 This lecture is part of an online mathematics course on group theory and will be about Burnside's Lemma. So for motivation for Burnside's Lemma we will look at the following problem. Arrange eight rooks on a chessboard with one rook in each row and column. In other words the rooks have to be not attacking each other so we've got an eight by eight board looking something like this and what we have to do is just select a square from each row and column so that none of these are in the same row or the same column so we might do it something like this so that would be one possible arrangement and the problem is how many ways are there of doing this and this is rather easy because there are eight ways we can choose the rook in the first row and once we've chosen the rook in the first row the rooks in the other row can't be in the same column so there are now seven ways to choose the rook in the second row and again we have to delete this column so now there are six ways to choose the rook in the third row and you can see how this is going we get eight times seven times six times five times four times three times two times one which is four oh three two oh possibilities so that's rather easy. A more interesting problem is how many ways are there of doing this optosymmetry so if we rotate the board we get a different arrangement but we don't want to count that as different we want to count it as being the same so we can ask how many optosymmetry and this problem was solved by Lucas in his book theory of numbers it's volume one page 222 if you want to look at top um so um the the key point is to use group theory so your first guess might be just to take the number of arrangements and divide by the order of the group of symmetries so so the number of symmetries of an of a square is just eight so you might try dividing that by eight because you might think each arrangement is isn't a group of eight the trouble is there are some arrangements that are symmetric under some symmetries for instance if we took all the rooks lying down the diagonal then we wouldn't get eight different arrangements from that we would only get two so you have to somehow account for the fact that some of the arrangements are symmetrical so to do this we're going to use something called Burnside's lemma so what we have is we have a group uh the dihedral group of order eight acting on a set of size 40320 and we want to know how many orbits you remember two elements of a set actually run by the grouper in the same orbit if there's an element of the group taking one to the other so asking for the number of different arrangements optosymmetry is just the same as asking for the number of orbits or generally if we've got a group g acting on a set s we can ask how many orbits there are and Burnside's lemma says the number of orbits is equal to the average number or fixed points so this is the average we we sum over one over the order of the group g of um s g where s g is the number of elements of s fixed by g fixed by the element g in other words this just means g s equals s um so Burnside's lemma um is named after um this guy Burnside and comes from his book theory of groups of finite order you can see the his statement of the lemma here let me just magnify it a bit so you can read it so here he's saying the sum of the number of symbols left unchanged by each of the permutations of a group of order n that that's taking a sum over all elements of the group of the number of fixed points and he's saying that's equal to t times n where t is the number of orbits and n is the number of elements of the group um on Burnside's lemma what was not originally due to Burnside it seems to mean one of these lemmas that was discovered by absolutely everybody working on group theory in the 19th century so it was certainly known to people like Cauchy and Frobenius before Burnside used it anyway um it's fairly easy to prove so let's just look at the proof what we do is we look at at the number of pairs g s with g s equals s so here g is in the group g and s is in our set s and we can count this in two different ways first of all um we can count by summing over um elements g so we just have to sum over all elements of g and for each elements of g the number of possible elements s is just the number of points of s fixed by g so remember this is points fixed by the element g on the other hand we can count the number by first summing over all elements s so we sum over all s and for each element s um we take g s where this is the um a g s is the subgroup fixing fixing fixing the element s and um g s is just the um size of g divided by the size of the orbit of s remember if if g acts on a set um then g is equal to the size of the set times the subgroup fixing one element so we can write this as sum over all orbits um of the size of the orbit times the order of g that should be the order of g divided by the size of the orbit so instead of summing over elements of s we're going to um sum we're going to divide the elements of s into orbits and we just sum over the orbits and for each orbit we take the number of elements in the orbit and multiply it by this element here and this is just the sum over all orbits of the order of g which is equal to g times the number of orbits and if you compare this expression with this expression all you have to do is to divide both sides by the order of g so the number of orbits is equal to one over g times sum over g of s to the g which is the average number of fixed points um so now we're going to apply this to our problem so first of all we have to work out what are the elements of the group d8 so we said this has eight elements and we can just write them all out as follows first of all there's the identity element which does absolutely nothing to the square and then we've got an element that sort of flips it like this there's another element which is very similar which kind of flips it about a horizontal instead of a vertical axis so these are flips about horizontal vertical axes and then we can also flip it around a diagonal axis so we can either flip these two corners or we can flip these two corners and we can also rotate it so we can rotate it clockwise or anticlockwise and finally there's one more thing we can do we can just rotate it by 180 degrees which means we flip these two corners and these two corners well so that gives eight elements but we don't really need to do all eight elements because if you look at these two elements here you can see that these are really equivalent they're going to have the same number of fixed points and similarly these two have the same number of fixed points and these two have the same number of fixed points so we can divide the elements up into well these are called conjugacy classes so what is a conjugacy class well an informal definition means we say two elements are conjugate if they sort of look the same okay this is this is a sort of obviously not a strict mathematical definition but it works most of the time if two elements kind of look the same then they're called conjugate well let's have a more precise mathematical definition because we obviously can't prove theorems about things that sort of look the same so why do these two sort of look the same well um so the first one is flipping around a vertical axis and the second one is flipping around a horizontal axis and you can turn the vertical axis into a horizontal axis by applying one of these transformations here so if I if I pick an element suppose I call this element g and this element a and this element b you can see you can get a by first rotating by 90 degrees and then applying b and then undoing the 90 degree rotation you had so so these two elements look the same because they're related like this and this is this is the official definition of conjugacy so we say a and b are called conjugate if a equals g bg to the minus one for sum g in the group g and you can see this is an equivalence relation using the axioms for a group and the equivalence classes are just the conjugacy classes so this reduces the amount of work we have to do instead of looking at eight elements we only have to look at five um reducing from eight to five isn't that big a deal but for bigger groups the number of conjugacy classes can be far less than the number of elements of the group so this can save a lot of effort um so now let's um solve our problem by summing over all the elements of g so let's write out the elements of g clean piece of paper so here are the conjugacy classes and here are the number of elements in the conjugacy class so first of all we've got the identity element and there's one element in the conjugacy class and here's the number of fixed points so the number of arrangements fixed by the identity we've worked this out before it's just eight factorial which is four zero three two zero um next we have the conjugacy class consisting of things like this or this and here there are two elements in the conjugacy class and the number of fixed points is pretty obviously zero because if we've got a rook in the first row then under this transformation there'd have to be another rook in the first row which isn't allowed so we just get zero there that's nice and easy what about this one so i'm going to flip two opposite corners and now there's one element in this conjugacy class what's the number of fixed points well we can choose any we can we can choose a rook in the first row that gives us eight possibilities well then that will also determine a rook in the bottom row so so we've we sort of determined two rooks and those will knock out two columns for the second row there are then going to be six possibilities and the rook in the second second row will then determine a rook in the seventh row and that will again knock out two columns so in the third row we have four possibilities and in the fourth row we have two possibilities so we get 384 ways of doing that um if we rotate by a quarter of a revolution or back the other way there are two of these and now how many well we first put a rook in the first row well it can't be in the corner because then there would have to be another rook in this corner which would be in the same row so they can't be in the corners but they can be anywhere else which gives us six possibilities and if we choose one of these we get four other rooks which leaves us four rows and four columns and again we can't choose a rook in the corner of those four columns so we have two possibilities so altogether there are 12 ways of arranging like this final one is a little bit trickier to work out so there are two conjugacy classes and now let cn be the number of arrangements on an n by n board then we see that cn is equal to cn minus one plus n minus one cn minus two the reason for this is we can choose the first rook to be either in this corner let's suppose we're doing we're doing this arrangement in which case we then have to arrange c we then have to arrange n minus one rooks in down here or we can choose it not in the first corner so somewhere else so there are n minus one other ways to put it in the first row we then it then determines another rook in the first column so we knock out two rows and two columns and we have cn minus two ways of arranging the rooks in the remainder so from this we can work out what cn is we have c0 c1 c2 c3 c4 c5 c6 c7 and c8 and c0 and c1 are easy they're both just one so we can work these out as one two four ten twenty six seventy six two thirty two and seven hundred and sixty four so here we have seven hundred and sixty four possibilities and now we just multiply the number of elements in each conjugacy class by the number of fixed points so we get 40320 there we get zero there we get 768 here we get 24 there and we've got 1528 there now we add these up and we get 42256 then we take 42256 divided by the order of g which is eight and the final answer we get is five two eight two so this is the number of ways of arranging eight rooks on a chessboard optosymmetry so the next lecture we will look at groups of order nine