 This lecture is part of Berkeley Math 115, an introductory undergraduate course on number theory and this will be about products of groups and its applications in number theory. So we recall that a group G is just some set with a binary operation written as additional multiplication and suppose you've got two groups G and H. Then we can form a new group which is the product of G and H and its elements consist of pairs with G, little g in the group big G, little h in the group big H. We define the product of these by G1H1 times G2H2 is equal to G1G2H1H2. Or if we want to write it additively, we would have G1H1 plus G2H2 is equal to G1 plus G2H1 plus H2 and there are masses of examples of these everywhere. For example suppose you've got a vector space which is two dimensional, suppose we're working over say the reals. Then a two dimensional vector space is just pairs G1H1 with G1 and H1 both real and this group operation is usually just the same as the addition of two vectors in a two dimensional vector space. Of course we can have a vector space of dimension greater than two and in the same way we can take the product of more than two groups. Then some other examples of products of groups are, well first of all there's the real numbers under multiplication. Now any real number under multiplication can be written as either plus or minus one times a positive real. So now we've got two groups here, the group of positive reals under multiplication and a little group of order two with just two elements here. Now you see that the fact that any real number can be written uniquely as an element of this little group of order two times an element of the group of positive reals is more or less the same as saying that the group of real numbers under addition is or is isomorphic to the product of the group minus one. Plus one with the group of positive reals. If you want you can do the same thing with complex numbers. So the group of none zero complex numbers is isomorphic to the complex numbers of absolute value one which form a sort of the unit circle times the group of positive complex numbers mean positive reals. So the non zero complex numbers can be written as the product of the group, the circle group times positive real numbers. So now we come to some examples of products of groups in number theory. First of all, we have the Chinese remainder theorem. So the Chinese remainder theorem says that the integers modulo mn are the same as the group of integers modulo m times the group of integers modulo n whenever m and n are co-prime. So the previous statement that we had was that if we've got an integer modulo mn it can be written uniquely as an integer modulo m pair consisting of an integer modulo m and an integer modulo m. And in group theory this just says that z modulo mnz is isomorphic to z over mz times z over nz. And we can do the same thing with the if we take these two sets and look at them under multiplication instead of addition of course to get a group under multiplication we need to take non zero elements. So we find the group z over mnz star that's the elements co-prime to mn under multiplication is isomorphic to z over mz star times z over nz star. So there's an additive form of the Chinese remainder theorem and a multiplicative form of the Chinese remainder theorem. So let's have a couple of examples of this. If we take z over 6z this is just isomorphic to z over 2z times z over 3z. Or if we take the group z over 12z star so this is four elements one five seven and nine this is isomorphic to z over 2z star times the group z over 3z star and this is so that should be a four. This is two elements one and three under multiplication mod four and this is two elements one and two under multiplication mod four. So so under multiplication mod three. So we can very often split groups into a product of of simpler groups. In fact this is for a Boolean groups this is very common if a group g all groups in this lecture will be a Boolean in case I forget to say so. So if she has order mn with m and n co prime then g is equal to a product a times b where a is the elements with. Elements a with m a equals zero here I'm going to write the group additively rather than multiplicatively and B is equal to the elements. And be with and be equals zero. So we saw this when say G was of order six. So if if G is the group Z over six C then that is order two times three and the elements of order two of the other elements zero three and the elements of order dividing three of the element zero two. And four so this is isomorphic to the group of integers mod two and this is isomorphic to the group of integers mod three and G is a product of these two groups. Now to see G splits as a product like this and we have to check two things first of all the intersection of a and B is just the trivial group and that's kind of obvious because if. If G is in the intersection a intersection B then this means that mg equals zero and mg equals zero. And this implies G equals zero because one is equal to mx plus ny for some x and y that's because m and n are co prime. On the other hand, we can also check that a plus B is equal to G. What this means is that any elements of G can be written in the form a plus B with a in a and B in B. And to see this we just use Euclid. We can write one is equal to ms plus nt for some s and t. So a so G is equal to G times ms plus G times nt. And this element is in the group B and this element is in the group A because if we multiply this by n we get mn times G which is zero and if we multiply this by m then similarly we get zero. So from the fact that a and B have intersection the trivial group and the fact that every element in G can be written as a plus B. It's very easy to see that G is just isomorphic to the product of a and B. So any finite group, any finite abelian group is a product of groups of prime power order because we can just write the order of G as a product of prime powers and these are co prime. So we can decompose group G to be a product of all these products of groups of all these prime power factors. In fact, we can go a bit further. If we've got any finite abelian groups, then G is a product of cyclic groups. This is where I'm taking G to be a finite abelian, as usual in this lecture. So you remember a cyclic group is one that's generated by one element G. So it's just isomorphic to G over nz where n is the order of the element generating it. And to see this, what we do is we suppose that G is generated by some elements G1, G2, Gn. That means every element of G can be written in terms of these elements. We could take G1 up to Gn to be all the elements of G if we wanted, but it doesn't really make any difference. And then we have some relations between them. So we might have a11G1 plus a1nGn equals 0, a21G1 plus a2nGn equals 0, and so on. And we write down all the relations between these elements we can think of. And what we want to do is to manipulate all these relations until they become a particularly simple form. So what we do is we put all these relations into a big matrix a11, a12, a1n, a21, a22. We can even give it an infinite number of rows if we like. It doesn't really make much difference. And think what we can do with these. Well, first of all, if some generator, if we've got two relations like this, we can add a multiple of the first relation to the second relation without really making any difference. That means we're doing a row operation on the matrix. We're allowed to add any multiple of one row to another row, and we've still got an equivalent set of relations. We can also do column operations if we change G1 to G1 plus NG2. You can see the difference that this makes to the relations is we're just adding a multiple of the second row to the first row. And again, this just gives us an equivalent set of generators and relations. So we can also do column operations. That means we can add a multiple of one column to another without changing group G. And now what we do is we do row and column operations to make a11 as small as possible, but still positive if we can do it like that. By the way, one thing we can do with row and column operations is we can also permute rows and columns. You can either just notice that you can permute them because you can permute the generators or you can more or less get it up to sign just by using the row operations we've already said. So we suppose we've got a11 as small as possible, but positive. And if we can't make it positive then all the entries are zero and we're sort of finished. And now what we can do is we can do row operations. We can subtract the first column from all these other columns in order to make we have a division with the remainder algorithm. So if we take a12, for example, you write a12 equals q times a11 plus remainder. And that means we can replace a12 by the remainder and we have nought is less than r is less than a11. And since we assumed a1 was minimal, we must have r is equal to zero. So we can subtract the first entry from all the other elements of this row and make them zero. And similarly, we can clear out the first column. So we can turn our matrix into something like this. It does an a11, lots of zeros here, lots of zeros here, and an a22, a23, a32 and so on. And then we can do the same trick with this bit of the matrix. We can rearrange the rows and columns using row and column operations to make a22 as small as possible. And then we can use it to kill off all the entries in the row and column. And if we continue like this, we end up with a matrix looking like this. It does a11 and lots of zeros and an a22 and lots of zeros, a33 and lots of zeros. And if we take a look at this, it's saying our group is given by generators g1, g2, g3 and so on. And the only relations that matter are a11, g1 is equal to zero, a22, g2 is equal to zero, a33, g3 is equal to zero. And you can see that this just means we've got a cyclic group of order g1 and then we've got a cyclic group, z over g2, sorry, a11z, g over a22z generated by g2 and so on. So we've managed to write this group as a product of cyclic groups of order a11, a22, a33 and so on. And if we want, we can mumble something about the Chinese remainder theorem and can take each aii to be a prime power order. So in fact, we've sort of got a complete list of all finite abelian groups. Any finite abelian group is a product of groups of order, cyclic groups of order power of p, the sum prime p. And we've actually had lots of examples of this. So let's see, let's look at the group z modulo 41z star. Okay, this is a finite abelian group of order 40. So and 40 is equal to 2 cubed times 5. So this should be a product of cyclic groups of prime power order. And what are they? Well, the cyclic group of order 5, there's only one cyclic group of order 5. So we worked out what this was last lecture. The group of order 5 was generated by an element 10, if I've remembered correctly. And we also had a cyclic group of order 8, which was generated by an element 38. So we've managed to split this group. So this is isomorphic to a group of order 5 times a group of order 8. Cyclic group of order 5 and order 8. Well, here we only got a, it was a little bit easy because we ended up with a cyclic group. If we look at something like z over 8z star, so this is order 4. Again, this is a product of cyclic groups of order 2. So one of the groups could be 1 minus 1, and the other could be generated by 1 and 5. And you can see that everything co-prime to 8 modulo 8 can be written as a product of one of these two elements times a product of one of these two elements. In fact, we saw that z over 2 to the nz star can be written as a product of groups of orders 2 and 2 to the n minus 2. That's as long as n is at least 4 or something. And that's because the group of order 2 has two elements 1 and minus 1. And we saw that the group of order 2 to the n minus 2 is all powers of 5 modulo 2 to the n. So this group isn't quite cyclic, but it's a product of a cyclic group of order 2 to the n minus 2 and a product of a group of order 2. You notice, by the way, this is sort of formally very similar to what happens if we take the real numbers under multiplication. This is again a product of a group 1 minus 1 of order 2 and a group of times a group of positive reals. And the positive reals isn't quite the set of powers of some number like 5, but if you choose a number which is very slightly bigger than 1 such as 1 plus epsilon, then all the powers of 1 plus epsilon look sort of a little bit like the group of positive reals. They're sort of almost dense in it. So these two groups are sort of in some ways formally quite similar. Next we have the following problem. Let's find the smallest integer n such that x to the n is equivalent to 1. Modulo a million or x co-prime to a million. And if you're not thinking very clearly, you may just sort of mumble something about Euler's theorem and work out what phi of a million is. And phi of a million is this is phi of 2 to the 6 times 5 to the 6. And this is going to be 2 to the 5 times 5 minus 1 times 5 to the power of 5. And you may think that's the answer. Well, this is much too big. What you should really do is you take this group Z over a million Z under multiplication and you write it as a product of cyclic groups of prime power order if you like. And we can do this using the Chinese remainder theorem. So by the Chinese remainder theorem, it's isomorphic to Z modulo 2 to the 6 Z star times Z modulo 5 to the 6 Z star. And now we can figure out what each of these two groups is. First of all, we know that since 5 is an odd prime, this group here has a primitive root. So it's actually Z modulo 5 to the 6, which is 5 minus 1 times 5 to the 5 Z. Well, we can continue further using the Chinese remainder theorem because this is isomorphic to group Z over 4 Z times Z over 5 to the 5 Z. So we've written as a product of cyclic groups of prime power order. And for powers of 2, we need to be a little bit more careful because as we saw, this isn't cyclic. It's isomorphic to group of order 2 generation by 1 minus 1 times a little group of order cyclic group of order 16. So here we've managed to split this group into this product of 1, 2, 3, 4 different groups. Now we've got to find the smallest number such that if we take that multiple of any element of this, we get 0. And obviously it has to be the least common multiple of 2, 2 to the 4, 4 and 5. So the least common multiple is 2 to the 4 times 5 to the 5, which is quite a bit less than what we get from Euler's theorem, which would say it's 2 to the 7 times 5 to the 5. So Euler's theorem is wrong by a, well it's not wrong, but it's inefficient by a factor of 8. Okay, next lecture I'll be continuing to discuss applications of rings and fields to number theory.