 In this lecture, I'll be showing how to calculate the character tables of some small finite groups. So the first one we're going to do is the dihedral group of order 8, which is just a group of symmetries of a square. So the first thing you do is you write down its conjugacy classes. The dihedral group of order 8 is five conjugacy classes. There's the identity element. Then you can swap both pairs of diagonal corners. So well, that's a 180-degree rotation. Next, you can swap the left and right-hand sides, or you could also swap the top and the bottom, and these are conjugate. Then you can swap, reflect in the diagonal of the square, and there are again two of these. And finally, you can rotate by 90 degrees in one direction or the other. So there are the five conjugacy classes. And in order to write down the characters, the first thing to do is to write down the one-dimensional characters. So we know that dih modulo the center is isomorphic to z over 2z squared, the four group. And we know the character table of the four group. It's just got four one-dimensional characters. So we just write those down. So the four characters will all be trivial on the center because that's in the kernel of this map. And then we will get the character table of the Klein four group, which sort of looks like this. So we know there are five conjugacy classes, so there should be five characters. And we found four of them with no effort at all. So there's just one more, and there are several ways to get the last one. We could just use the fact that dih is a two-dimensional representation, but it's even faster to use the orthogonality relations. So we know the sum of the squares of the degrees must be eight. So we have one, two, one, two, three, four plus something squared is eight. So this must be two. These two comms are orthogonal, so that must be minus two. These two are orthogonal, so that zero and similarly these two are zero. So here is the character table of the dihedral group of order eight. The next example we will do is the alternating group A4 of order 12. So this is just the symmetries of a tetrahedron. And again, we start off by writing down its conjugacy classes, and it's got four conjugacy classes. It's got the identity element. Then it's got a conjugacy class consisting of a product of two transpositions, and there are three of these. Next, there's a cycle, one, two, three. And you might think there are eight of these. Well, actually only four in one conjugacy class, and there's another cycle, one, three, two, which is not conjugated to it. So these are actually conjugated in S4, but they're not conjugated in A4, and it's a little bit confusing trying to work out what the conjugacy classes and alternating groups are. So as before, we start off by finding the one-dimensional characters. And you know A4 has a subgroup, consists isomorphic to z modulo 2z squared, consisting of these three elements and that element. And the subgroup is isomorphic to z modulo, sorry, the quotient is isomorphic to z modulo 3z. So there are three one-dimensional characters of this, which become three one-dimensional characters of A4. So we, these characters are all equal to one symbol of this map. So we just get ones there and then we get the trivial character and we get omega, omega squared, omega squared, omega, but omega is a cube root of one. So omega cube equals one, omega is minus one plus root minus three over two. And now we check and see what we've got. There are four conjugacy classes. So there should be four characters and we found three of them. So as before we can find the fourth one just using orthogonality relations. So the sum of the degrees is 12. We've got one, two, three. So that must be a three. The first two columns are orthogonal. So that must be a minus one. These two columns are orthogonal. So that must be a zero. And these two are orthogonal. So that must be a zero. So here's the character table of the alternating group A4. Incidentally, another way of getting this fourth character is to write down the permutation representation of A4. So you remember the character of this is just given by the number of fixed points. That would be four, zero, one, one. And the permutation representation is not irreducible. You can see it as norm two if you take four squared plus four times one squared plus four times one squared. It's twice the order of A4. So it must split as the sum of two representations. And you see it is in a product one with the trivial representation. So it's the trivial representation plus something else. And you can find this by just subtracting the trivial representation from this representation. And we get three minus one, zero, zero. So we could calculate this directly rather than just using the orthogonality relations. So the next example we'll do is S4. And here there are five conjugacy classes. So you remember conjugacy classes of symmetric groups are just given by cycle shapes. So the cycle shapes are these elements of order one, two, two, three, and four. And we should just write down how many there are in each conjugacy class that we can do orthogonality relations. And as before, we first look at one dimensional representation. So there's the trivial one. So let's call this trivial. And there's the sign representation. You remember symmetric groups have a map to the group of order two depending on whether elements are even or odd. So that gives us a representation one minus one, one, one minus one. So this is the sign representation. Next, we can take the permutation representation, which I'm going to draw in orange because it's reducible. So we look at the number of fixed points of these and we get four, two, zero, one, zero. And then as before, we notice that this has norm two. So it's the sum of two representations, one of which is the trivial representation. So we subtract that and we get three, one minus one, naught minus one. So this is perm minus one dimensional representation. There's another one we can find by taking tensor products. So if we take the tensor products of this representation and this representation, we get three minus one, minus one, naught one. So this is three tensor with the sign representation. So let's stop and see what we've got. We've got four irreducible representations and we should have five. So we can get one more by using the orthogonality relations. So we should have the degree here. Well, we have one squared plus one squared plus three squared plus three squared plus something squared is 24. So this is two. And by using orthogonality relations is before we see the final representation as this. There's another way of getting this representation. We can map S4 onto S3. And S3 has a two-dimensional representation. You remember the character table of S3 looks like this. I worked it out in an earlier lecture. And this representation here of S3 becomes a two-dimensional representation of S4, which is this one here. Notice, by the way, that you can compare the character table of S4 with the character table of A4. So you see every representation of S4 restricts to a representation of A4. So these two representations restrict the same one-dimensional representation of A4. Similarly, these two representations restrict the same three-dimensional representation of A4. However, this representation, when you restrict it to A4, it becomes reducible and splits into the sum of these two one-dimensional representations. So when you restrict representations, sometimes two representations become the same and sometimes a representation becomes irreducible and splits. Before going on to S5, which is our next one, I want to just have a quick look at the symmetric square and the alternating square of a vector space. So you remember that if you take the tensor product of a vector space with itself, it splits as a sum of the symmetric elements of the form A spanned by elements of the form A tends to be plus B tends to A and anti-symmetric ones spanned by things of the form A tends to B minus B tends to A. And if V is a representation of a group, then both of these will be representations and we want to work out what their characters are. Well, first of all, we have the obvious relation to the character of S squared of V plus the character of lambda squared of V is obviously the character of V tends to V, which is the character of V squared. In order to look a bit further, suppose we've got an element g acting on V with eigenvalues lambda 1 up to lambda n. Then the eigenvalues on V squared are just lambda i, lambda j for i less than or equal to j. This is on the symmetric square of V. On the alternating square of V, the eigenvalues are lambda i, lambda j for i less than j. And the trace of g squared on V is the sum of its eigenvalues, which are lambda i squared for integers i. So we see that the trace of g on lambda squared V and the trace of g on symmetric square of V and the trace of g squared on V are related because the trace of g on S squared of V is equal to the trace of g on lambda squared of V plus the trace of g squared on V. So we get the relation that the character of the symmetric square of V at g is equal to the character of the alternating square of V at g plus the character on V of g squared. So we've got two linear relations relating the symmetric square and the alternating square. And by adding and subtracting these, we can find the following formulas for the character of the symmetric square of V. So this is just the character on V of g all squared minus the character on V of g squared plus over 2. And the character of the alternating square of V is the character of V on g all squared minus the character on V of g squared all over 2. By the way, this thing here isn't actually the character of a representation in general. It's something called an Adams operation. And in case you're wondering, you can also write down higher symmetric powers and higher alternating powers of V in terms of various other Adams operations. But this is by far the most useful one. Okay, having done that, we can now do the symmetric group with five elements, which I'd better be a bit careful about so that I don't run out of room. And that's before we start off by writing down its conjugacy classes. There are seven of these. I'll write them down in order of their order. So the number of these is 1, 10, 15, 20, 30, I think 24, 20. I hope that adds up to 120. And we can start in the same way we did with S4. So there's the trivial representation. We can take the sine representation, which looks like that. And we can take the permutation representation minus the trivial representation. So we get 4, 2, 0, 1, 0, minus 1, minus 1. And then we can tensor this representation with this representation. So we get 4, minus 2, 0, 1, 0, minus 1, 1. So what have we got so far? We found four irreducible representations, and there are seven conjugacy classes. So we need to find another two. Well, we can find the alternating square of the first four dimensional representation. And we can work it out using the formula I gave, provided we know what the square of every conjugacy class is. So what I'm going to do is I'm going to do a little orange line by showing you what the square of each conjugacy class is. So this conjugacy class squares to itself, and this one squares to that, and this one squares to itself, and this one squares to that. So now I can work out the alternating square of the four dimensional representation. So we get 4 squared minus 4 over 6, which is over 2, which is 6. Here we get 2 squared minus 4 over 2, which is 0. And then we get 0 squared minus 4 over 2, which is minus 2. And here we get 1 squared minus 1 over 2, which is 0. And we go on like this, and we get 6 naught minus 2, 0, 1, 0. And we can also do the symmetric square in a similar way, where we get 10, 4, 2, 1, 0, 0, 1. And now let's check to see whether these are irreducible. Well, you can check that they're irreducible by checking the norm, and you can see this is norm 6 squared plus 4 times 15 plus 1 times 24, which is 120. So this is irreducible. However, this one is not irreducible. The norm is 360. So it must be the sum of three irreducible representations. And we can find two of them because it's in a product with this representation as 1, and it's in a product with this representation as 1. So what we do is we take 10 minus 4 minus 1, and we get a representation 5, 1, 1, minus 1, minus 1, 0, 1. So we found six of the seven representations. And finally, the seventh representation, there are about half a dozen ways of doing it. We could use orthogonality relations, or we could tensor this representation with the sine representation to get 5 minus 1, 1, minus 1, 1, 0, minus 1. So we get the symmetric square of this and so on. There are all sorts of lots of ways of doing it. So here in black, we've got the character table of S5. And we should cross out this one because it was just an intermediate piece of scaffolding. Now, finally, let's do the representation of A5 and then compare it with S5. S5 is a subgroup of index 2 of S5. So any representation of S5 becomes a representation of A5. So let's first write down the conjugacy classes. When we take the even conjugacy classes of S5, so there's 1, there's 1, 2, 3, 4, there's 15 of these, and there's one of these, and there's 20 of type 1, 2, 3, and then we have type 1, 2, 3, 4, 5, and you might think there are 24 of these. Well, you have to be a little bit careful because there are actually 12 in this conjugacy class and this conjugacy class splits into two different conjugacy classes in A5. Splitting of conjugacy classes of symmetric groups into conjugacy classes of alternating groups is a bit messy. It's sometimes split and there's a rule for this which I can never remember. So as before, we can now take the irreducible representations of S5 and just restrict them to A5 and this makes it trivial to write down the characters. We get the trivial representation. We get a four-dimensional representation and we get a five-dimensional representation and we get a six-dimensional representation. Okay, well, that gives us four representations and we should have five. However, there's a bit of a problem. Some of the squares of the irreducible representation should be 60 which is the order of A5 but if you take the sum of the squares of these, it's bigger than 60, it's 78. Something has gone wrong. So you can see this one is irreducible. It's got norm 1, this is norm 1. This is norm 1 and is irreducible. You can see this one is not reducible because I've written it in orange. Plus it is norm 2, so it must be the sum of two irreducible representations. If you take its inner product with these three representations, you find it's orthogonal to these three, so it doesn't involve these three. It must be the sum of two other representations and you can figure out what these are quite easily by using the orthogonality relations together with the fact that the two missing representations must sum to this. So the degrees must add up to six and their squares must be 18. So the degrees must be three and three. And then here the sum of the squares of the elements in this column must be 60 over 20 which is three and we've already got three there. So these entries must be zero. These entries have to be integers adding up to minus two and some of the squares must be 60 over 15 so they have to be minus one. And now we run into a slight problem with these entries. So let's provisionally call them A and B. And I'm going to call these B and A because there's an automorphism switching these conjugacy classes. There must be an automorphism switching these columns which must therefore also switch these rows. And if you use the orthogonality relations you find that A plus B equals minus one. A times B conjugate is equal to minus one and A A bar plus B B bar equals three. And you can solve these equations for A and B and you find one of them is minus one plus root five over two and the other one is minus one minus root five over two. So little two by two blocks and character tables with a quadratic irrational are very common. In fact, we saw an earlier example of them with the group A4. You see there's another little two by two block of quadratic irrationals. Quadratic irrationals seem to be in practice the most common ones you've come across. You occasionally come across larger blocks of algebraic numbers of degree greater than two but at least if you're doing interesting finite simple groups two by two blocks are the most common. Okay, I think that's enough examples of character tables for the moment. Next lecture we'll be discussing real representations of groups.