 this representation theory taught will be about Frobenius groups. So what I'm going to do is first recall the definition of a Frobenius group and then prove a basic theorem of Frobenius about these groups, about the existence of the so-called Frobenius kernel. So Frobenius group G is just a transitive permutation group on some finite set so that any element fixing two points is the identity. We also add an extra condition that the subgroup fixing a point is not the trivial subgroup because otherwise any group would be a Frobenius group just acting on itself by left translation. So here are some basic examples. We could just take G to be the symmetric group S3 acting on three points and you see if any element fixes two of these points it obviously fixes the third so that it's the identity. Another one might be G, might be the dihedral group on four N plus two points so here we might have the group of automorphisms of a pentagon and you see if you fix any two vertices of the pentagon then the entire pentagon is fixed so that the dihedral group on four N plus two points is a Frobenius group. If you look at the dihedral group on four N points such as say the symmetries of an octagon this isn't a Frobenius group because if you take two opposite points there's a non-trivial automorphism fixing those so that's not a Frobenius group. Another example might be the so-called AX plus B group. Here we take A to be in some finite field say Z modulo PZ and A to be a non-zero element of this field and B would be an element of this finite field and the group of all automorphisms of the affine line taking X to AX plus B as then you can check it's a Frobenius group. So Frobenius groups are fairly common. If H is the subgroup fixing a point then Frobenius groups have the following property that G H G to the minus one intersection H is just the identity if G is not in the subgroup H. That's because anything in these two subgroups actually fixes two points because it fixes the point fixed by H and it also fixes the conjugate of that point by G. So this is more or less another way of defining a Frobenius group. So from this we can get a sort of picture of what a Frobenius group looks like. And all the elements of G sitting inside some set like this and then we might have a subgroup H and inside H is the identity element and H is all these conjugates. So this might be G H G to minus one and this might be some other conjugate. So G prime H G prime to minus one and all these conjugates intersect only in the identity element. So here we have elements fixing one point. The identity element of course fixes all points and all the remaining elements that are not in any of the conjugates of H and not the identity must fix no points. So this is a sort of rough picture of what a typical Frobenius group looks like. For example if we look at S3 we get exactly this structure. So we have a subgroup H consisting of say the identity and the element 1, 2 and we have three conjugates of H containing the other three transpositions and then we have two further elements of G 1, 2, 3 and 1, 3, 2 which fix no points at all. And now we come to Frobenius's theorem which says that the set of points fixing zero points together with the identity is a subgroup. That's kind of a bit surprising because there seems to be no obvious reason why if you multiply two elements fixing no points that should either be the identity or something else fixing no points. So in other words you can think of this subgroup as being everything not in one of these subgroups H together with the identity element and this will be a subgroup K called the Frobenius kernel and it's often denoted by K so it's called the subgroup K, the Frobenius kernel. Notice that if it's a subgroup it's automatically a normal subgroup because it's kind of obvious that the union of all conjugates of H is closed under conjugation so the complement is also closed under conjugation. So this is an example of a result that's fairly easy to prove using representation theory and very difficult to prove without using representation theory. So the idea of the proof is to use induced representations to turn representations of H into representations of G and we will construct enough representations of G to show that the set K is actually the kernel of all these representations we construct and is therefore a subgroup in fact a normal subgroup. So what we're going to do is recall that if we have a subgroup H of a group G then if we have a representation of H we have an induced representation of G so we just write in for the induced representation sometimes people write in from H to G but I'm feeling lazy so I'll just call it the induced representation. Conversely if we've got a representation W of G we can restrict it to H so we have a restricted representation the restriction of W. And these two operations as I mentioned in the last lecture are sort of a joint. So what does this mean? Well if we translate this into the language of characters it means that if we've got an induced character chi and we take the inner product with the character psi then this is equal to the inner product of chi with the restriction of psi. So here psi and the induced character of chi are representations of G and these two are representations of H and this inner product is just the usual inner product of characters. Incidentally you can see that here the operations end and restriction on characters look just like a joint linear operations on a vector space. So this is one of the reasons why the term a joint functor is used that a joint functors really are rather closely related to a joint linear maps. So let's start by taking chi to be the character of an irreducible representation of H and we're going to take this irreducible representation not equal to the trivial representation 1. And we're going to look at the induced character of chi. Now if we draw a picture of G to see what's going on you remember we have this subgroup H and the chi is some sort of function on the subgroup H. And in order to form the induced character you remember we just sort of take all conjugates of H and conjugate chi onto these. So we also have chi on all the various conjugates of H. So the induced character of G more or less looks like taking the values of chi and just sort of spreading them out over all conjugates of H. The slight exception is the identity element where we have to do something rather more complicated because the identity element occurs in all the conjugates of H. Well it's a bit annoying having to deal with the identity element so instead of inducing chi let's induce chi minus D times the identity character of H where D is the dimension of chi. Now this is a bit better because this character is not on the identity element of G so here I'm using 1 for the trivial character and for the identity element of a group just to confuse everybody. So if we induce this it looks like this. What we do is we take all the conjugates of H and on most of the conjugates of H the character looks like chi minus D or the conjugate of chi minus D and it's zero here and it's also zero on the identity element. So we have very good control over this induced character here and we can translate this into working at things like the inner product of chi minus 1 dot, sorry D dot 1 H so we want to work at the inner product of this with itself and that's easy because it's just the restriction of this induced character chi minus D dot 1 and we take the inner product of that with chi minus D dot 1 H and here we're taking the inner product of H and here we're taking the inner product of G and we can easily work at what this is because this is just equal to chi minus D dot 1 on H that's because all the conjugates of H are destroyed from H except that they overlap at the identity element. So this is just equal to chi minus D dot 1 chi minus D dot 1 worked out on H and this is equal to 1 plus D squared because chi is not equal to the identity character. On the other hand we can also work out end of chi minus D dot 1 H we can also work at the inner product of this with the identity character of G and again we can use the adjunction formula and find this is chi minus D dot 1 H and then we have to restrict this to H which is just the identity character of H so this is just minus D. So if we look at this character here it contains minus D copies of the identity character and its norm is D squared plus 1 so the D squared is accounted for by the D copies of the identity character so the leftover 1 means that there must be exactly one other irreducible character in this in other words the induced character of chi minus D dot 1 H is equal to D times 1 G plus some character psi where this is irreducible so we've got a funny way of going from irreducible characters of H to irreducible characters of D with G we're not quite inducing it we're first constructing we're first subtracting several copies of the identity representation then we're inducing that then we're adding back some copies of the identity representation this time of G and if we do that we end up with an irreducible character so that should be a minus there of course and we can also work out what the character psi is psi looks like this well on the group G let's look at the subgroup H so if we take the character of psi it will be the same as chi on the subgroup H or rather the same as conjugates of chi and it will even be the same as chi on the identity element because we did this trick of subtracting the trivial character and then adding back a trivial character meanwhile on the rest of G psi is just equal to D so what you notice is that first of all the kernel of psi contains K because any element such that the character value is the same as the identity element must be in the kernel of the character and secondly we notice that psi is equal to chi on H now this means that K is equal to the intersection of the kernels of all the characters psi that's because the characters psi run through all characters of H other than the identity character so their common kernel is just the doesn't contain anything in H but it does contain absolutely anything in K so this implies K is a normal subgroup so as an example we can just see what happens for the symmetric group S3 so let's just take G to be S3 and H to be 1, 1, 2 and see how this works out so here's a picture of S3 we've got 1, 2, 3, 1, 3, 2 as the two elements that are going to be K and here we've got the subgroup H which contains 1 and 1, 2 and we've got its conjugate so this contains 2, 3 and this contains 3, 1 so these are the three conjugates of H and now we're going to take a character chi of H which is non-trivial well, there's only one non-trivial character so it's going to be 1 on the identity and minus 1 here so its values look like this it's going to be 1 here and minus 1 here now if we induce it as we did last lecture its value is 3 on here and minus 1 on these three points but that's not what we want to do we first take chi minus D dot the 1 times the trivial character of H well D is just equal to 1 of course so this has values 0 minus 2 so this character now looks like this so here are the three conjugates of H and let's write this character in pink so this is minus 2 here and it's 0 here and now let's induce it up to the whole of G well if we induce it up to G we just get 0 there and we get minus 2 here and minus 2 there and we just get 0 at the identity element again so this is what happens if you induce chi minus D times 1 H and now we're going to add back the identity character of G and if we do that what we get looks like this so here are the three conjugates of H and now we're going to take this character where we take the induced character of chi minus D dot 1 H and we're going to add D times the identity character of G sorry that should be the identity character trivial character of H and if we do that all we're doing is we're adding 1 to everything there so we get 1 minus 1 minus 1 minus 1 1 and if we recall what the character table of S3 looks like you remember it looks like this 1 1 1 1 minus 1 1 2 0 minus 1 so our character is now this character here and you see in this particular case this character is a representation whose kernel is exactly the subgroup K consisting of the identity and the elements that don't fix anything so that proves Frobenius's theorem that the elements that don't fix anything together with the identity in a Frobenius group form a normal subgroup there's quite a lot more known about this normal subgroup there's a rather difficult theorem by John Thompson I think it was actually his PhD thesis which says that the Frobenius kernel of a Frobenius group is always a nilpotent group and there's quite a lot known about the subgroup H fixing a point as well it's more or less described all such groups okay that's all about Frobenius groups