with coefficients in Z. And there are various Tate co-homology groups for all integers N. And for all of these Tate co-homology groups you get a map from the Tate co-homology group of G to the Tate co-homology of H. However, using this definition requires defining Tate co-homology which takes rather a long time. So what we're going to do is going to forget about Tate co-homology and come up with a sort of elementary ad hoc definition which looks a bit mysterious. This is sort of the definition you get if you unwind the definition of Tate co-homology. So we define it as follows. This is the definition of the transfer from the abelianization of G to the abelianization of H. And what we do is we write G is a union of left cosets of H. So it's a union of elements A i H. And then we're going to look at what, let's pick some G in G and look at the action of G on all these left cosets. So you can, you remember G is really acting on the set of left cosets by left translation. So let's look at the action. Well, suppose these left cosets are A 1 H, A 2 H and so on. Then we have G A 1 is equal to A, well it's going to be A something, let's say A i 1 times H 1. G A 2 will be A i 2 times H 2 and so on. So the action of G on these left cosets is taking A 1 to A i 1 and A 2 to A i 2 and so on. And we have these undetermined factors of H on the left. And then the transfer of G is simply defined to be the product H 1, H 2, H 3 and so on. And this definition looks really fishy. So we better check that it is actually a well-defined definition. First of all, the order of the H i does not matter. You see we've got H 1, H 2, H 3. Well why don't we take H 2, H 1, H 3 and get a different element? Well the order doesn't matter because we're considering this to be an element in the abelianization of H. So whatever order you multiply these in, it gives the same element in here. So as we work in H, abelianization of H. So the order doesn't matter. Well the next problem is A 1 and A 2 and so on aren't uniquely determined. So what happens if we change A 1 to say A 1 H? Because A 1 H is equal to A 1 H H. So the definition of transfer seems to depend on our choice of left coset. Well it doesn't, actually it doesn't. Because we know that G A 1 of H is now going to be A i 1 H 1 times H. Sorry there shouldn't be a one there. And we've also got a G A J for some J is going to be A 1 times H J. And this will be equal to A 1 H times H to the minus 1 H J. So if we replace A 1 by A 1 H, then we replace H 1 by this and we replace H J by this for some J where J is the thing mapping to A 1. So we replace H 1 H J by H 1 H H minus 1 H J. And this is just equal to H 1 H J. So the choice of coset representatives doesn't matter. So T does not depend on the coset representatives A i. So it is actually well defined. Finally we want to check that T is a homomorphism from G 2 H abelianised. And for this we just calculate what it does. So suppose G A 1 is equal to A i 1 H 1 G A 2 is equal to A i 2 H 2 and so on. So that T of G is equal to H 1 H 2 and so on. And suppose we've got another element G prime of A 1 which is going to be a J 1 H 1 prime G prime A 2 is equal to A J 2 H 2 prime and so on. So T of G prime is equal to A H 1 prime H 2 prime and so on. Now we have to work out what G prime G does to things. Well this takes A 1 to A i 1 H 1 which is now going to be A something or other whatever G prime takes A 1 2 times H i 1 prime times H 1. And similarly G prime G of A 2 will be A something or other H H i 2 prime H 2 and so on. So T of G prime G is the product of all these. So it's H i 1 prime H 1 H i 2 prime H 2 and so on. And now you notice that these things here are just a permutation of these things here. So these things there not that one. So T of G prime G is just the product of all the H primes and all the H's. So it's the product of T of G and T of G prime. So this equals T of G, T of G prime. So we verify that this rather peculiar definition of transfer is indeed a well-defined homomorphism from the abelianization of G to the abelianization of H. Incidentally we define it using left cosets and you may wonder what happens if we define it using right cosets. If you define it using right cosets you get the same thing but I don't need that so I'm not going to bother proving it. Well the next question is how do you calculate it? So if G is in G let's try and find out what is T of G and the definition makes it not easy to see what this is. Well so let's try. Let's take G to be in H. So this makes it a little bit easier to calculate and then we calculate G A1 is going to be A2 and what we're going to do is just choose A2 to be equal to G of A1. You remember we could vary the choice of cosets without changing the transfer homomorphism and similarly G A2 then we can just choose A3 and it goes on like this until we get back to A1 but then we can't choose A1 because we we already fixed A1 so there has to be some element H here. Let's call this H1 and then we can carry on like this GAN plus one is equal to AN plus two all the way up to GAN plus K is equal to AN plus one times some element H2. So what we're doing is we're splitting up the action of G on the set of cosets into cycles and for each cycle we just get one element of H and then the transfer of G is going to be H1, H2 and so on. Well how does that help? Well what we do is we notice that G squared A1 equals A3 G cubed A1 equals A4 so G to the N A1 and you go all the way this equal to A1 H1 how does that help? Well this says that H1 is equal to A1 to minus one G to the N A1 and similarly for H2 and so on. So H1 is actually a conjugate of G to the N. Well how does that help? Now suppose H is the following special property. Suppose that in H no two distinct elements are conjugate in G. Well then from H1 equals A1 to minus one G A1 you notice these are both in H and they're conjugate so this gives us H1 equals G and similarly H2 equals G and so on. So sorry G to the N and this would be G to the whatever the next number is so that should have been G to the N. So the transfer of G is just equal to H1 H2 and so on which is equal to G to the N G to the K and so on which is equal to G to the number of cosets which is the index of H in G. So we've actually found a formula for the transfer of an element of H provided H has this slightly funny property here. Well when does H have this property? Well first all if G is abelian then the transfer of an element of G for G and H will just be G to the G the index of G and H so it's not really a terribly exciting map. This also occurs if H is in the center of G and another case that occurs if H has order two because if H has order two then any two elements in H that are conjugate must obviously be the same because they're either the identity or they're the unique element that's not the identity. So in other words if H is order two this means that the transfer from G to H acts on H as G goes to G to the index of G in H. Now if this is even then that's rather uninteresting it just says the transfer is the trivial homomorphism but if G modulo H is odd this means that the transfer map from G to H which is equal to the abelianization is onto. So in other words so if the order of G is of the form two N with N odd then there is a homomorphism given by the transfer from G to H where this is a seal of two subgroup. So this is a homomorphism T onto H. So G is a semi-direct product K with H where this is index two and this is a group Z modulo two Z. So we found the structure of all groups of order two times an odd number. They're all semi-direct products of a group of order two and some group of odd order. So in particular G is not simple so there are no simple groups of whose order is congruent to two modulo four. And of course in particular we can apply this to the group of order 30 that we start with and see that it has a normal subgroup of order 15 as we said earlier. So we can do a bit more than this. So suppose P is the smallest prime dividing the order of G. If P squared does not divide G then the seal of P subgroup is Z over PZ. And you notice that no two distinct elements can be conjugate because if they were conjugate they would have to be an inner automorphism of G acting on this non-trivially and the group of inner automorphisms of this is only divided by divisible by primes less than P but G doesn't have any elements of order less than P. So the transfer map from G to Z over PZ acts as G goes to G to the order of G over P for G sorry little G in Z modulo PZ. And since this is co-prime to P this transfer map is onto. So G has a normal subgroup of index P we call that a normal P complement. So G is not simple. In fact if you look at this argument a little bit more carefully you see that it also applies even if P squared does divide the order of G provided the seal of P subgroup is cyclic. So in particular we found that if so for any simple group its order must be divisible by the square of some prime in particular it must be divisible by the square of some prime not dividing it. So let's give one final application of the transfer so suppose G is simple and suppose that the seal of two subgroup of G is well it can't be Z modulo two so the next simplest possibility is Z modulo two Z times Z modulo two Z. So it's got three different elements of order two and the question is are these conjugate in G or not? Well then all elements of order two in G are conjugate well suppose not then look at the elements in the two seal of subgroup we've got an element one and we've got three elements A, B and C. If these aren't all conjugate then either no two of them are conjugate or or maybe A and B are conjugate we don't really care but there's going to be some element C not conjugate to anything else. So then some element C not equal to one in Z over two Z squared is not conjugate to anything else. Now we look at the transfer map from G to Z over two Z times Z over two Z and the transfer of this element C as before is equal to C to the order of G modulo divided by four which is must be equal to C because this is odd. So we've got a homomorphism from G onto a group of order two or four so G has a normal subgroup of index two so G is not simple. So you see that's that the transfer map has given us a small amount of information about simple groups. If the seal of two subgroup is the Klein four group then all involutions in G are conjugate. Okay the next lecture will be about what a ridiculously large number of groups of prime power order there are.