 This lecture is part of an online mathematics course on group theory and will mainly be about Cauchy's theorem in group theory. So to motivate it, we've classified all groups of order at most nine so far, so let's try and classify groups of order 10, which is two times five, or more generally groups of order two p for p prime. Now it would be very useful in classifying these groups to know that there's an element of order p and an element of order two in the group. Fortunately this is given by Cauchy's theorem which says that if p is prime divides the order of a group g, then g has an element of order p. This only works if p is prime, so for example if g is equal to z modulo 2z times z modulo 2z, then g has order four, so four divides the order of g, but g has no element of order four because they're all of order two or one. So this is a sort of partial converse of part of Lagrange's theorem. Lagrange's theorem says that if a number divides the order of a group, then sorry, it says that if g is an element of a group, then its order divides the order of the group. Conversely you can ask if a number divides the order of a group, does the group have an element of that order? In general the answer is no, but Cauchy's theorem says that this is true if the number is a prime, and we're going to give two proofs of Cauchy's theorem. One is a sort of general purpose method of proving things in group theory and the other is a kind of slick proof that uses a clever idea. So first of all, so we want to prove, so suppose p prime divides the order of g, and we want to show that there is an element of order p. So we assume this result is true for all smaller groups. So when I say assume the result, I mean we want to prove g has an element of order p. So what we're going to do is to assume that if p divides the order of any smaller group, then g has an element of that order, and by induction this will be enough to prove the result is true for all groups. So first of all, assume let's first do the case where g is abelian. Well we pick g and g of some prime order q. If q equals p, we're done. If not, we look at g modulo, the group generated by g. Now since g is abelian, all subgroups are normal, so this is a perfectly good group. And this is order less than g. So by induction we can find some h in g over little g. You notice this has ordered divisible by p because this is ordered divisible by p and this has ordered co-prime to p. So where h is order p. Now let's lift h to some element a in g. So we're finding some element in the group big g whose image is h. And then a has order a multiple of p because its image in this group here has has ordered p. In fact, you can easily see that a has ordered either p or p times q. If it has ordered p, we're done. If it has ordered p times q, then a to q has ordered p. So in either case, we found an element of g of order p. So that does the case when g is abelian. Well, what about if g is not abelian? So suppose g is not abelian or not necessarily abelian. I mean, we don't really mind if it happens to be abelian. Well, every proper subgroup, so we can assume every proper subgroup has order co-prime to p. So proper subgroup means a subgroup not equal to g itself. So this just says that this is strictly less than g. And that's because if some proper subgroup had ordered divisible by p, then we could pick an element of order p in that subgroup by induction and we're done. So we can assume that every proper subgroup has ordered co-prime to p. So every subgroup, every proper subgroup, every subgroup not equal to g has index divisible by p because the index of a subgroup is just the order of g divided by the order of the subgroup. The order of g is divisible by p and the order of the subgroup isn't. And now let's look at this equation here. Let's look at all elements of g and we know this is equal to the size of the center plus the sum overall conjugacy classes of size greater than one of the size of the conjugacy class. So this is just exactly the same formula we had last lecture. You remember we're looking at the action of g on itself by conjugation and looking at the orbits and the orbits either size one in which case they're in the center or they have size greater than one in which case they are some conjugacy class size greater than one. And now the size of each conjugacy class is equal to the order of g divided by the order of subgroup fixing some element. So this fixes some element of the conjugacy class. And we notice that gs has order less than g because we're assuming the conjugacy class has size greater than one. So now we see this is divisible by p because we assume the order of g is divisible by p and these terms here are all divisible by p because we said that every subgroup has index divisible by p. So that follows from this fact here. Well if this term is divisible by p and this term is divisible by p we see that the center of g is divisible by p. Well the center of g is abelian. I mean the elements of the center commute with everything in g so they certainly commute with each other. So the center of g is an abelian group whose order is divisible by p. So by the previous sheet of paper where we proved this result for g abelian the center of g has an element of order p. So we've proved our theorem that every group finite group of order divisible by p has an element of order exactly p. This is a fairly typical example of how you prove things in finite group theory. If you've got some result you assume it's true for all smaller groups and you sort of nibble away at your group trying to reduce it to a smaller group where we can apply induction. For instance we first nibble away by doing the abelian case and then we nibble a bit further by showing that either there's a smaller subgroup we can of order divisible by p we can use or we can find an abelian subgroup of g of order divisible by p. So that's the first proof of Cauchy's theorem. There's actually a slick proof of it where all you do is you look at the solutions of this equation. So this is a second proof of Cauchy's theorem. So we ask how many solutions are there to this following equation g1 g2 up to gp is equal to one. Let's count the number of solutions. Well the number of solutions is obviously the order of g to the p minus one because we can choose g1 up to gp minus one to be anything and then there's a unique solution for for the pth term. And we notice this is divisible by p because we assume the order of g is divisible by p. And now if we've got any solution abc up to x say if this is equal to one then x abc up to what's the number before x w is equal to one and similarly w x abc up to that is equal to one because this expression here is just we just multiplied by x on the left and x inverse on the right and if we do that to one we just get one. So what we see is all the solutions of this form into groups of p solutions unless all these terms abc and so on are the same. If they're all the same we only get one solution. Notice we have to assume p as prime here for instance if we assumed if p was say four then we would get the term abab and if we permuted this we would get abab and if we cited it again we'd just get abab again and abab again. So we wouldn't actually get a group of four solutions to this we'd only get a group of two of them because when we got through two of these we'd get back to where we started and that depends on the fact that four is not prime so if we go through to if we just cycle this two times we get back to where we started. So we're assuming p is prime in order to show that the number of solutions like this is divisible by p. So we find g to the p minus one is something divisible by p plus the number of solutions of g to the p equals one and now we see this is divisible by p and this is also divisible by p because it's a collection of groups of p solutions all of which are kind of cycles cyclic permutations of each other. So so this is divisible by p so we see that for any finite group so if p prime divides g then the number of solutions of g to the p equals one is a multiple of p well there's one solution which is just to take g equals one and all others have order p so the number of elements of order p is congruent to minus one mod p that means it's a multiple of p plus minus one and furthermore in fact you can do a little bit more than this because each element of order p if g is of order p it generates a group of order p and this group contains p minus one elements of order p so all the elements of order p in g form into sets of p minus one sets so the number of subgroups of order p is something that is minus one mod p divided by p minus one and you notice this is also minus one mod p so this must be something that is one modulo p so the number of subgroups of order p is always one plus a multiple of p quite often just one of course in any case the number of subgroups of order p is non-zero so there must be elements of order p okay now we can go back to groups of order 10 or rather groups of order 2p so suppose g is equal to 2p then we pick some element a of order p b of order 2 and then look at the subgroup a which is generated by a is a subgroup of order p and this is index 2 so it is normal in g and so we see that g is a semi direct product of a by b where this is this is a group z modulo 2z and this is a group z modulo pz so this is just like the case of groups of order 6 we now have to figure out always that a group of order 2 can act on a cyclic group of order p well well the cyclic group of order p has an automorphism group z over pz star and this is only two elements of order 2 and the reason for this is that if p is a prime then z modulo pz is a field under the usual addition and multiplication modulo p now if you've got a field a polynomial of degree n has at most n roots so if we're trying to solve the equation x squared minus one is equivalent to zero mod p it has at most two roots and these roots are obviously plus one and minus one unless p equals two in which case there's only one root so there are exactly two automorphisms of order two of z modulo pz unless p is equal to two which is a kind of case we've already done and by the way you have to be a bit careful here because z modulo nz is not necessarily a field may have more than two solutions of x squared equals two for instance z over eight z we have x squared equals one if x is equivalent to one three five or seven mod eight so so z modulo eight z actually has four square roots of one which is causes endless problems z modulo eight z is of course not a field and eight isn't prime so if something isn't a field a polynomial of degree two kind of more than two roots anyway to get back to our group um we've got a is of order p and it can be is acted on by z over two z which is our group b of order two and the action um um that there are two possible actions so a is generated by an element a with a to the p equals one and b by an element of order two and we can either have the trivial action where b acts as the trivial automorphism of a so b a b to the minus one equals a and this implies that our group of order two p is just a times b which is z over pz times z over two z which is isomorphic to z over two pz and is a cyclic group on the other hand we could have the non trivial action where b a b to the minus one is now a to the minus one so this gives us the um um um dihedral group of order two p so the dihedral group is just the group of all symmetries of an n sided polygon so if we take n to be five just for definiteness the um dihedral group of order two times five has a rotation by um one nth of a revolution so this is the element a and we can also reflect in some line through the diagonal so the element b is going to be reflection like this and you can check that um um have i got a and b the right way around yes um b a b to the minus one is equal to a to the minus one so if we reflect and then rotate and then undo the reflection we've actually rotated in the opposite direction um so next lecture i'll be saying a bit more about dihedral groups