 This lecture is part of an online mathematics course on group theory and will be mostly about the structure of finitely generated abelian groups. So in previous lectures we've done groups of order up to 15 so the next thing to look at is order 16 and as we will see next lecture there are a rather a lot of groups of order 16 and we're just going to look at the abelian groups of order 16 and let's write down the abelian groups we can think of of order 16 well there's z modulo 16z and then we can take products like z modulo 8z times z modulo 2z, z modulo 4z times z modulo 4z, z modulo 4z times z modulo 2z times z modulo 2z and z modulo 2z times z modulo 2z times z modulo 2z times z modulo 2z times z modulo 2z. So these are the obvious abelian groups of order 16 and the question is are there any others and if we think about the abelian groups we've had earlier in the course they can all be written as products of cyclic groups for example the groups we had in the previous lecture z over nz star all turn out to be products of cyclic groups so it's not true that every abelian group is a product of cyclic groups for instance the the rational numbers q so these are the rationals under the group operation of addition is not a product or sum of cyclic groups which is fairly easy to check so it's certainly not true that every abelian group is a product of cyclic groups however and we have the following theorem every finite abelian group is a product of cyclic groups of prime power we can even take them to be a of prime power order so they're cyclic groups the form z over p to the n well it turns out that the proof of this works just as well if we instead of using finite groups we talk about finitely generated groups so a set of elements generates a group if a group if every element of the group can be written as as products of elements in this set well if we allow finitely generated groups then we have to allow not only cyclic group z over p to the nz but we also have to allow the group z see any cyclic group of order z over kz for some integer k can be written as a product of groups of prime power order but if if we're allowing infinite groups we also have to allow z so what we're going to do is to show how to prove this there's actually a more general result which we won't prove but is proof that the proof is almost the same as this which says that any finitely generated module over a Euclidean ring is a sum of cyclic modules so this ring is r the cyclic module is going to be r over a times r okay i'm not going to explain what a module over a ring is because we're not actually going to use this result but the case we're doing is where if you take the ring r to be the ring of integers then a final then a module over the integers is just the same as an abelian group so this result just says that a finite abelian group is a sum of cyclic groups anyway the main property of the integers we use is the following one if we've got integers a and b with b greater than zero we can divide a by b getting a quotient and a remainder with the remainder being less than b so this is so q is the quotient if we divide a by b and we get a quotient of q and we have a remainder left over and that the key point is that the remainder has to be less than the number we divided by a Euclidean ring is one which has a similar division with remainder where the absolute value may not be the absolute value on the integers will be something else like say the degree of a polynomial or something like that so this is the property we're going to use now what we do is we take some generators for the abelian group so let's take some generators g1 opt to gk and we're going to write the group additively because it's abelian and we look at some relations between these generators so relations between generators are just some group operation on these generators that produces zero so the relations might be somewhere like n11 g1 plus n12 g2 plus plus n1 k gk equals zero so this would be a typical relation it says that some multiple of g1 plus some multiple of g2 and so on is zero so all these numbers n i j are going to be integers and then we might have another relation n12 g2 plus n22 g2 plus n2 k gk equals zero n13 g what's that actually but g1 g1 and so on n14 g1 and so on so we write down lots of relations you can even write down an infinite number of relations if you like the point is we should write down enough relations to force to force the group generated by these to be the abelian group we're interested in we can always arrange this by writing down every possible relation between the generators of our group which will give us an infinite number but as i said that's that will actually be harmless in the following proof so we've we've we've we write down enough relations to specify our group and now we're going to manipulate these relations and get them into a nicer form it's rather tiresome writing all these relations out so instead of writing out all the g's and the n's we just write down a matrix n11 n12 up to n1 k n21 and so on up to n2 k so on so this is a possibly infinite matrix and we're going to think what can we do to the matrix without changing the group so operations on this matrix so first of all we can add multiple that's an integer multiple of one column to another and the reason is that if we've got generators g and h so if the generators contain g and h and other things then we can change to g and g plus n times h for any n in the integers because if if these two and some other elements generate the group then obviously these two and vice versa and the effect on the relations will be to some do some operation to the to the columns containing g and h you'll either add or subtract n times one of these columns for the other i'm not going to tell you which because i i almost certainly get it the wrong way around if i tried secondly we can add a multiple of one row to another and this is again fairly obvious because if r1 and r2 are relations then r1 and r2 plus nr1 are equivalent relations so if we know r1 and r2 are zero then we know that these two are zero and if we know that these two are zero then we know that those two are zero so we can do all the usual row and column operations on matrices without affecting the group without affecting our abelian group and now what we do is we do these row and column operations to make the top left entry n11 as small as possible when we say as small as possible we mean that the absolute value of n11 should be as small as possible but greater than zero so we've now got n11, n12 and so on and n21 and so on now we can subtract now it's multiples of the first row and column from the others to make all entries n12, n13 and so on and n21, n22 and sorry n31 and so on equal to zero now we can do this because n12 must be a multiple of n11 if it wasn't we could subtract multiples of n11 from n12 in order to get a remainder that was smaller than n11 and then we could switch um we could just switch the columns by rearranging the generators and find a smaller number positive number n11 so n11 divides all these and therefore we can just clear them out by subtracting the first column from all the other columns and the first row from all the other rows so we get n11, 000, 000, 000, 000, 000, n22, n23 and so on so we can turn our matrix of generators and relations into this form and now we can just repeat this so we forget about the first row and the first column and just repeat the same same thing on this smaller matrix here and what we end up with is something that looks like this we've got n11, n22, n33, up to nkk and all the other entries are zero then we can throw away all rows of the form 000, 000, 000 because a relation that telling you that zero is equal to zero is not really terribly useful and you may as well discard it so what have we got we've got our group has generators g1, octa, gk and relations n11, g1 equals zero, n22, g2 equals zero and so on and this is just the same as saying so the group is isomorphic to z over n11z so this is going to be generated by g1 times z over n22z generated by g2 and so on and now we can just sort of finish up by replacing z over nz is isomorphic to z over p1 to the n1z times z over p2 to the n2z and so on where n is equal to p1 to the n1 p2 to the n2 and so on so we can assume that all these groups are actually cyclic of prime power order except that n might be zero so z over zero z is just isomorphic to z so any finitely generated abelian group can be written as products of cyclic groups of prime power order and products of z and of course if the group is finite we don't get this case we just get groups of prime power order so um well this is quite nice on the other hand it means there's not really it's kind of difficult to think of anything interesting to say about finite abelian groups because they're all known and have such an easy structure um you can also ask is the when are two finite abelian groups isomorphic so I suppose g is isomorphic to z over p1 to the n1z times z over p to the n p2 to the n2z and so on we can ask are p1 n1 p2 n2 and so on determined well obviously they're not because you can just change the order so let's say determined up to order and if we don't insist that these are prime powers then this obviously isn't true because z over 6z is isomorphic to z over 2z times z over 3z so we've got to be a bit careful that we're not getting a complication like this turning up and the answer is yes um for example we can see this as follows um if we've got z over p1 to the n1z times z to the p1 the n2z and sometimes p1z over p1 the n3z and so on so the number of times p1 to the something occurs is determined and the reason it's determined is determined by the number of elements of order p p1 in g you can see the number of elements of order p1 plus 1 is is going to be p for this factor times p for this factor times p for this factor so we know how many times some power of p1 occurs and similarly um the number of elements of order p squared then determines the number of times we get p to the 2 plus something in the denominator if we know the if we know the number of times some power of p occurs in the denominator and we know the number of elements of order p squared by thinking about it but you'll see you can figure out the number of times we get one of these denominators with the n i at least two and by continuing like that you can figure out the exponents of all prime powers in the group so the group is uniquely determined by these prime powers um so we can now determine the number of groups of order p to the n so how many abelian groups of order p to the n are there well if we write n is equal to a plus b plus c and sum with a greater than or equal to b greater than or c greater than zero then we get a group um z over p to the a times z over p to the b times z over p to the c and so on and this gives a one-to-one correspondence between abelian groups of order p to the n and ways of writing n like this so the number of groups is equal to the number of partitions of n so the partitions of n are just the solutions of n equals a plus b plus c with a greater than or equal to b greater than or equal to c and sum greater than zero just the number of ways of writing n as the sum of positive integers where we ignore um order um so we can look at the first few cases of this so let's write out the number n and look at the number of partitions of it so if n is zero there's just one partition if you think about it because nought is actually an empty sum of integers n equals one there's one partition which is just one two we get two two or one plus one three we get three three two plus one one plus one plus one four we get five three oops four three plus one two plus two two plus one plus one and one plus one plus one plus one and you see these corresponded to the groups of order 16 we had z over two to the four z z over two cubed z times z over two z and so on for five we get seven which is getting five four plus one three plus two three plus one plus one two plus two plus one two plus one plus one plus one one plus one plus one plus one plus one one two three four five six seven and um it's obvious how this goes on it's getting a bit boring writing these out so i'm not going to go any further um the number of partitions of an integer is a um rather remarkable functions heavily studied in number theory but this is a group theory not a number theory course so i won't say much more about it um i just finished by mentioning that there is a very close correspondence between the theorem that says that every finite abelian group is a product of groups of the form z over p to the n z and the following theorem it says every matrix over the complex numbers has a georg normal form so you remember georg normal form means you write it as a whole lot of blocks down the diagonal where each block has an eigen value and one's up there so there might be a lambda there and a lambda i guess these should be different eigen values but anyway so you remember georg normal form of a matrix has blocks down the diagonal and each block has the eigen value down the diagonal and one's just above it and at first glance this theorem seems to have nothing to do with the structure of abelian groups but in fact they're almost the same theorem so this theorem works for modules over z that means abelian groups acted on by z which is just the same as abelian groups for this you look at um modules over the ring of polynomials over the complex numbers well what is a module over the ring of polynomials over the complex numbers well a module means an abelian group acted on by this in a nice way so a module is just a vector space with a linear transformation so saying it's actually on by c means it's a vector space over a c and saying it's also actually on by c of x means it must have a linear transformation which is just the action of x now we said that finite abelian groups are sums of cyclic ones z over p to the n z where z is prime and if you uh the corresponding theorem for modules over polynomials just says that every module is a sum of modules of the following form you take c of x over p to the n times c of x where p is now an irreducible polynomial so an irreducible polynomial in the ring of polynomials are just like primes in the ring of integers well the irreducible polynomials over the complex numbers are just of the form x minus alpha for alpha a complex number and if you work out what these modules are they're just vector spaces with a linear transformation such that x minus alpha to the n equals one and it's not very difficult to show that if you do that you can find a basis which looks like alpha alpha alpha one one one one where there are n rows so the Jordan decomposition of a complex matrix is almost exactly the same as writing a finite abelian group as a product of cyclic groups you can the proof of these two theorems is almost identical you just change the ring from the integers to the ring of polynomials um so next lecture we will discuss the non-abelian groups of order 16 without actually classifying them because as we will see there's so many of them that