 So, yeah, so last time we talked about group operations, so that, so we have a group G acting on a set X, so this is a map from G times X to X, so G pair of G and X is mapped to something called G times X, which satisfies two actions, namely that if we kind of twice pick these two elements in this way, we compose the action twice, it's the same as if we multiply the elements in the group and act by that, and then the identity element should act as the identity, so this is for all X and all G and H. And we had introduced two important concepts connected with group operations. One is the stabilizer and the other one is the orbit. So the stabilizer of an element will just be the set of all elements in the group which map it to itself. So this is GX, which is the set of all G and G, such that G times X is equal to X, and we also have the orbit which is just all the elements of X, so the orbit of an element X and X will be the set of all elements of X which we can reach by acting by elements of G on it. So this is G of X, which is a set of all G times X is G and G. So this is a subgroup of G and this is a subset of X. And we had seen that if we have a group operation, then we get a decomposition of X into this joint subset, so then the different orbits form a decomposition of X into this joint sets because they are actually equivalence classes of an equivalence relation. And we finished by the orbit stabilizer theorem which said that where we have a group G acting on a set X and we take R, a system of representatives for the orbits. So in other words, this is a set which contains one element out of each orbit and then we find that the number, so I want G X on X, and in this case the X is finite, then the number of elements of X is equal, this is kind of trivial, sum over all these representatives of the number of elements in the orbit of X. This is just the statement that X is the union, the disjoint union of the orbits and we can rewrite this as the sum over all X and R of the index of G in GX because one finds that the number of elements in the orbits is precisely the index of GX in G. So this is actually a very simple result but it's useful and therefore it's a theorem. So now we want to look at the special case of an action, namely the action of a group by itself by conjugation, group G on itself. So first what does it mean, conjugation? So we have G a group and we see that G and H in G are conjugated if there is an element A in, if there exists element A in G such that say H is equal to A G A to the minus 1 and you can easily see that this is an equivalence relation, this is a trivial exercise. And so we can look at the equivalence classes which are called conjugacy classes of elements are called conjugacy classes of G. So we want to say something about these conjugacy classes and we want to see that this actually is an operation of G on itself. So let's first define something more. So for an element X in the group the centralizer of this element consists of all elements commuting with it. So let X in G, so the centralizer of X is just, so I write it with Z of X to be the set of all G and G such that GX is equal to XG. So these are all elements which commute with our given element X. I have some simple remark. First the centralizer of any element X is a subgroup of G. That's very easy. Then I can look at the center in the exercise. I use the different notations C of G but I want it now with a Z. C of G is, we know it's a normal subgroup of G and this will always be a subgroup of the centralizer of any element X. That's obvious by definition. The center consists of all elements in the group which commute with all elements in the group. So in particular that's a subset of all those which commute with a given element X. And it's also obviously a subgroup. And finally we have an element X in G is in the center if its centralizer is the whole of G. Again this is trivial because the elements in the center are the elements in the group which commute with all elements in the group. And that's precisely what's being said here. Okay so this is really a remark. There's nothing to prove. Everything is directly from the definition. So it turns out that the conjugacy classes are somehow very important in variant of the group and if one wants to understand the group one has to understand the conjugacy class. We will not really do that. Maybe in the next semester you will also hear something about representation theory of groups and you find for instance there's a close relation between the representations of the group and the conjugacy classes of the group. But now we want to describe this and we want to describe the conjugacy action of a group on itself so that the conjugacy classes will actually be the orbits of the definition. So the conjugation or action of G on itself is just the obvious thing. This just is defined by G times H is equal to say G H G to the minus 1. Okay so the element G acts on the element H by this conjugation. So by definition the orbits are precisely the conjugacy classes. These equivalence classes are precisely the conjugacy classes. So the conjugacy class of an element G is the same as the orbit because you know after all this is precisely how it was defined. And we note also that the stabilizer of an element say X in G with respect to this conjugation action is precisely the centralize of X. No, because by definition the centralizer of X is equal to the set of all G and G such that G X G to the minus 1 is equal to X and this is obviously equal by multiplying by both sides by G and multiplying on this side by G to the set of all G and G such that G X is equal to X G which is and so now I said it precisely wrong so this is the stabilizer is by definition this and this is the centralizer of X. So this statement is also obvious. So now we just basically we just want to reformulate the orbit stabilizer theorem for this conjugation action and then it gets a new name. It's called the class equation. So it's just another way to count in some sense the elements in the group but it's actually just a special case of the orbit stabilizer theorem but still it's a useful result therefore it gets its own name so theorem class equation. So we take G a final group and so we take R to be a set to a subset of G which contains precisely one element from every controversy class so such that every element in G model of the centralizer of minus the centralizer of G every element in G which does not lie in the center of G is conjugated to precisely one element in R okay then the number of elements in G is equal to the number of elements in the center of G plus the sum over all elements in R of the index of the centralizer of X in G. So in some sense note that in some sense I have here it's formulated in some sense in an unnecessarily complicated way as we will see in a moment but this is usually the way one wants to use it so that's why we write it like this. So note we can say it also as follows so maybe I can prove it and prove it and then we can see. So we take a system of representatives of the orbits of the conjugation action of G on itself then maybe then we can apply the orbit stabilizer theorem in that says that so by the orbit stabilizer theorem we have that the number of elements in G is the sum of all X in R of the index so of R prime was our system of representatives the index of the stabilizer with respect to the conjugation action of this X in G and we remember that we had just seen that this stabilizer was actually the centralizer of X so we get this formula still with the prime and now we have to we see that if our element X lies in the center of G then it commutes with all other elements so its centralizer is the whole of G so this index is equal to 1 and so we put just so it means that on other words we also have that if it lies in the center of G then it's also clear that the orbit G of X is just equal to X because after all by conjugation because it commutes with every element G in G so if I make G X G to the minus 1 then this is equal to X so it means that the center so therefore we can put just put minus the center of G so if you have an element in the center then it is automatically equal to its own conjugacy class so in R prime all the elements of the center are represented and we just remove these and in that case if we have an element in the center then this index is equal to 1 because these two are equal and so this formula is just equal to the number of elements in the center of G plus the sum over all X in R the index of G C X so we see that this is just a reformulation of the orbit stabilizer theorem where we also kind of have singled out the elements in the center because we have there the orbit contains only one element okay so now we want to use this to show that any group which has a number of elements the square of a prime number is a B so we know so let P be a prime number we know we proved it some time that if we take so if the number of elements in a group is equal to P then P is cyclic in particular it is what yeah I mean what it wouldn't make much sense otherwise yeah then G is cyclic so now we want to see what happens for the next case when we have a square so now we want to show if the number of elements in G is P squared then G is a billion or commutative and for this we want to use this result so the first statement is so maybe do I give the definition yeah so first I want to give general definition so if I have a group so P is still a prime number a group P G is called a P group if the number of elements in the group is a power of P if there exists a number N bigger than zero that number of elements in G is equal to P to the N we'll see that when we do the seal of theorems that they also have something to do with P groups contained in given groups so and now the first statement we want to make that the center of a P group has at least the number of elements in the center of P group is at least P so for position let G be a P group so the number of elements in the group is some power of P then the number of elements in the center of G is at least P okay so we want to prove that so we assume that the number of elements in G is equal to P to the N for some N bigger than zero and we know that the center I mean we know the center of G is a subgroup of G so we know that the number of elements in the subgroup always will divide the number of elements in the group for finite groups so we know thus the number of elements in Z of G divides the number of elements in G but obviously the divisors of P to the N are just the powers of P so one P, P squared under P to the N minus one so it follows thus the number of elements in G is a power of P so it would be P to the M where M is bigger equal to zero but it could also be that M is equal to zero so that the center has only one element in which case we would not have proven our proposition so we only have to see so therefore it is enough to see that the center of G is not equal to one that there is at least one non-chival element because we know that the number of elements in the center is a power of P so it must be positive power of P so it's at least P okay so we assume by indirect argument we assume that the center of G is equal to one so in particular the center of G has precisely one element so then we apply the class equation so what does it say? it says that the number of elements in G which we happen to know is P to the N is one one is the number of elements in the center plus the sum over all X and R of the index of the centralizer of X in G where R was a system of representatives for the non-chival conjugacy classes it means that so X are a system of representatives for the conjugacy classes with more than one element or in other words for the conjugacy classes of elements which do not lie in the center of X so here the sum is precisely over those elements for which the centralizer is not equal to the whole of G the centralizer is a subgroup so maybe I can also write this so this is equal to the set of all X so representatives for anyway so I just say it again so if I have an element here we take precisely the representatives for the conjugacy classes for which the centralizer for which the conjugacy class is more than one element so where the element is not does not lie in the center which is equivalent to the fact that the centralizer of X is not equal to G so for X and R we have that the centralizer of X is not equal to G but the centralizer of X is a subgroup so but the Z of X in this case is a subgroup of G so we know that the number of elements in a subgroup is a power of P and so we take P to the N divided by a smaller power of P so the number of elements here will be this index is a positive power of P so we find that for all X and R we have that this index is divisible by P so to say it again we know that this is a subgroup of G which is not equal to G so its number of elements is a power of P which is smaller than this N and so the quotient is a positive power of P this is divisible by P our contradiction we have that P to the N is 1 plus something divisible by P that contradiction P to the N is equal to 1 plus a number divisible by P okay and so with this we find that indeed it's not true that the center consists only of one element and so therefore it is a positive power of P and therefore it's bigger or equal to P and now we want to prove this statement which I announced here that if I have a group such that the number of elements is P squared for prime number then it's a B so the proposition every group of order P squared where P is a prime is a B you know for instance that all groups of order 4 are B and on the other hand we know that the symmetric group in three letters which has six elements is non-Abbian so we take such a group so we have to show we want to show the group is commutative so we can reformulate this in a different way so G is a B and if for all elements X in G we have that the centralizer of X is equal to the whole of G this is just by definition the centralizer of X is the set of all elements in G which commute with X and so if for every element in X the set of all elements which commute with X is everything it means every element commutes with every element so let's see so we want to show that so obviously if X so now if we have such so let's take an element in X in G so if X is an element in the center of G then obviously the centralizer of X is equal to G essentially by definition on the other hand if X does not lie in the center of G it actually will finally not be able to happen if G is a billion but if we assume we have an element X which does not lie in the center of G then it means that if I take the centralizer of X this certainly contains X because obviously X commutes with itself and it also contains the center of G because the elements in the center of G commute with every element but X does not lie in the center of G so this contains P elements so it means it contains these elements all the elements in the center plus also X itself but we know the number of elements of the center of G is bigger equal to P and we so the number of elements in the centralizer of X is strictly bigger than the number of elements in the center of G because we have also the element X so that means the number of elements in Z of X is bigger equal to P plus 1 but the centralizer of X is a subgroup of G and so G has P squared elements and we have a subgroup which has at least P plus 1 elements as the number of elements of a subgroup must be a divisor there is no divisor which is bigger equal to P plus 1 of P squared except for P squared itself so it follows that the number of elements of Z of X is a divisor of P squared so the number of elements of G equal to P squared it follows that Z of X is actually equal to G so we have found that for every element X in G the center is equal to G so it means if you think this to the end obviously we find that this case actually does not occur because the group is commutative we have proven it is commutative so X is always in the center of G because all elements are but we are making an indirect argument so that's okay so this was this result so now are there any questions about this comments or time so now I want to look at something a bit more special I want to look at the symmetric group again and I want in particular to understand the conjugacy classes of the symmetric group so we have talked about the symmetric group in the letters before and now I want to look at it slightly more carefully so talk about the symmetric group as in so we want to describe this group more precisely we will introduce new notations for elements in the symmetric group in terms of so called cycles and then we want to see that the basis of the symmetric group can be described in terms of these cycles the length of the cycles so we call that Sn this was the set of all bijections from the set 1n to itself with the as a group operation the composition sigma tau was equal to sigma composed with tau we have introduced the notation so for sigma we write sigma as some kind of a matrix where we write on the top the numbers 1, 2 and so on until n in the bottom we write the images so sigma of 1 this is certainly a way how one can describe all permutations now we want to introduce a new notation in terms of so called cycles which many of you might also know so you can just say it definition so say let we take some elements a1 the distinct elements of our set 1 to n so we take r numbers from 1 to n different numbers and then the cycle a1 to ar is the following permutation well it's what one might think if one understands the word cycle it means that it should be something which maps a1 to a2 a2 to a3 and so on ar minus 1 to ar and ar back to a1 and all other elements which are not among a1 to ar should stay fixed so this is the map permutation sigma in sn such which is given by sigma of ai is equal to ai plus 1 if for i equals 1 to r minus 1 and sigma of ar should be well maybe just say that and sigma of ar so a1 for i equals r so sigma of ar is equal to a1 so you really have like if you would kind of put them all on a circle you kind of turn it once around and obviously and sigma of b is equal to b for all b which do not belong to the set a1 to ar so this is this cycle so we call r the number r is called the length of the cycle of the cycle a1 to ar and we say that if I have a cycle of length r I also call it r cycle and in particular the two cycles are somewhat more important and they are called transpositions two cycle transposition so you know if you have a two cycle it just means you have two elements a1 a2 and the corresponding permutation kind of exchanges them at least all other ones fixed and now maybe call so we call a1 to ar the support of the cycle and we call some cycles disjoint if their supports are disjoint so cycles a1 to ar b1 to s are called disjoint well if the corresponding sets of elements in one n are disjoint so if and only if if I take the set a1 to ar then this does not intersect set b1 to bs we don't have any common elements so we can first we have an obvious remark cycles are after all these are permutations and it is I claim it is obvious to see that disjoint cycles commute so that means if a1 to ar and b1 to bs are disjoint cycles so this set of these elements does not intersect there is no common elements here are disjoint cycles then a1 to ar times b1 to bs is equal to the other way around that's kind of obvious because what these things do to elements has nothing to do with each other so if I apply this to any element k then if k is among db1 to bs then I will apply this part to it and this will not do anything and this will do the same here and the same if it is one of the a's and if it doesn't belong to any of these it will anyway be the identity so in any case they do the same thing so now we want to show that every element in the symmetric group can be decomposed I mean written as a product of disjoint cycles any unique way every element say sigma in Sn is the product sigma equal to sigma1 times sigma whatever s of disjoint cycles so if we require that so disjoint means pairwise disjoint so no two of them have any element in common no whatever so if we require that if I take the union of the supports of the cycles that this is equal to the whole of the set 1n then this decomposition is unique if we require that the union i equals 1 to s the support so that the union of the supports of the cycles is 1n then this decomposition is unique up to obviously reordering obviously as the disjoint cycles commute I can also order them but I claim that the number of cycles and besides the which cycles they are is unique it's determined by sigma so this second part I will not actually prove so I kind of make this an exercise although it essentially follows from what I prove but and so this decomposition so in this so under this assumption so the maybe I don't need that now it's coming later where is it yeah so we will call so definition so if we have such a decomposition into disjoint cycles such that the union of the supports is everything we call this the cycle decomposition of sigma so in case so composition sigma equal to sigma 1 to sigma n into disjoint cycles with whose union is 1n so such that the union of the supports is called the cycle decomposition so by the buff it is unique so we can look at an example if I have it somewhere well whatever so we can look at for instance let's look at the so cycle decomposition of say the element 1 6 what is it 2 3 4, 5, 6 5, 3 4, 1 so how we do this so we can say we start with the element 1 it is sent to 6 element 6 is sent to 1 so we have a cycle so we have element 6 so if we take the element 2 it is mapped to itself so that is also a cycle now we take the element 3 it is mapped to 5 5 is mapped to 4 and 4 is mapped to 3 so back to 3 and that would be our cycle decomposition so one thing that one should obviously notice is that the one cycle it is just the map which sends one element to itself and all other elements to itself it is just the identity so one cycles are just the identity and so this is why I had to assume that the support is everything in order to get uniqueness otherwise I would have had to assume that it doesn't contain any one cycle that would also be possible and then you can also see that this notation the notation is not precisely unique if you have element 1, 2, 3 this is the map which sends 1 to 2 and 2 to 3 and 3 to 1 but I can also write it as 2, 3, 1 because it is the same thing 1 is mapped to 2 2 is mapped to 3 and 3 to 1 so we can so the the decomposition into cycles is unique but the notation is not unique you can always kind of rotate them element okay so now I want to prove this theorem maybe I first maybe I can first prove the theorem so that's so as I said I wanted to only prove the existence but the uniqueness is actually quite easy so prove so well we will find that so how does one do this what are these cycles actually find is that if you take the element sigma it has the property that it sends an element in the cycle to the next element in the cycle that permutes them so you find that under the operation of the group generated by sigma the supports of the cycle will be precisely the orbits and so that's therefore the way how we want to do it so let h equal to the cyclic subgroup generated by sigma so if I take the map so we have this map of multiplication of elements in 1n by powers of by elements in h so we have the multiplication from h times 1n to 1n which is just we take an element in the subgroup generated by sigma and we apply it to this so tau i is mapped to tau of i so this is obviously an operation of h on 1n basically it's just we have the operation of the symmetric group on the set 1n by just applying the element in the symmetric group to the corresponding set and now we restrict it to the subgroup h so we have an operation and so we can look at the orbits so we choose some elements b1 to b we want it s such that that such that if I apply h to b1 and so on to h applied to bs these are the different orbits of this action so we know that the set b1 to bs is just a system of representatives for the orbits for this action and we know that that that the set 1n is equal to the disjoint union of the set so now I mean I've called h is just all the powers of this one element sigma so let's see we want to so for the i equals 1 to s we choose the minimal power of sigma such that if I apply this power to bi I will get bi back so let mi be equal to the minimum k positive integers such that if I take sigma and apply it k times so sigma to the k of bi is equal to bi you know this element sigma is an element in the finite group so there's some positive power of it which is the identity element and so certainly there is some minimal positive power which is the identity on this one element bi so then it is easy to see that if I take this element bi sigma of bi and so on sigma of mi minus 1 so the mi power of sigma apply to bi these are disjoint distinct elements they are pair wise distinct it's like you are doing some exercise so if if you would have that you know for some smaller number so if two of these would be equal then you could multiply with the corresponding power of so sigma to the L of bi with then it follows that if I take so forth let's say 0 is more than equal then you see that you can just multiply with sigma k to the sigma to the minus k and you would get that bi is equal to sigma of L minus k to the bi and so we have found a smaller power which is equal to bi so we can so these will be pair wise distinct and how does sigma act on them so they are pair wise distinct and we also see if I take the orbit of our element bi I claim this is just this element this is just these elements bi sigma of bi sigma m i minus 1 of bi because in fact what you find is that if you take sigma so in fact by definition if you apply sigma to any element here it gives you the next element and if you take this one and you apply sigma to it you get this so in fact if I look at this set here bi sigma of bi until sigma mi that's one of bi this is a cycle because we see precisely by definition if I apply so let's see maybe I can explain it so if I so I get anyway so if I obviously this is by definition if I write it like this it is a cycle I can write down any distinct elements I get a cycle what I have is that if I take sigma i of so if so if x is an element in the set bi sigma of bi sigma mi minus 1 of bi then if I take sigma of x this will be as I said the next one here so this is the same as sigma i of x if sigma i is the cycle because the sigma precisely acts by you know rotating this by one putting everyone to the next and the last one to the first and this is precisely what sigma i does by being a cycle so so thus we get sigma i for equal to distinct bi sigma of bi until to sigma mi minus 1 of bi for i equals 1 to s are disjoint cycles I mean they are disjoint because the elements here are the elements in the orbit and we know the orbits are disjoint and the union of the orbit is the whole of x so who's such that and the union of the orbits union of their supports is 1n and now if we take any element in 1n I claim that it acts by the product of these cycles so we want to claim that sigma is equal to sigma 1 times sigma s so that this is the cycle decomposition and so we take an element in 1n and so we have to show that these two permutations do the same thing to it so obviously if you apply sigma to it we get sigma of x and if you apply this we get something else so let's see what we do get so let x so we know that the union of these sets here is the whole of so it means that x lies in one of them so then there exists and there's a unique i in so then there is an i in the set 1 to s such that x is an element in the set sigma of bi sigma mi minus 1 of bi well in fact if we take all these elements the union of all these elements is x so we precisely have that x in fact x is equal to bi to the power j for some j with j so not j so now you should protest if I write nonsense so sigma to the j of bi with for some j which lies between 0 and mi minus 1 because the elements here are precisely this bi and then what does sigma do to it then sigma applied to x obviously is equal to sigma to the j plus 1 of bi which will either be equal to bi if this was mi if this j was equal to mi minus 1 or otherwise it's just the next one so the index is like this okay so that's what the sigma does and what does sigma if I take sigma i of bi of x so if I take sigma j of x for k for j different from i then this is equal to x because so this the cycle only permutes the elements which lie in the only the elements which are here so if our x is none of these elements it is not moved by this so sigma i sends bi to sigma of bi and so on but if I have an element which is none of these here it will just stay fixed and on the other hand if I take of x this is equal to sigma i of sigma j of bi and this is just we apply the cycle to it and this precisely says we go one step further in the cycle so this is bi if j is equal to mi minus 1 and sigma j plus 1 of bi if j is different from mi minus 1 and so we see that and so if I take sigma 1 times sigma s of x then these things commute so x is not so the sigma k which are different from the sigma i will not do anything so this will just be equal to sigma i of x and this is which is as we have seen the same as sigma of x so it's kind of I mean I made it maybe a bit more complicated than it is somehow these different cycles have to do with each other so our element x will lie in one of these cycles and then only the corresponding cycle acts on it the other ones don't do anything and therefore we get the same result so this will prove the result so we see in particular that actually in the cycle decomposition we find that the supports of the cycles are actually the the orbits of the of the elements under the powers of the thing we want to decompose so as a corollary we find that every element in the symmetric group can be written as a product of transpositions we see where this plastic is every element sigma in Sn is a product of transpositions so we have so this way for this we just have to see that every cycle is a product of transpositions so assume we have because every element in Sn is a product of cycles so if every cycle is a product of transpositions we are done so we write down any cycle A1 to AR is a cycle and I claim I can write it as a product of transpositions in fact I can just write down what it is so I take A1 A2 A1 A3 and it goes on until A1 comma AR this cycle is in this way a product of transpositions so transposition after all was a two cycle well and you know you can just that's kind of clear so here you want to say that this thing maps A1 to A2 so if you take A1 A1 is mapped to A2 then A2 never occurs again so A1 is mapped to A2 now A2 is mapped to A1 here and A1 is mapped to A3 but then A3 never occurs again so A2 is mapped to A3 and so on so it's easy to see so this is easy to check so this shows that is a product of transpositions and as every element in the symmetric group is a product of cycles we are done how much time? basically none let me see so I maybe just give a preview so now we want to use this to describe the conjugacy classes I mean as I said I will only say so we will find that if I have two cycles of the same length then they are conjugated to each other and if you have a if an element in the symmetric group has a cycle decomposition such that the lengths of the cycles are the same for both if I have two elements which have cycle decompositions with the same lengths then they are conjugated to each other and vice versa so that means conjugacy class of an element will be determined by precisely by the lengths of the cycle in the cycle decomposition of the element and if you look here so such a cycle decomposition is a decomposition into the wet set one you know in particular if you take a support of the cycles this is a decomposition set one n into this joint subsets so the number of elements in each cycle if you take the tuple of the number of elements in each cycle this is set of numbers which add up to n and you do this up to ordering so you find that the conjugacy classes are in one one correspondence to the partitions of the number n so of the ways how you can write the number n as a sum of numbers and so this we will explore the next time and the other time the other thing we will explore the next time is that we also look at the sign of a permutation so if you have a permutation then I had told you here that it can be written as a product of transpositions and this writing this permutation as part of permutations of transpositions is not unique there are many ways but you find out that you can if you take minus one to the power the number of transpositions you need to write this thing this is something which only depends on the permutation itself the sign of the permutation we will prove this and okay that's maybe the beginning and then after that we'll start preparing ourselves for proving the silo theorem