 So this lecture is part of an online mathematics course on group theory, and today we're just going to be discussing a few examples of groups. So in previous lectures, we more or less looked at all groups of order up to about 31, and we're going to stop looking at all groups of every order because nothing new happens to most orders, and instead we're going to skip ahead to the orders where something more interesting goes on. So what we're going to look at first is G of order 48, and we're not going to attempt to classify all groups of order 48 because this is worse than the groups of order 16, which were already rather a mess. And what we're going to do is to look at two of the most, two of the examples that turn up most often. First, we're just going to check that G is not simple. In other words, it must have a non-trivial normal subgroup. And to prove this, let's look at the seal of two subgroups of G. So the number is 1 mod 2, and it must divide 48, so the only possibilities of the number of them must be 1 or 3. And if the number is 1, then the seal of two subgroup is normal and has indexed 3, so we're done. If the number is 3, then we get a transitive action of G on the set of three seal of two subgroups. Well, if we've got a transitive action of G on a set of three elements, this gives us a non-trivial homomorphism from G to the group S3 of all permutations of three elements. And S3 is order less than G, so the kernel is a normal subgroup not equal to 1 or G. In fact, if you look a bit more carefully, you see that G must have a normal subgroup of index either 2 or 3. So there aren't any simple groups of order 48. So what groups are there? Well, there are two obvious groups of order 48. First of all, we have GL2 of F3, where F3 is the field of three elements. So this is just 2 by 2 matrices of non-zero determinant. And how many are there? Well, that's quite easy because these are just automorphisms of the two-dimensional vector space over the field with three elements. And the number of automorphisms is 3 squared minus 1 because if we fix the basis, we can take the first element of the basis to any non-zero vector. And then we can take the second element of the basis to any vector that's not a multiple of the first basis element. So we get that it has order 48. Another one is the binary octahedral group. So you remember the octahedral group has ordered 24 and is isomorphic to S4. And the binary octahedral group is obtained by looking at the map from S3. This is the unit quaternions onto SO3R. This is rotations of three-dimensional space. And inside this group is the octahedral group. And we can take the inverse image of the octahedral group in S3. And the kernel of this map has ordered 2. So we end up with a group of order 48. Now, both of these groups are quite similar. For instance, GL2 of F3 acts on a set of four points, which I mean to take to be 0, 1, 2, and infinity. So this is the projective line over the field with three elements. And the action is given as follows. So if tau is one of these two points, one of these four points, and we've got an element a, b, c, d, an a, b, c, d of tau is going to be a tau plus b over c tau plus d. And you can check for any field, this gives a group action of the projective line over the field where the line is the elements of the field together with infinity and arithmetic operations on infinity are defined in the obvious way. So we've got an action of GL2, F3 on this set. And in fact, its image consists of all permutations. So this gives us a homomorphism from GL2 of F3 onto S4. And the kernel is just the diagonal matrices that are both one or both minus one. So the binary octahedral group maps onto the octahedral group, which is S4, and is a kernel in the center of order 2. And the group GL2 of F3 has exactly the same property. So you can ask, are these groups the same? And in fact, they're really amazingly easy to confuse, but they are in fact different because this has only one element of order 2, because the group S3 has only one element of order 2. But this has several elements of order 2. For example, we have the element 1 minus 1 or the element minus 1, 1 and so on. So in spite of the fact these groups look very, very similar, they are in fact subtly different. Next, we're going to move on and look at groups of order 60. So let's list some groups of order 60. So first of all, there's rotations of an icosahedron, let's draw an icosahedron. So there's an icosahedron and we saw earlier that this had 60 elements. Secondly, we've got the group SL2 of F4. This is the field with four elements, and it has order 4 squared minus 1 times 4 squared minus 4. And we then have to divide by 4 minus 1 because we're taking the set of elements of determinant 1. And this is just 60. Thirdly, we could take the projective special linear group over the field with five elements. So projective means we quotient out by the centre consisting of two elements, the element minus 1 minus 1. In case you're wondering why we didn't quotient out by the centre in this case, the answer is minus 1 minus 1 equals 1 1 because we're working over a field with a characteristic 2. So quotient out by the centre has no effect. And this is order 5 squared minus 1, 5 squared minus 5, then we divide by 5 minus 1 and then we divide by an extra factor of 2 because we're quotient out by the centre. Fourthly, we can look at the group A5, which is the subgroup of even permutations of S5. And this is 120 over 2 equals 60 elements. So all these four groups of 60 elements and they're all isomorphic. So let's see why they're isomorphic. So first of all, if you can map any of these groups to a group of permutations of five elements, that gives you a map to the symmetric group S5 and it's then really easy to check that the group will be isomorphic to A5. So for some of these groups, that's quite easy to do. For instance, SL2 of F4 acts on F4 union infinity, the project of line over the field with four elements. And this is five points. So this gives us a homomorphism SL2F4 to S5 and you can easily check the image is just A5. If we've got the group of rotations of an icosahedron, then the group of rotations of an icosahedron acts on five objects and you can see this as follows. So this isn't actually an icosahedron, but it's a dodecahedron. So they have the same rotation groups essentially because they are dual. So for example, you can see a cube and an octahedron have the same rotation groups because if I take a cube and take a point in the center of each face, I get the vertices of an octahedron. So they have the same rotation group. And I can do the same with an octahedron. If I take a point in the center of each face, I get the vertices of a cube. And similarly, an icosahedron and a dodecahedron have the same rotation group. So I might as well use that. Now this dodecahedron contains a cube. You can see the cube. If I look at these four vertices and these four vertices, you can see there the vertices of a cube, which unfortunately, these structure the same color as the dodecahedron vertex. So perhaps it's not so obvious. And you can see there are five ways to embed a cube in a dodecahedron. So each rotation of a dodecahedron gives you permutation of these five cubes. Another way to see that is to take an icosahedron and put some tetrahedra around it. So if you look very carefully, you can see there are five tetrahedra. And you can either think of all these tetrahedras being contained in a dodecahedron, or you can take that intersection if you've got a really good spatial imagination. You can see the intersections in icosahedron. So again, rotations of an icosahedron will give you a permutation of these five tetrahedra. In any case, we get a map from rotations of an icosahedron to S5. And again, you can check the image is A5. For the remaining one, PSL2F5, I don't know a really easy way to see that this is isomorphic to A5. Most of the proofs involve a fair amount of computation, which I'm just feeling too lazy to do. The main property of these groups is they are all simple. So let's show that the group of rotations of an icosahedron is simple. So for that, we're first going to look at the conjugacy classes. So let's find all conjugacy classes of the tetrahedron. Well, first of all, it's got rotation by one-fifth of a revolution about one of these axes. And we can also have rotation by two-fifths of a revolution. And you may think we should have rotations by three-fifths of a revolution because we don't, because if you think about it, rotation by three-fifths of a revolution is the same as rotation by two-fifths of a revolution if you look on the other side of the icosahedron. So how many of these are there? Well, there are 12 vertices. And for each of these, we can put the vertex at the front and rotate by a fifth of a revolution. So we get 12 of these and 12 of these. Then if we fix a face, we can rotate by a third of a revolution. And again, rotation by two-thirds of a revolution is the same if we look at it in the other direction. So there are 20 of these. Then if we fix an edge, we can rotate by half a revolution about that edge. Now, there are 30 edges. You might think we get 30 of these. But if you notice, opposite edges are giving you the same rotation by half a revolution. So we get 15 of these. And finally, we have the identity. And that gives one. And are there any other rotations? Well, if we add these up, we get 60. So we found all rotations of the icosahedron. And these are all the conjugacy classes. Now, let's look at a normal subgroup. First of all, the order divides 60. Secondly, the order is the sum of one plus sum of these numbers 12, 12, 15, 20. Because any normal subgroup is closed under conjugacy classes. So it must be a union of conjugacy classes. And it's now very easy to check the only way to take one plus a subset of these numbers that divide 60. The only solutions are just one or 60, which is the trivial subgroup and the whole subgroup. So there are no normal subgroups of A5 other than the obvious ones. So A5, the group of rotations of the icosahedron is a simple group. You notice this is the first simple group. We've come across that isn't cyclic. There's a traditional exercise show that there is that any simple group of order less than 60 is simple. And how do you do this? Well, the idea of this exercise is you just go through every possible order less than 60 and check it's simple. We've actually done most of these already. I mean, we've done prime order, we've done all drop to 30 and so on. There are only a few cases left that take more than a few seconds thought. So the orders 48, 56, 36, I guess, all take perhaps slightly more thought than average, but none of them are particularly difficult. So we can also show that any group of order 60, that is simple, is isomorphic to A5 or the group of rotations of the icosahedron. And I'll sketch the proof of this without giving all details. First of all, we look at at the seal of five subgroups. Well, the number of them is one mod five and must divide 60 and there can't be one because that would be normal. So the number is six. And each of these contains four elements of order five and they must all be disjoint. So this gives us 24 elements of order five. We do the same for seal of three subgroups. So the number must be one, four or 10 because it divides A5. And it can't be one because that subgroup would be normal. And it can't be four because that would give us a homomorphism to the permutations of four elements. And the kernel of that would be a normal subgroup. So there must be 10 seal of two subgroups. And as before, this gives us 20 elements of order three. Now, if you look at elements of order two, there must be one, three, five or 15. If there were one, three or five, then we would have a homomorphism to a symmetric group on three or five points. And that would either give us a normal subgroup or show we were isomorphic to A5. So the number of elements of order two must be 15. And if we adopt, we found 59 elements. There's one element of order one. And we've sort of found the orders of all elements. There must be 24 of order five, 20 of order three, and 15 of order two. Now, if you remember, when we were talking about the transfer map, the seal of two subgroup can't be cyclic. So the seal of two subgroup is Z modulo two Z times Z modulo two Z. And we also pointed out when we were discussing the transfer that in this case, all elements of order two must be conjugate. Here, this uses the transfer map we were discussing earlier. So the centralizer of an element of order two must have order as order 60 over 15, which is equal to four. So it must be the seal of two subgroup. This means that any seal of two subgroups must be disjoint. Because if they had an element in common, then that element would commute with everything in both seal of two subgroups and would have centralized of order bigger than four. Now, each seal of two subgroup has three elements of order two. And there are 15 elements of order two altogether. So there must be exactly five seal of two subgroups. I just noticed a slight error I made on the previous thing. Any two seal of subgroups are disjoint except for the identity element. Of course, they have the identity element in common. So there are five seal of two subgroups. So we get a homomorphism to the group of permutations of these five seal of subgroups. So we get a homomorphism of G to S5. And from that, it's very easy to reduce G must be the alternating subgroup of S5. Next, we will show that an is simple for n greater than or equal to five. So this gives a series of simple groups. And we want to show the idea is any normal subgroup n not equal to the identity contains an element fixing a point. And to see this, we pick some element G in the normal subgroup. And we're going to look at G, H, G to minus one, H to minus one, where H is a three cycle. So three cycles are all in a n. And we want to arrange that this is non-trivial and fixes a point. So this is G multiplied by the conjugate of G inverse. And now we just have to, this is one of the points where you sort of have to get your hands a little bit dirty and do some calculations. Suppose G contains a cycle one, two, three, something or other with at least three points in, then we can conjugate by a three cycle to turn this three into something else. So we can take H, G, H to minus one to be one, two, something else. Now to do that, we need a three cycle that doesn't mess up these two elements here. So we must have these numbers one, two, and at least three other points that we're acting on. So this only works if n is greater than or equal to five. So if we take H, G, H to the minus one to be one, two, something with something not equal to three, then the quotient of these two elements will fix one and it won't be the identity. So we're done. On the other hand, if G doesn't contain a cycle of length at least three, then G might be one, two. And then if it fixes an element we're done, if it has another cycle of length bigger than two, we're done. So we may as well assume it looks like this. And then it's quite easy to find a three cycle H with that property. For example, we might take the three cycle to mess up these three elements somehow. And again, we find an element in this normal subgroup that fixes a point. So if n contains some G fixing a point, now the subgroup fixing a point is a n minus one. And this is simple if n is greater than or equal to six because we've already proved that a five is simple and we're using induction. So the intersection of n with this group here must be a normal subgroup of this group and this group, the only normal subgroup other than the identity is a n minus one. So n contains the a n minus one fixing a point. But then n contains all the a n minus ones fixing all points because all the subgroups fixing a point are conjugate and n is normal. And these a n minus ones all generate a n, which is very easy to check. So n is equal to a n. Okay, so that's all I want to say about subgroups of order 60. Next lecture, we'll look at subgroups of order 120 and 168.