 This lecture is part of an online mathematics course on group theory and will be about the Jordan-Holder theorem So if we've got a group G We can break it up Into simple groups as follows. So we can find a chain of subgroups one is equal to G naught contained in G1 to contain in Gn By the way, I should say I'm considering all groups to be finite in this lecture such that GI is normal In GI plus one and GI plus one over GI is simple Meaning it's got no normal subgroups other than the whole group in itself And this is quite easy because if G is simple We're done if it doesn't we pick a normal subgroup to be one of these groups GI and then we just sort of continue Applying this to the normal subgroup and the quotient by the normal subgroup. So this is called a composition series and We can give some examples of composition series. So from the previous lecture We studied several groups of order 120 and these have composition series as follows first of all If we take the binary icosahedral group Then it has a composition series that looks like this and the composition factors Which mean these quotients well, this factor is obviously a group of order two and this group is the group of rotations of an icosahedron or the Quotiently the group a five Another group was the symmetric group s5 and this has a composition series which looks like this So this time this quotient is the group a five and this quotient is order two So you see we have two groups With two composition series and each composition series has a group of order two and the group of order a five only they're put together in different ways We can have some more like this for instance We can take the group z modulo 2z and this is contained in a five times z modulo 2z Or we can take one contained in a five Contained in a five times z modulo 2z and again this quotient here is order two And this is a five and this quotient here is a five and this quotient here is order two So we see that the composition factors don't determine the group here We've got two and a five as composition factors and here we've got two as a five and the group doesn't determine the order of the composition factors because in this group we can first do two then a five or do them in the other order for these two the order is fixed so There's a there's a lot of variation in what composition series can look like By the way, I should point out that each GI is normal in the next one This doesn't imply that G1 says normal in GN for instance if we look at one contained in Z over 2z contained in Z over 2z squared which is contained in the tetrahedral group a four Then each of these subgroups is normal in the next one But Z modulo 2z is not normal in in a four It's an example of something called a subnormal subgroup if you've got a chain of subgroups such that each is normal in the next Then we say that the one at the bottom is subnormal in the top So So there's a lot of variation in the composition series However, you see in this composition series, although the two and the a five occur in a different order They each both occur once and this is illustrates the Jordan-Holder theorem so these were two 19th century group theorists, which says that any two composition series of a group G Have the same number of each simple factor and We're going to prove this theorem by sort of driving a taxi cab around in a in a rectangular array So the idea is as follows supposed to have got two composition series So one is equal to a naught contains and contains an a m contains and g equals G with a i plus one over a i simple and We take one is equal to b naught contained in b1 Contained in bn equals g with each of the bi plus one over bi simple And we want to show these quotients are the same as these quotients possibly in a different order so what we do is we form this big array we take a M intersection bn, which is just equal to g and then we have a M intersection bn minus one go all the way down to a m Intercept b naught and then we have a naught intersect b naught here This is just one of course, and this is just the whole group G and Then we have a naught. So a 1 intersect b naught And so on so so we have this sort of two-dimensional array of groups wherever I got to this should be a naught intersect bn and We notice that each of these groups is a normal subgroup of the two groups above above it So what we have is this sort of two-dimensional array of groups as a sort of think of this as a rather unimaginative if you planned town where each of these vertices is some group and each of these lines is a quotient which is so which is a quotient of two groups and Each quotient is either a simple group or or one so so line is simple or Or one so it's so it's quite common for for these two groups here to be the same for example in this diagram here all these groups on the left hand column are the Trivial group so they're all the same as each other and we can imagine a taxi cab wandering through this town and The taxi cab is going to go from the this corner to this corner. So for instance, it could go along there And as the taxi cab goes along we look at the different Quotients it passes through so this taxi cab is going to pass through one one one one And then it's going to pass through these quotients b1 over b naught b2 over b1 and so on So this taxi cab route gives us the second composition series on the other hand we could also drive like this and this would give us Quotients one one one one and then we would get a one over a nought and so on and We can have other routes for instance the taxi cab could do something like this and We would again get some other quotients Along this route and what we want to do is to show that every taxi cab route has the same quotients Possibly in a different order and that would be enough to show that These two composition series have the same factors So how can we do this? Well, you notice that we can get from any taxi cab route to another one just by Repeatedly flipping it on some squares. So if we look at something like this We can have two taxi cab routes. So it might come along like this and go like that And we might have another one Which goes Like this so it's just the same except we've changed this square here and Obviously by repeating this operation we can get from any taxi cab route to any other taxi cab route We're assuming the taxi cab driver is honest and goes by the shortest route and doesn't stop going around in circles or something Which wouldn't make a whole lot of sense So let's look at what happens here well What we're getting is a group a i into sec b j a I minus one intersect b j a I Intersect b j minus one And a i minus one intersect b j minus one and We may as well quotient out by this group here since it's a normal subgroup of everything so what we have is is a group x and Certain subgroups a and b Which intersect in the identity? So we've got a particularly simple arrangement and as before each of these quotients is either simple Or is the identity and now let's look at what happens Well, if a equals b Then we can look at what the quotients are Well the quotients here are just a And the quotients here are just x over a So what happens if we go by these two different routes? Well, if you go this way We get x and x over a and if we go this way We again get a and x over a so the factors we get don't change what happens if a is not equal to b Well a and b a simple normal subgroups of x From which it follows easily that x must actually be equal to a times b and what we're getting now Is this quotient is a this quotient is b this quotient is b and this quotient is a So if we look at these two routes See this route we do a followed by b and this route We do b followed by a so Every time we change the taxi cab route We either keep the order of two factors the same or we switch them so so any two taxi cab routes have the same factors In possibly different order So this proves the Jordan holder theorem Which says that a number of times any given simple group occurs in the composition series doesn't depend on the composition series so By the way, this also works for groups with Operators That means it works for groups with some extra automorphisms on the groups and the main example of this is modules over a ring So we can think of a module over a ring as being a group with various operators So what this says is if you've got a module over a ring It can be broken up into simple modules and the number of times each simple module occurs is independent of how you break it up So This kind of in some ways reduces the study of groups to the study of simple groups Well, not entirely because as we've seen if if you've got various simple groups There can be many ways to stick them together into a group But some problems can be reduced to the problem of Solving simple groups. So now we want to know how do we classify Simple groups Well, the simple finite groups have been classified. There are 18 infinite series and two typical examples of this are the groups alternating groups a n for n greater than or equal to five and the Projectors special linear groups n over fq. This is provided n is greater than or equal to two and n is greater than or equal to three if Q is two or three and Various other series and then there are 26 sporadic groups Which don't fit into this classification. So the smallest is m11 which has about 7 9 2 0 elements and the biggest is the monster Which has about 10 to the 54 elements The proof of this is Maybe the most difficult Published proof in mathematics. There are various Computer proofs which are a good deal longer where a computer checks a gazillion different cases but as far as journal pages is concerned the classification of Simple groups is the most difficult and longest theorem. How long it is nobody really knows I've seen estimates of about 10,000 pages Or 20,000 or something that the trouble is this proof is scattered around so many journal pages that nobody really Knows quite how long it is If you want to see a summary of it There's a nice summary of it in these three books. There's a book by David Gorinstein the finite simple groups and introduction to their classification and then he had Follow-up he was going to publish three volumes on this so he had classification for non characteristic to type Unfortunately, he died before he could finish this but ash back a lay in Smith and Solomon finished it for him and Did the ones of characteristic to type so these three volumes? Total more than a thousand pages and they basically just give a summary of the proof without giving the detailed proofs so that the proof itself is I think there are very very few people who know most of it So how does the proof work? Well, you can't really summarize a 10,000 page proof in a sentence But I'm going to try it anyway. It works by looking at the central isers of an involution So this is the key theme in the whole proof so you remember an involution is an element of order to and Central is it is just the subgroup of things that commute with it and Brower pointed out that if you know the central is of an involution of a simple group Then you can determine it up to a finite number of possibilities So very roughly the idea of this proof is you pin down the possible structure of the central iser of an involution of a simple group And then given the central iser of an involution you try and identify the group In order for this to work a group actually has to have an involution If it doesn't then you can't really pin it down by looking at its the central iser and a group has an Involution if and only if it is even order so the very first step in the proof is to show that any simple group has even order and This is by itself one of the most difficult themes in mathematics It was originally proved by phyton Thompson in this rather famous paper solvability of groups of odd order So if a group of odd orders solvable then it can't be simple. So this shows that every Simple group has an involution and this single paper is it's about 250 pages long and it's Incredibly difficult to read in parts. So here's a typical page of it. You can see it Contains all these rather complicated generation of relation calculations I've no idea what's going on here I've tried reading parts of this and my experience is it takes me a good hour to understand each page of this paper So it would take 300 hours of hard work to understand proof completely Okay, that's all I want to say about simple groups for the moment So the next lecture will be studying outer automorphisms or more precisely the outer automorphisms of the symmetric group s6