 So this lecture is part of an online course on group theory and will be mostly about the groups of order 24. Now, there are 15 of these and what we're going to do is just quickly look through all of them and pick out a couple of the most interesting ones and look at those in more detail. So let's first of all see how to classify them. First of all, there's a seal of three subgroup and there are two cases. There might be one seal of three subgroup, which is normal or there might be four because the number of seal of three subgroups is one mod three and divides 24. So that's the only possibility and these are not normal. So let's first look at the case when the seal of three subgroups are normal. So that means the group G would have the seal of three subgroup as a normal subgroup and would be a direct product with a group of order eight, which would be the seal of two subgroup. And there are five possibilities for this. So this group can be Z modulo eight Z, Z modulo four Z times Z modulo two Z, Z modulo two Z cubed. It could be the quaternion group or it could be the dihedral group. And we have to figure out what is the action of this group of order eight on this group of order three. Well, this group of order three has an automorphism group of order two. So we're just looking at homomorphisms from these groups to a group of order two and there's one obvious possibility. So the automorphism could be trivial. Sorry, not the automorphism. The action on Z modulo three Z could be trivial or non-trivial. And if it's trivial, the groups is then G is just a product of these two groups. So we get five cases. If the action is non-trivial, we have to count how many essentially different ways this group can match a group of order two. And the answer is there's one way, two ways, one way, one way and two ways. So these two groups can act in two ways on a group of order three and they can be distinguished by the fact that the kernel of the map from this to Z two can be either a cyclic group of order four or a cyclic group of order two times a cyclic group of order two. And I'm not going to go through these in detail. Leave it as an exercise for the listener or something. The only one really worth commenting on is this one here which turns out to be the binary dihedral group of order 24. So it's a sort of double cover of the dihedral group of order 12. Other than that, I can't really think of anything much to say about these groups. So that's done. The case is when the seal of subgroup is normal and it gives us altogether 12 groups. And we're now going to do the three cases when the seal of subgroup is not normal. So altogether we're going to see 15 groups. So now let's look at the case when the seal of three subgroup is not normal. So there are four seal of three subgroups. And now G acts by conjugation on these four groups on the four seal of three subgroups. So we just remember the action by conjugation means an element G acting on a seal of subgroup S is just G S G to the minus one. And it acts transitively on these. So we get a homomorphism from G to the group S4. So this is the symmetric group of all permutations of the four seal of three subgroups. And this is on, so it's not onto necessarily. We can look at the kernel. So what is the kernel? Well, the kernel, so the kernel must have ordered dividing six because we've got a transitive action of G on a set of four elements and G is 24 elements. So the kernels order one, two, three or six. Well, it can't have ordered three or six because if the kernel was had ordered three or six order three subgroup of the kernel would be normal in G. As you can easily see, group of order six has a subgroup of order three that would be fixed under all automorphisms of this group of order six. So it'd have to be a normal subgroup of G. So the kernel has ordered one or two and both of these cases can occur. First of all, the case when the kernel is order one is rather easy because if the kernel is order one then we've got a map from a group of order 24 to a group of order 24 with trivial kernel. So this implies G must actually be equal to the symmetric group. If the kernel is ordered two, we've got to think about it a little bit more. So if the kernel is ordered two, the image has ordered 12 and the image has more than one seal of three subgroup because otherwise G would have a normal seal of three subgroup. So is isomorphic to A4 because if we look at all groups of order 12, A4 is the only group with more than one seal of three subgroup. And now A4 has a normal subgroup of order two and its inverse image in G will be a normal subgroup of order eight. So in this case, G has a normal subgroup of order eight and this means carrying on with the orange case. So G is a semi-direct product of an order eight group with the group of order three acting non-trivially because if it acted trivially, then we would be back in the case when it had a normal S3 seal of three subgroup. And we can check through the groups of order eight and we find there are just two possibilities. It could either be Z modulo two Z cubed semi-direct product Z modulo three Z or it could either be Z modulo two Z or it could be the quaternion group of order eight semi-direct product Z modulo three Z. And this case, you can check just gives us A4 cross Z modulo two Z and this case gives us one of the more interesting groups it's called the binary tetrahedral group. So you see that we've found three ways to join the groups A4 and Z modulo two Z. So we found these three groups without of order 24 without a normal seal of two subgroup. And I want to sort of try and draw pictures of them. So I'm going to draw an A4 sitting on top of the Z modulo two Z. So this is going to be the binary tetrahedral group. And what happens is it has a center of order two and the quotient by the center is the tetrahedral group or alternating group A4. On the other hand, we could have the group A4 with the group of order two sitting on top of it. So this is the symmetric group S4 which has A4 as a subgroup and the quotient is a group of order two. And finally, we can just have the product A4 times Z modulo two Z, which doesn't really have either of these groups sitting on top of the other. They're just sort of sitting next to each other in a sort of equal fashion. And this is a product of two groups. And this similar thing happens for alternating groups A, N for N greater than for other values of N. You usually find there are three ways to join them together. I mean, except sometimes when we discuss the alternating new base six, we'll sometimes find this occasionally more than there are some extra things you can do. But the alternating groups A, N all have three ways of gluing them together with a group of order two and either put the group of two on order top or on the bottom or by the side of the A4. So now we're going to look at the binary tetrahedral group a little bit. So we recall that we had the group S3 of unit quaternions mapping to SO3 of R, which is all rotations of three dimensional space. And inside the group of rotations, we have the group A4 of unit quaternions, so not of unit quaternions of rotations of a tetrahedron and the binary tetrahedral group, which is, you can denote it by A4 with a little hat on top, maps onto this group and has a center of order two. So this is an example of a central extension of the group A4. Central extension just means a group mapping onto A4 with the kernel being in the center. Central extensions turn up quite a lot in mathematics and they're usually rather difficult to see. So we can actually write down the binary tetrahedral group explicitly. Suppose we take all integer quaternions. So we take plus or minus A, plus or minus BI, plus or minus, if I'm doing this plus or minus is four, so I'm getting confused, it's A plus BI plus CJ plus DK for ABCD integers. So this is one obvious guess for what the integer quaternions are, but it's not actually a terribly good guess. Hurwitz found a slightly bigger ring of integral quaternions where you take either ABCD in the integers or ABCD are all in the integers plus a half. So for example, a half plus a half I plus a half J plus a half K is also one of these integer quaternions. And you can check these are closed under multiplication. And now we can look for the units in these integral quaternions. And these units are just the quaternions with Z times Z bar equals one, in other words, A squared plus B squared plus C squared plus D squared equals one. So we can write down the integral Hurwitz quaternions. Let's write down Hurwitz quaternions. So the integral Hurwitz quaternions, the following we have plus or minus one, plus or minus I, plus or minus J, plus or minus K. So these are the obvious ones. These just form the quaternion group of order eight, but then we've got 16 more because we have plus or minus one, plus or minus I, plus or minus J, plus or minus K, all over two. So we get eight here and another 16 here. And these form the binary tetrahedral group. So we'll just finish off this electron groups of order 24 by saying a bit more about the symmetric group S4. What we can do is look for its normal subgroups. It doesn't actually have all that many. It's got a normal subgroup Z over 2Z times Z over 2Z and it's got a normal subgroup A4 and this is containing S4. So you remember A4 is just the group of permutations that fix X, W minus X, W minus Y, W minus C times X minus Y, X minus C, Y minus C. And the group Z2 times Z2 is a normal subgroup of A4 and consists of the three permutations, one, two, three, four, one, three, two, four, one, four, two, three together with the identity permutation. So S4 is the first group with no normal steel of subgroups. If you look, every group we've had so far apart from S4 always has a normal steel of subgroup for some prime, but here we're starting to get more complicated groups that you can't construct like that. However, S4 is still an example of a solvable group. So group G is called solvable if we can find a series of subgroups, one contained in G naught. So one equals G naught, which is contained in G1, contained in G2, contained in GN equals G so that GI is normal in GI plus one and the quotient GI plus one over GI is cyclic of prime order. Actually, we could just let this be a billion because you can break any a billion group up into extensions of cyclic groups of prime order. So roughly speaking, a group is called solvable if you can sort of build it out of cyclic groups of prime order, which are the easiest sort of groups. And we can see that all nil potent groups are solvable. It's easy to check by induction because a nil potent group has a center so you can start off here. So all groups we've seen so far are solvable and the only one that's a little bit tricky to do is S4, but we see that S4 is this chain of subgroups and we can break this one up into two copies of Z2 so we can get a chain like this. In fact, the first, the smallest group that isn't solvable is A5 which we will get to a bit later. That the reason why groups are called solvable comes from Galois theory where if you've got a polynomial you can form a Galois group which is some sort of permutation of the set of all the roots of the polynomial. And if the Galois group is solvable then you can solve the equation by radicals. You can write down an explicit formula for the roots only using the usual field operations together with taking roots. And the fact that S4 is solvable corresponds to the fact that any polynomial of degree at most four you can write down an explicit solution by radicals. Over the fact that A5 is not solvable corresponds to the fact that some polynomials of degree five cannot be solved by radicals. It's actually, as I said, A5 is a little bit odd in that it has this normal subgroup. Most other alternating groups don't have this normal subgroup here. The quotient of S4 by this normal subgroup here gives a map from S4 to S3. So we actually have a homomorphism from S4 to S3 which is unusual. Usually a symmetric group SN has no homomorphism to SN minus one. You can see this as follows. Suppose we take four points. So suppose these are the four points acted on by S4. Then we can construct a set of three things that are acted on by S3. And the three things are always adjoining these points up in two pairs. So if I take these four points, I can join them up in pairs like this, or like this, or like this. So we have a set of three objects which are ways of joining up four points in pairs and any permutation of these four points is going to give a permutation of these three objects. So any permutation of four points will give us a permutation of three objects. So this gives us the homomorphism from S4 to S3. Well, there's quite a lot more to say about S4 but I think it would be better to have a general discussion about symmetric groups rather than just S4. So next lecture will be on symmetric groups.