 So this video is part of an online mathematics course of group theory. This lecture will cover products of groups. So in the previous lecture, we classified groups of order up to three and showed they were all cyclic. So the next case to look at is groups of order four. Now, if a group is order four, then by Lagrange's theorem, all elements of order dividing four, so they must have order one, two, or four. Now, if there is an element of order four, say G, then the group consists of the elements one, G, G squared and G cubed. So is cyclic and isomorphic to Z modulo four Z. So we may as well assume that all elements have order one or two. So this means that X squared equals one for all X in our group G. So here we're taking G to be a group of order four. And now if the square of any element is one, then the group is abelian. So this implies XY equals YX for all Y and that's very easy to see because we know XY squared equals one and this is just XY times XY. So now we multiply both sides of this equation on the left by the inverse of X, which is just the same as X because X squared equals one and we get YXY equals X. Now we multiply on both sides on the right by Y inverse, which is equal to Y. So we get YX equals XY. So we're using the fact that Y equals Y to the minus one. So we use the fact that XY squared equals one and X squared equals one and Y squared equals one and we deduce that the group is commutative. And what we're going to do is let's classify all commutative groups G with X to the P equals one for all X where P is some prime. So here we're just doing the case P equals two but the case of all primes is no more difficult and will be useful later on. Well, what we do is we write the group operation as addition rather than multiplication, then G is a vector space over the field FP with P elements. So you remember there's a finite field which can be regarded as the integers modulo P considered as a field with addition and multiplication. And it's a vector space because it's got this commutative operation of addition and you can multiply by elements of the finite field of order P because P times X is equal to naught for all X. We're now writing the group operation additively. So instead of saying X to the power of P equals one we say P times X equals one. Well, we know how to classify vector spaces over a field. Any vector space over a field is just a sum of one-dimensional vector spaces which are isomorphic to the field. So G is isomorphic to FP to the N for some N. Let's take G to be finite for the moment. I don't want to worry too much about infinite groups. So we just have an n-dimensional vector space and you know an n-dimensional vector space for N finite is just determined by its dimension. So there's only one. So the commutative groups with this property there's one for each prime power order. So G has order P to the N for some N and G is isomorphic to FP to the N where this just means the additive group underlying the vector space. So these are called elementary Abelian P groups. And next to cyclic groups they're some of the simplest possible groups. Anyway, let's go back to our group of order four. So for order four there are two groups. Z modulo four Z the cyclic one and Z modulo two Z times Z modulo two Z where this is just a sum of two vector spaces. We should just check they're really different that there isn't some mysterious isomorphism between them we've overlooked. So how do we do that? Well, if two groups are isomorphic then they have the same number of elements of each order. So this has one element of order one and two of order four and one of order two. Whereas this group here has one element of order one and three of order two. So the number of elements of various orders is different. So these groups definitely different. It's possible for groups to be not isomorphic but have the same number of elements of each order but that's quite rare. So this is usually a good way of distinguishing groups. This group is called the Klein four group sometimes and you can think of it as being the group of symmetries of a rectangle. So a rectangle there are four symmetries and you can easily check they all have order two. We should also take a quick look at the subgroups of these two groups. Well, the subgroups of a cyclic group are easy to find because they're just generated by powers of some power of the generator. So we've got Z over four Z and it contains a subgroup of order one and it also contains a subgroup of order two. So it's got just three subgroups. The trivial one, the whole group and one group here. Z modulo two Z times Z modulo two Z, you can check it actually has three subgroups of order two and these all contain the subgroup of order one. So any element of order two generates a subgroup of order two. So these correspond to the three elements of order two. So these give pictures of the subgroup structure of two groups where we've drawn each lines between each lines between subgroups to show which subgroups are contained in other subgroups. Next, we should talk about non-split extensions. So you sometimes write things called exact sequences that look like this for groups. And we say this is exact at B if these arrows are homomorphisms of groups and the kernel of this map from B to C is the image of this map from A to B. So what does it mean for this to be exact? Well, it means the kernel of the map from A to B is the image of this map, which is zero. So the kernel of the map from A to B is just the trivial group. So this map is injective. So if you have nought goes to A goes to B, it means it's a diagrammatic way of saying this map is injective. Similarly, if B goes to C goes to nought as exact, it means the image of B is the kernel of this map, which is the whole of C. So it means this map is surjective or onto. So when you draw a diagram like this, it means A is more or less a subgroup of B, B maps onto C, and A is more or less the kernel of the map from B to C. You write zero here if the group is written additively. You sometimes write one goes to A, goes to B, goes to C, goes to one, where this means the trivial multiplicative group if the groups are written multiplicatively. So we can write nought goes to Z modulo two Z, goes to Z modulo four Z, goes to Z modulo two Z, goes to nought because Z modulo four Z has a homomorphism onto Z modulo two Z and the kernel is just a group of order two. We can also have a map Z goes to two Z, goes to Z mod two Z times Z modulo two Z, goes to Z modulo two Z, goes to nought. And this is because whenever we've got groups A, B and C, we always get an exact sequence like this, where the map here is just projection to B and the map here is just the obvious inclusion. So we've got two different exact sequences with the same groups on the end and a different group in the middle. And it's a very common problem in mathematics that you sometimes get an exact sequence and you know the groups A and C and you want to know what the group B is. And as we see from this example, you can't necessarily figure out what B is just knowing C and A. Here A and C are both a group of order two and there are two possibilities for B. So if the exact sequence looks like this, this is called split. So this is a split exact sequence and this is non-split. You can't write this group here as a product of these two groups in any reasonable way. And one of the most common mistakes in mathematics is sort of absentmindedly assuming that exact sequences are split. So sometimes this mistake is kind of hidden, but it's definitely not true that all sequences like this are automatically products. This is the simplest counter example. Now we're going to give some examples of products. So here we have the first non-trivial example of a product. There's a group of order four that is a product of two other groups. I'd better say what a product of two groups is. If you've got two groups G and H then you can make them define a product group in the obvious way. You just take the product of the underlying sets and if you've got an element G1 times H1 in this and you want to multiply it by G2 times H2 then you just multiply it in the obvious way. This is G1 G2 times H1 H2 by definition. So we can always form a product of any two groups and that will be a group. Conversely, it's often useful to be able to identify a group as a product. So suppose G has subgroups A and B and suppose A and B commute and every element of G is of the form A times B with A in A and B in B. And B and B in a unique way. Then G is isomorphic to A times B, the product of A and B. This sort of squiggly line means two groups are isomorphic. And the reason for this is we have a homomorphism from A times B to G which just takes A and B to the product of A and B. And you can check this is a homomorphism because A and B commute with each other and it's a bijection because every element of G can be written like this in a unique way. So we've got a homomorphism that's a bijection. So G is actually isomorphic to a product. And we'll be using this quite a lot to identify various elements of G. Quite a lot to identify various groups as products in the following examples. So the first example is just the group of non-zero real numbers. So this is isomorphic to the product of the group one and minus one with positive reals. And we can see this just by using the criterion we had before every element of the reals positive, every non-zero element of reals can be written as a positive real times one or minus one and these obviously commute with each other. The complex numbers are kind of similar. So the complex numbers as a group is isomorphic to S1 times the positive reals as before. Here S1 is just the circle group of the product positive numbers of absolute value one. And the positive reals are just the positive part of this axis. And you can see this is just essentially the polar decomposition of a complex number. And so S1 sort of indicates the argument of a complex number. And this corresponds to the absolute value of a complex number. So this is really just a product of groups. An example from number theory is the Chinese remainder theorem which says that the group Z modulo MNZ is isomorphic to a product Z over MZ times Z over NZ provided M and N a co-prime. And again, this is easy to prove. We have a homomorphism from Z modulo MNZ. To Z over MZ times Z over NZ because we can just map the element one here to one here and one here. And the kernel is the multiples of the things that are multiples of M and N. And if M and N are co-prime, this implies it must be a multiple of MN so is not in Z over MNZ. So if M and N are co-prime, the kernel of this map is zero. And these two groups have the same order and this map is injective. So it must be bijective. So we get an isomorphism of groups whenever M and N are co-prime. Notice by the way, we get for example that Z modulo six Z is isomorphic to Z modulo two Z times Z modulo three Z. We saw earlier that Z over four Z is not isomorphic to Z over two Z times Z over two Z. And this example here shows that we shouldn't be too casual about saying it's obvious these two groups are not isomorphic because at first sight, it looks as if these two groups are not isomorphic, but they are in fact isomorphic. In fact, and the founder of group theory, Kayleigh, is kind of famous for having accidentally made a mistake of assuming these two groups are different. You know, he sort of said there are three groups of order six when they're actually two because he accidentally confused these. Similarly, Z over M and Z star is isomorphic to Z over M Z star times Z over N Z star where M and N are co-prime. So you remember putting a star here means we just take the elements co-prime to M N and we use multiplication rather than addition as the group operation. And this can be proved in much the same way that we did for the additive version of Z modulo M and Z. For example, let's take M N equals 15 then we find Z modulo 15 Z star is isomorphic to Z over five Z star times Z modulo three Z star. And from this, we can work out the structure of this group. See this group is order eight. And this group, as we saw earlier, is cyclic of order four and it's isomorphic to Z over four Z. And this is cyclic of order two, isomorphic to Z over two Z. So we see at this group here, we can now work out what its structure is. It's isomorphic to a product Z over four Z times Z over two Z. So this gives us a good picture of what this group looks like. Incidentally, we remember Euler's theorem says that X to the power of eight is congruent to one mod 15 if X is co-prime to 15. But we see from this decomposition that that's actually not the best possible because every element of this group has order dividing four. So in fact, X to the power of four is congruent to one modulo 15 whenever X is co-prime to 15. So using this decomposition, we can actually tighten up Euler's theorem and get a slightly sharper version of it. Now we come to some examples of groups that are products, although it's not actually obvious. So suppose we take the rational numbers Q. So this is going to be the rationales. I'm gonna take the non-zero rationales under multiplication. And we know that every rational number, every non-zero rational number can be written as a product of powers of primes in a unique way. What this means is the rational numbers is isomorphic to a sum of copies of the group Z. So let me explain what this is. So if we have a lot of groups A, B, C, then we can form a sum of groups. This is usually only used when A, B and C are abelian. So I'm going to write the groups additively. And this is going to consist of all elements where we take an element A from A and B from B and so on with almost all of A, B and C are zero in the group. If we allow an infinite number to be non-zero, then we get a different group called the product of these groups. So for finite sums and products, finite sums are the same as finite products because if we're saying almost all elements are zero, that's the same if you've only got a finite number. Almost all by the way means all but a finite number. Anyway, the group of non-zero rationales is isomorphic to the direct sum of copies of Z. And this is done as follows. If we've got a rational number R, we just write it as two to the N2, three to the N3, five to the N5 and so on where, sorry, I should have said it's equal to this times a group of order two, I always forget that. So we can write it as plus or minus this number here. And this gives us a group of order two because we can have a group of two elements, plus one and minus one is just cyclic of order two. And this is isomorphic to Z because we can just make this isomorphic to Z by mapping an element N of Z to two to the N here. So we get, it's a product of a group of order two and sum of lots of groups isomorphic to Z where this is powers of two, this is powers of three, this is powers of five and so on. So this gives a complete description of the non-zero rational numbers as sum of groups, most of which are Z and one of which is Z modulo two Z. Another example of some groups that aren't quite obviously products are the rotation groups of various plutonic solids. So not the rotation groups, the full group of symmetries. So here, if I take an octahedron say, the group of rotations is order 24 but it's full group of symmetries is order 48 because I can also reflect in various hyperplanes and I can also have a symmetry mapping every point to its opposite point. So altogether if you count up carefully you find there are 48 symmetries. So the full group of all symmetries of the octahedron is order 48 and this group splits as a product of two groups. The reason is we have a subgroup of order 24 which consists of rotations and we also have a subgroup of order two which consists of two elements. One is the identity and the other takes every point to its opposite point. And you can see that every symmetry of the octahedron can be written as a product of a rotation and one of these elements here. And these also commute with each other. You can see this by writing these symmetries as matrices and this then just becomes a three by three matrix with minus ones down the diagonal which commutes with everything. So we've got two commuting subgroups of the group of symmetries and every symmetry can be written as something in here times something in here. So the group of all symmetries of an octahedron which is order 48 actually splits as a product of these two subgroups. And exactly the same thing happens for an icosahedron. It has an automorphism minus one taking every point to its opposite point and similarly for a cube and a dodecahedron. If we try this on a tetrahedron, it fails. So a tetrahedron, the full group of symmetries is order 24 whereas rotations has order 12 but it no longer splits as a product. The point is a tetrahedron no longer has a symmetry minus one that commutes with everything else. So four of the platonic solids that their full symmetry group splits as a product and the tetrahedron doesn't. So for the final example of a group that unexpectedly splits as a product, let's look at the group of rotation. So the group of all roots of unity in the complex numbers. So if you want to get a picture of this, we draw the unit circle and all roots of unity lie on the circle. So there's the point one, there's a square root of unity, there are fourth roots of unity, there are some cube roots of unity and sixth roots of unity and eighth roots of unity. And if we go on like this, you see we get a root of unity for every point whose argument is a rational multiple of a full rotation. And this group sure looks like a circle. I mean, we've got a sort of dense subset of a circle and the circle certainly doesn't split as a product. So it doesn't look as if this group is a product, but in fact, it splits as a product of various subgroups. So for each prime, we take all the roots of order two to the power of something. And here we take the roots of order three to the power of something and so on. So we've got all these subgroups and these subgroups all commute with each other. And in fact, every root of unity can be written uniquely as a product of elements in these groups. And this just follows from the Chinese remainder theorem. For instance, if we take the roots of order N, of order dividing N, this is just isomorphic to a cyclic group of order N because we can take the one of smallest argument and the others will all just be powers of it. And the cyclic group of order N splits as a product Z to the two to the something Z times Z to the three to the something times Z and so on, where these are the biggest powers of two and three and so on dividing N. So every root of unity can be written uniquely as a root of unity of order of power of two times the root of unity of order of power of three and so on. So this, the group of all roots of unity splits. Sorry, I should have said a sum of these groups not a product of these groups. The sum of the subgroups of roots of order of prime power. So the next lecture will be mostly on groups of order six because all groups of order five are just commute, are just cyclic and there's not much to say about them.