 So this lecture is part of an online mathematics course on group theory and will be about semi direct products. So let's start by looking at the example of a group S3 of order 6, which is just the group of symmetries of a triangle. Now this group is not a product of two smaller groups, but it's quite close to being a product of two smaller groups. So let's take A to be the subgroup consisting of the identity and switching to elements and B to be the subgroup of order 3 consisting of the identity and the rotations. Then we see that the group S3 is equal to AB in the sense that every element of S3 is of the form AB with A and A and B and B and this is unique. So we've got a map of sets from A times B to S3 taking A and B to AB and this is an isomorphism of sets. However, it's not an isomorphism of groups. So we have a product of groups A times B, but the map from here to here is not a homomorphism of groups, it's just a map of sets. And the problem is that the subgroups A and B of S3 don't commute. In other words, the product of A and B is not necessarily the same as the product of B and A. So what actually happens if we try and multiply two elements like this? So suppose we take the element A1 B1 and try and multiply it by the element A2 B2 and try and get an element of the form A times something. Well, this is equal to A1 A2 times A2 to minus 1 B1 A2 times B2. And this thing here is in B. So here we've got an element of the group A times an element of the group B except B1 has sort of been twisted by A2 in some sense. So it's convenient to write x, y, x to minus 1, y, x equals y to the x. So this is the definition of y to the x. And the reason for writing this as an exponent is this actually satisfies the usual rules for exponents. So y1, y2 to the x is equal to y1 to the x, y2 to the x as you can easily check, and y1, so y to the x1, x2 is equal to y to the x1, x2. So this funny notation actually does satisfy the usual rules for exponentiation that you're used to. So what we've got is, so we have a subgroup A and we have a normal subgroup B and the group G is equal to AB uniquely. We mean every element of G can be written uniquely as something in A times something in B. And we also have an action of A on B. This is actually going to be a right action. In other words, for any element B and A, we've got an element B to the A, and this action preserves products of B. So B1, B2 to the A is equal to B1 to the A, B2 to the A. So this says that it preserves the product of B and B to the A1, A2 is equal to B to the A1 to the A2. This says it is an action of A on B. In other words, you can think of this as being a homomorphism from A to the group of symmetries of B. And we can ask conversely, suppose we didn't have this group G, so suppose we only had a group A and a group B and an action of A on B. Can we reconstruct G? Given a group A and a group B and a right action of A on B. And the problem is to construct a group from these so that B is a normal subgroup and A is a subgroup and this right action is given by conjugation. Well, it's kind of obvious how to do this. We just take G to be the set A times B and define A1, B1 times A2, B2 to B. It's really easy to get confused by this. A1, A2 times B1 to the A2, B2. So this is what the product of these would have to be if we'd started with the group G and had the normal subgroup A and the normal subgroup B and the other group A. So the problem is, does this make G into a group? And the only problem is to check the associative law. So we should check existence of inverses and identity and so on, but those are very easy to check. The one we have to do a bit of work on is the associative law. So what happens is we take A1, B1, A2, B2, A3, B3. So we've got three elements of G and we can multiply them in two different ways. We can first multiply these two or we can multiply these two and we've got to check we get the same. So let's multiply these two. We get A1, A2, B1 to the A2, B2. And now we want to multiply this element by this element here and we get A1, A2, A3, B1 to the A2, B2, or action on by A3, B3. On the other hand, we can first multiply these two and we get A2, A3, B2 to the A3, B3. And then we can multiply these two and we get A1, A2, A3, B1 to the A2, A3 times B2 to the A3 times B3. So the problem is, is this equal to this? And you can see those are equal because this expression here is just equal to B1 to the A2, A3 times B2 to the A3 by one of the rules for exponentiation and this is equal to B1 to the A2, A3, B2 to the A3 by the other rule for exponentiation which is very convenient because it's this element here. So we do indeed get a group and this is called the semi-direct product of A and B. So summary, given A, B and a right action of A on B, we get a semi-direct product which is denoted by a quiddle, it's this sort of time sign with a line there. And this is B is a normal subgroup, A is a subgroup and the action taking B to A to the minus one B, A in a semi-direct product B is the same as the given action of A on B. Of course, if we've got a left action of A on B, we can also form a semi-direct product which is usually noted by B with this thing going the other way around. So this time the left hand thing is the normal subgroup. So this sort of line is on the opposite side of the normal subgroup and this is almost impossible to remember. So, for example, we see that S3 is isomorphic to Z over 2Z, semi-direct product Z over 3Z. So here this would be the subgroup 1, 1, 3, 2, 1, 2, 3 and this could be the subgroup with elements 1 and 1, 2. So we can use this to classify subgroups of order 6. So let's classify groups G of order 6. And what we're going to do is to show that G has to be some sort of semi-direct product. So first of all, G has an element of order 3. Well, this follows because all elements of G must have order 1, 2, 3 or 6. So if it's got an element of order 6, it's 6 and certainly has an element of order 3. If it has no elements of order 3 or 6, then all elements would be of order 2 and we've classified groups such as all elements of order 2 and they all have a power of 2. So let's call this element G, then 1 G squared is a subgroup of order 3. And it is normal as all subgroups of index 2 are normal. Remember we had this in a previous lecture. So we found a normal subgroup of order 3. Secondly, G has a subgroup of order 2. And if it didn't have a subgroup of order 2, then all elements other than the identity would have to have order 3. But however, all elements of order 3 come in pairs because we've got an element of order 3 and it's inverse. So G has an even number of elements of order 3. But since there are an odd number of elements that aren't the identity, there must be some element that doesn't have order 3 and isn't the identity. So it must have an order 2 or 6. And if it is order 6, we can just cube it and get an element of order 2. So we've got a normal subgroup of order 3 and we've got a subgroup of order 2. This means G is a semi-direct product of A and B where A has order 3 and B has order 2. And the only thing we don't know is what is the action of B on A. So this action is going to be given by conjugation in the group G. But we don't yet know how this group of order 2 acts by conjugation on order 3. So how can a group of order 2 act on a group of order 3? Well, what we do is we look at all automorphisms of the group of order 3. So the automorphisms, well, if this is generated by an element G, well, any automorphism must take G to an element of order exactly 3. So we either map G to itself or we map G to its inverse because these are the only two elements of order 3. So this is the identity automorphism and this is the non-trivial one. So there are two ways Z over 2Z can act on this because the group of automorphisms is a group of order 2 and there are only two homomorphisms from a group of order 2 to a group of order 2. So case 1, Z over 2Z acts trivially on Z over 3Z. In this case, we get a direct product Z over 2Z times Z over 3Z, which as we've seen is Z over 6Z. Case 2, Z over 2Z acts non-trivially. So if Z over 3Z is generated by G and Z over 2Z is generated by H, then we're putting G of H equals G to the minus 1. And as we've seen, this is just isomorphic to the group S3 because S3 is a semi-direct product with this non-trivial action. Now we're just finished by giving a couple more examples of semi-direct products. So the next example is the AX plus B group. So this is the group of all transformations of the real numbers of the form X goes to AX plus B. So it's all linear transformations, all linear maps from the real numbers to itself. Here A is not equal to 0, of course. And you can represent this as matrices of the form AB01. Now you see if you, this has a normal subgroup of all translations of the form X goes to X plus B. And it also has a subgroup of the form X goes to A times X. And these two operations don't commute with each other. If you add B to something and multiply by A, it's not the same as multiplying by A and then adding by B. And we get an action of this normal subgroup, which is isomorphic to the reals. Sorry, we get an action of this subgroup, which is isomorphic to the non-zero reals on this normal subgroup, just as multiplication. So we would have the action of A on an element B is just multiplication of B by A. So we would have the rather confusing notation that BA is just B times A. Sorry about the fact that this rather conflicts with the normal notation for exponentiation. Another example is an n-dimensional version of that, just the isometries of R to the n. So isometries just mean maps from R to the n to itself at preserved distance. And this is a normal subgroup of translations, which is isomorphic to R to the n, of course. And it has another subgroup of rotations. Sorry, not rotations. What's the word? Orthogonal transformations. These are just isometries that fix the origin and are given by elements of the orthogonal group. The orthogonal group is usually narrated by O, n of R. This means the group of n by n orthogonal matrices where an orthogonal matrix is just a matrix that preserves distances. And this group here acts on R to the n. And if we take the semi-direct product of this orthogonal group by the group of translations, we get the group of all isometries of R to the n. Notice that the previous example was very similar to the case n equals 1, except we weren't just doing isometries. We're also allowing things to be rescaling. A similar example is the Poincaré group in physics. So the Poincaré group is the orthogonal group of spacetime, which is a bit like four-dimensional Euclidean space except the inner product on it is allowed to have negative vectors. So again, it has a normal subgroup of translations of spacetime and a subgroup of Lorentz transformations. So the Poincaré group is a semi-direct product of translations of spacetime by the Lorentz group. So that's more or less classified groups of order six. Groups of order seven are all cyclic because seven is prime. So the next interesting case to do is to classify the groups of order eight, which we will do in the next lecture.