 This lecture is part of an online mathematics course on group theory, and we'll be about the coxota Todd algorithm So Last lecture where you were discussing the symmetric group SN and we saw that it was generated by the transpositions one two two three Three four and so on to n minus one n Where this was the element that swaps one and two and fixes everything else and so on And what we want to know or what are the relations between these transpositions? So let me call this a b c and so on just for simplicity And then we notice that a squared equals one b squared equals one c squared equals one and so on it's pretty obvious if we look at a b Um So I'm up a b if we look at the product of a and b. This is a three cycle So a b cubed equals one and similarly b c cubed equals one Cd cubed equals one and so on if we look at a c This is a product of two two cycles that are disjoint. So their product has ordered two. So ac squared is equal to one And similarly ad squared equals one Bd squared equals one and so on and coxster invented a very Compact way of writing down relations like this in the form of a coxster diagram So the coxster diagram of SN looks like this with n minus one points and there is one point for each relation so this is a point for the relation a b c d and so on and The coxster diagram indicates the following relations. We have a relation x squared for each point x and We have the relation x y cubed equals one if the point x is joined to the point y and x y squared Equals one if x is not joined to the point y so you can see that These relations here are just just correspond to the relations from this coxster diagram And and the question is are there other relations Well, of course, there are lots of other relations because we can just Derive lots of relations from these for instance a squared times b squared equals one But what we mean is are there any other relations that you can't derive from these relations here in other words if we take these generators and Force these relations on and no others. Do we end up with the symmetric group? So this is called defining a group by generators and relations I mean people sometimes prove that there is a group defined by generators and relations But it's this kind of obvious you just take all words in The generators and put the equivalent relations on them saying two are equal if if it's a consequence of the relations and so on So let's look at the group Generated by these elements subject to these relations Then there's a map from the group To the symmetric group and this is on to because we showed that SN is generated by these relations And it's well defined because SN satisfies the relations So we want to know what is the kernel of This map and what we'd like to do is to show the kernel is equal to one so that the group is actually isomorphic to the symmetric group and the coxtot odd Algorithm is a way of actually Proving things like this So so it's enough to show That the order of G is at most equal to the order of the symmetric group because Because that that since this map is known to be on to if they have the same orders. This must actually be an isomorphism So instead of doing this abstractly, let's do it for the particular case of s5 So s5 Has the following coxtot diagram. So it has four elements, which I'll try and color in because it's Let's call these a b c and d And what we're going to do is Let's take the group with these Generators and relations and we notice it's got a subgroup It's called that so this it's got a subgroup um h Which is presumably going to be s4 and we know by induction that the order of h Is at most the order of s4 Which is 24. So what we want to do is to show that um the order of Actually, we don't notice s5 yet. So we better not call it a s5. Let's call this group g generated by these we want to show um, is the order of g divided by the order of h at most equal to five And if we can show this then we will have shown that the order of g is at most 120 So how do we do that? Well the idea is um, we construct the permutation representation on cosets Of h by brute force So um, this this will be an action of g on something on a set And this contains a point fixed by h So we're just going to try and Um write down the representation So here's the point we start off by drawing the point fixed by h and um Now let's show the action of the various generators on it. I'm going to just write out again quickly what these generators are so this is the yellow one and the red one And the green one And the blue one and what do we know about it? Well, it's fixed by h And h was this subgroup here So we can write down the action of these three elements on it the yellow and the red And the green Point just map this point to itself. So these are lines with arrows on except i'm not bothering to draw the arrows Since because these elements of order two it doesn't matter which direction you go in Well, the trouble is we don't know what the action for blue element on this point is so let's just draw it So the blue element is going to take this point to another point And now we try and figure out What the action of the yellow red and green elements are on this point here? Well, how can we do that? Well The yellow commutes with blue. So if we do blue yellow blue, it's the same as doing yellow So yellow must actually fix that And similarly red commutes with blue. So red fixes that And green doesn't commute with blue. So green must map this point to another point And now we try and figure out what things do to this point. Well, yellow commutes with green So if we go green yellow green, that's the same as yellow. So yellow does that and Green Green doesn't commute with blue. But if we do green blue three times to get the identity So let's see we do green blue green blue green and that must be the same as blue. So blue does this And Then what does red do to this? Well red doesn't commute with green So we can't reduce that red does this so red must map this to another point like this And then we try and work out what blue does or blue commutes with reds. So it goes like that and yellow Um We haven't got to yellow yet. Um Um, which green green yellow. So green. Oh, yes. So um Green and red have ordered three. So if we do red green three times we get red green red green red green So green must go like that and yellow Doesn't commute with red. So yellow doesn't we can't reduce the yellow maps this point to itself. So yellow must map this to yet another point And now Yellow commutes with blue. So blue must map this point to itself And green commutes with yellow. So green maps that to a point and red yellow cube is equal to naught So we go yellow red yellow red yellow and that must be go to red. And so Our conclusion is that the permutation representation must look like this Um, because we've we've now found out what every generator does to every point and we can see that it has five points so g over h Must be equal to five and notice that we've got a completely explicit representation of g on five points We just say what what each of these generators do so, um So the order of this group g Is that is at most five times the order of h which is at most 120 So so because g maps on to s5 of order 120 We see that g is actually isomorphic to s5 And if you look at this you can see exactly the same thing Works for all symmetric groups instead of getting a line of Five points for the symmetry group s5 We would get a line of n points for the symmetric group s n and the the diagram would kind of look a bit like this um so, um I'll do a few more examples using the cops of ton algorithm to show slightly more complicated things that can happen so, um, let's look at the following Diagram, so here i'm going to take a yellow point a green point And a red point and this time i'm going to take a blue point and i'm going to put it here And i'm going to take our subgroup h to be generated by these elements here And now we can apply the cox to tod algorithm So we fix a point This is fixed by h So the yellow green and red things all map this to itself And now the blue the blue element has to take this to some other point And now blue commutes with red so Red must map that point to itself and blue commutes with yellow So we get something there, but blue doesn't commute with green. So green has to take that to some other point And now we try and figure out what what happens to this. Well, um Green doesn't commute with red or yellow Um and or with blue Well, we can figure out what blue does to this because green blue green blue green must be the same as blue So blue fixes that but red and yellow Take this to two different points. So if we've now got a sort of branching out going on So now what's going on? Well, um well blue commutes with yellow so We get yellow blue yellow must be the same as blue and blue commutes with red. So similarly that goes like that And yellow green has ordered three. So yellow green yellow green yellow must be green and similarly you must get that and Um, but yellow must take this to some other point And now yellow commutes with red. So yellow red yellow must be the same as red So we must get that there and now i'm going to um um, let me see, uh Blue again maps that to that And we've got a green line here And i'll just do the last ones very quickly because i'm getting all of this um, we find red fixes that and yellow fixes that and Blue must map this to another point and if we work out we finally finish because all the elements map this point to itself so We have found that if this big group is g and this little group is h then g Over h has order exactly 1 2 3 4 5 6 7 8 Since h is the group s4 Of order 24 This shows the order of the group g Gen with given by these generators and relations is 8 times 24 Is equal to 192 Um, we could if we want to do the coxs at a Todd coxster algorithm where we start with the h being the trivial subgroup But if we did that we would have a graph with 192 vertices, which is rather cumbersome So so that's why we try and fix a big subgroup h. It's to reduce the number of vertices Okay, I now want to do a couple more examples where more complicated things happen So first let's look at this coxster group. So I'm going to take three elements a Sorry yellow red and blue And I'm going to say that so each of these three generators has ordered two and the product of any two has ordered three And I'm going to take my subgroup h to be Generated by these two elements here. So let's see what happens um So we start over the point here and this is going to be fixed by the red And the yellow elements so the blue element um goes off like that and Then we can get a yellow element and a red element do that and yellow times red cubed is three So we actually get a sort of hexagon of points here So we can fill in these points here and now um yellow blue cubed equals naught from which we see that blue maps that to that and maps That to that And what happens here? Well blue has to take these off two different points and blue yellow cubed Is one so we get a hexagon there and blue Red cubed is one so we get another hexagon here So we get some more points and From the fact that blue yellow cube equals one we see that yellow fixes that and from the fact that blue red cube equals one We see that red fixes that But red has to take these points off to somewhere Sorry that shouldn't be red that should be yellow and Then we get more hexagons like this And you may be wondering how long this goes on for and the answer is it goes on forever because um this group H actually is infinite index in the group g So this diagram just goes on and on and on so this is one slight problem with the cox to tod algorithm it um Sometimes if the index is infinite it just keeps on going on forever And you never quite know whether it's Going on forever or stopping at some point so actually what this group is is um An example of an affine reflection group suppose I draw three mirrors a blue mirror A red mirror rather a blue mirror and a yellow mirror With angles of making an equilateral triangle in the middle Then you can see that reflections in these mirrors satisfy these relations they Each reflections order two and the product of any two has order six so um you can sort of see these reflections in these mirrors just form an infinite group because if you form a triangular lattice like this You can see the reflections are taking this Triangle here To any one of this infinite number of other triangles So so the group g generated by these three reflections is indeed infinite So that's one thing you have to watch out for a bit with the cox to tod algorithm. It's not guaranteed to terminate Um, there's some other things that can sort of go wrong and I'll try and illustrate them. Um first of all, um I've been assuming all generators have ordered two in the examples so far So I ought to do an example where some generators have ordered three So let's take b cube equals one c cube equals one and bc squared equals one and try and figure out what group this is Um, and I'm going to take my subgroup h to be this group here And let's try and do the cox to tod algorithm So we start off with a point Fixed by b. Um, let me call say b is going to be the red element and c is going to be the blue element so I've said b is going to fix a point so b acts like this and I better put an arrow on it because now it is order three So it matters which direction you go along it and c now does this So we've got two more points here And now what does b do to this will be now As order three so I get two more points like this And we can say bc squared equals one so Here we've got a b times c so um, we have to get another point like this Uh, sorry, that should be the blue line picked up the wrong pen Um, and Um, then we see that um, um Which we're around it blue red blue Red has to be the identity which red commutes with blue So now we see we've got a problem because if we if we start here we get blue red. Sorry. Where was I? blue red blue red Well, now we suddenly rose we made a mistake Because this point has to be equal to this point. So we get some collapse and um, the three examples I did first of all We didn't have any collapses because I was rather careful About my choice But in general when you were working at the cox to todda algorithm You quite often find that two points that you thought were different Actually turn out to be the same and this is a real pain when you're doing it because it means you have to start and redraw The whole diagram again from scratch More or less or else you have to cross out things and it gets to be a bit of a mess So we discover that really what was going on was this um and now we have a triangle like that and now um Red blue red blue has to be the identity. So blue has to actually take that to that So we find that there are in fact only four points in the Cox to tod algorithm. So g over h Is of order four. So g has order at most 12 You might think that it has ordered exactly 12 because h is order three But we don't actually know h is order three. We're guaranteed that b cube equals one But maybe b is actually equal to one Well, in fact, if we look carefully we find that b is not equal to one because this actually gives us um an action If you look at the permutations Generated by these we actually get a group of order 12 acting on these four points. In fact, the group g is turns out to be isomorphic to the group a four of rotations of a tetrahedron Um, so that's the cox to tod algorithm And what it does is if you're given a presentation A set of generators and relations of a group. It allows you to calculate the index of some subgroup Generated by subgroup of the relations in the whole group Um provided this index is finite if it's infinite the the algorithm just goes on forever um Obviously doing this by hand gets a bit tedious if there are more than a dozen or two dozen Points in the graph. So it's implemented on a computer um This collapsing is actually really tricky to get right on a computer because When when two of the nodes end up collapsing you have to join them But that might cause further collapses because when you join two nodes you get further relations between the nodes and so you get a sort of cascade of collapses Which is a real nightmare to keep track of if you're trying to do a computer implementation of it Actually, it's rather a nightmare to keep track of even if you're doing it by hand to be honest if you would like an example to practice on The following example is none trivial take the following cox to diagram and take the subgroup h to be this And try and work out what the index Of h in this big group is you need a reasonably big piece of paper to be able to do that Okay, next lecture. We're going to go on past groups of order 24 and look at the groups of order 27 Which are for some unfortunate historical reason