 Okay, good morning, so then maybe we can start so it's the first time we have lecture so early. Not such an early morning person, but anyway. So now we were aiming towards the sea law theorems, which will be what we end the part on group theory with. In the moment we were still talking about operations on subsets, so it was that we have some group G which acts on S, so then it follows that G acts on the subsets of S, so on the set of subsets of S and so if, so by just if U is a subset of S we have the action is by G U map store G times U which is just the set of all products G times U, so this is the action of G on the element U with U in U. And we had introduced the stabilizer G U, which was the set of all elements in the group which leave U fixed under this operation, which means that U is mapped to itself, not that it's the identity on U, so it's the set of all G and G such that G U is equal to U, and this is a subgroup of U. And so we finished by stating the following result is proposition, so if G acts on the set S and U subset S is a subset as we are here, then the stabilizer of U is equal to G if and only if U is the union of G orbits on S, okay, and so this was the last thing we said. Now we want to consider two special cases of such actions, namely in each case S is actually equal to G, we look at an action of G on itself and we want to see how it acts on certain subsets of S, for instance on subgroups. So the first case is just the left translation, so we want now S is equal to G and we want to look at the action by left translation and the action by transposition. So first let's look at the left translation, so proposition, let G act on itself by left translation, so I remind you that this just means we have a, so the action is just from G times G to G where the operation of G on U on G may be on A in G is just the multiplication in the group, okay, so in this case we take any subgroup of G, then the order of the stabilizer of U divides the order of, divides the number of elements of U, so maybe we assume also, we assume that G is finite for simplicity, so then the order of the stabilizer of U divides the number of elements of U, yeah, now it's a subgroup, I don't know whether I actually need that, no I don't think I need it, so I could also say, yeah it's I think okay even for a subset, I don't think I will need it in that case but in any case, so this is a straightforward consequence of this result, so by this result here, so maybe I can call it proof, so by this proposition here which I have restated here, we have that U is a union of G orbits for the left translation, no, no, no, sorry, so I may be right, I write H to be GU, okay, now if I look at the action of GU on U, then obviously as GU is a stabilizer of U it fixes U, no, so therefore this result applies, so we have that U is a union of H orbits on G, so we just restrict the left translation from G to H, no, just if we have a map from G times G to G we have a map from H times G to G, and it is, this result applies and it's a union of orbits, but recall that, so for any element say X in G, we have that map from H to the orbit of X which just sends an element H here to H times X is injective, because if we take the multiplication from the left, it will always be, so it means that the number of elements in HX is equal to the number of elements in H, but now, so it means, so now U is a union of H orbits, of a certain number of H orbits, each of these H orbits has precisely as many elements as the number of elements in H, so the number of elements is the sum of so and so many times the number of elements in H, so that means thus H divides number of elements in U, so it's basically a straightforward corollary of this, so just for applications, you know for instance one special case is when, so this poses some restrictions on how these actions can be, so for instance we have the following easy remark, so you know GU is a subgroup of G, the stabilizer is a subgroup of G, so we know that GU, the number of elements of GU divides number of elements in G, because we know that the number of elements in a subgroup divides this and now we also have, so as GU is a subgroup, so if the number of elements in U and the number of elements in G are relatively prime, then it follows, so that means they have no common divisor except for one, then it follows that the number of elements in GU must be number that divides both U and G, so it can only be one, so then GU must be equal to one, so by just counting the number of elements in the set, we can see that sometimes the stabilizer must be trivial, so anyways this is not very, these are all not very exciting results but they are used, and now the other case we want to look at is the operation of G on itself by conjugation, so G is still a finite group and we want now to actually use the conjugation on subgroups, so operation by conjugation, so in this case we let, so H be a subgroup of G, where this for every element G in G, we can define the conjugate subgroup by just conjugating every element with G, so we have the conjugate subgroup, so I write it G H G to the minus one, which is just sometimes short-hand for doing this for every element, so this is the set of all G H G to the minus one where H is an H, and it's straightforward to see that this is a subgroup of G, basically obvious, and it's also easy that this conjugation defines an operation of G on the set of subgroups of G, so the conjugation, so from G to the subgroups of G to the subgroups of G, which sends element G and H to G H G to the minus one is an operation of G on the set of subgroups, now this is kind of clear that if you take the identity element it leaves this thing fixed and if you take the composition of two elements then obviously you just compose here and this will be the same as doing it twice for the two different elements, so by definition this is an operation, so just want to make, so now again we want to say what the stabilizer is, so the stabilizer of an element H, so of such group H is obviously the set of all G which map H to itself under conjugation, so the stabilizer for this action, the conjugation action, so it's called the normalizer of H, so denoted N of H and so by definition it is the set of all G and G such that G H G to the minus one is equal to H, just a couple of remarks about this, so we have kind of seen this before, you know this says the same as that G H is equal to H G which is the same as that for every H in H, so this condition says that for every H in H G H G to the minus one is an element in H, so we know that the condition to be a normal subgroup is precisely that this is true for all G in G, so by definition we have that the normalizer of H is equal to G if and only if H is a normal subgroup and I mean again it's obvious that N of H is, so it's easy to see that N of H is a subgroup, easy N of H is a subgroup, now the question is whether I made this exercise anyway, okay and so but anyway maybe I should but I actually want to use this so we have seen it's easy N of H is a subgroup of G and it's also easy to see that H is a subgroup of N of H because obviously it certainly is a subset it is also easily I mean and it's a group so therefore it follows that the number of elements you know if we are on the situation where the group are finite as we have made the assumption here so we know that the number of elements of H divides the number of elements of N of H and N of the number of elements in N of H divides the number of elements in G, okay so the order of H divides that of N of H and the order of N of H divides that of G, okay this is there and I can also maybe as another simple remark so let the say C be the number of different conjugate subgroups to H so by which I mean these are the different subgroups of G which are conjugate to H for instance if H is a normal subgroup we see that if you make the conjugation get it back so in this case C would be 1 and then we find by the Orbitz Stabilizer theorem so this is the same the different conjugate subgroups is just the number of elements in the orbit of H under the conjugation action so by the Orbitz Stabilizer theorem we get that the number of elements in G is equal to the number of elements in the Stabilizer times the number of elements in the orbit, okay so this was what I wanted to say about this so as you see there's no real result nothing particularly interesting here the point is that we want to use these as you know as tools in what we want to do and now we want to start talking about the CELO theorems so these are so this will be somehow the most advanced result that we do about groups in this course afterwards we go to rings and fields and so what is it about so these describe subgroups so H in G but G is a finite group which have the property that the number of elements in H is a power of a prime so and in particular they will describe it for the maximal so we know if H is a subgroup of G then the number of elements so the order of H divides the order of G so if you have a subgroup then the number of elements of the subgroup must be a divisor of the number of elements in the group but it is not true that for every divisor of the number of elements of G there is necessarily a subgroup it's actually difficult to decide for which divisor there will be a subgroup and so one of the things so but the converse is not true so it's difficult to decide for which divisors of G there is a subgroup with this divisor as number of elements so there's so the first CELO theorem gives us kind of a very partial converse namely tells us the following so so the first namely it says that if we write so if P is a prime is so for P a prime number and if we write the number of elements of G as P to the M times what we want times S so maybe I write R where P does not divide so we take the maximal power of P which divides the number of elements of G then there's always one subgroup of G which has P to the M elements so we don't know it for all divisors but if you take the maximal power of a prime number which divides G there will always be such a subgroup and such a subgroup will be called a CELO P group subgroup and then the other two CELO theorems tell us a little bit more about the CELO P groups and about other P subgroups of G so P sub a P group is a group whose number of elements is a power of the prime P and so it turns out that you know this these groups tell us quite a lot about these subgroups tell us quite a lot about the group G for instance one finds out that one gets a lot of restrictions from these theorems in particular also the first CELO theorem about what the groups of with a certain number of elements is are so we will as applications of the CELO theorems give some kind of partial classification of groups of certain small orders so we will say for certain numbers what precisely the groups are that up to isomorphism which have so many elements and so it's not so clear from the statement that this is so powerful but we will see okay so maybe now I so after this kind of introduction I will try to do things precisely so we will first in first state the CELO theorems and first the first we will first state the first CELO theorem then give some applications then state the other CELO theorems and maybe also give an application and then we will in the end prove the CELO theorems so in the in the following we want that let G be a finite group and we always write so let P be a prime prime number and we write the number of elements in G as P to the m times r where m is a non-negative integer and P does not divide okay as I have stated here and now some definitions so first I think already so a group H is called a P group so P is still our prime number if the number of elements is equal to P to the K for some K bigger than zero so the P groups are the groups which have as number of elements a power of a prime now we are interested in in particular P subgroups of our group namely these P CELO groups so a subgroup H of our finite group G is called a P CELO subgroup if well so it's a subgroup if the number of elements is the maximal possible so for P group so if the number of elements is the highest power of P that divides G so if it's you know so it is a P group and it has the maximal number of elements for P group contained in G and so first the CELO theorems as I said is that such groups always exist so if I have a group which I can write in such a way actually I want that it's actually visible by P and then there is a P CELO subgroup so theorem this is which we called the first CELO theorem so I mean in the setup I have here so G is equal to elements in G is P to the M times R so we have there is a P CELO subgroup of G so there's at least one such subgroup okay so this is the first statement it's not a power and not at all obtuse why should this should be true but anyway that's the way it is we will see later how one can prove it first we do a corollary which is I think usually called Gushi's theorem so if a prime P if a prime number divides the number of elements in a group G then there is an element in G which has order P of order precisely P so it means that if I multiply X with itself P times I get one and if I take any smaller power of X I don't get one okay so this is the first statement so let's see so we want to prove this as a corollary to this CELO theorem so we take H in G is CELO a P CELO subgroup so by the theorem we have such a thing and we want to find action element in H which has this order which is somehow much easier so let maybe I called Y in H be an element different from one from the neutral element so then Y will so the sub the subgroup generated by Y will be a subgroup and so the number of elements in the subgroup will have to divide this the number of elements in H so maybe I write that the number of elements in H is equal to P to the M so we have that so if I look at the subgroup generated by Y this has the order of Y elements we know that the subgroup is just con just consists of one Y squared and so on until Y to the order of Y minus one and we know that the number of and therefore the order of Y divides the number of elements in H because Y is a subgroup of H and so the order of Y is equal to P to the K for some number K between one and M no because Y is different from one so it's the order of Y is not zero is not is not one and on the other hand you know it's a subgroup of this thing so it's at least most okay whatever the K is we can take we can put X equal to Y to the P to the K minus one so then we know that this is not equal to one obviously because the order of Y is P to the K but if we take Y if we take X to the P this is Y to the P to the K minus one to the P so it means we multiply this so this is equal to Y to the P to the K which is equal to one so we see that the order of X is P okay so we have found an element of order P which is just this we take any non-zero element of the non-neutral element of the P0 subgroup and take it to the appropriate power then it has order P okay now we want to give some more applications so I'm I will slightly differ from the from the notes because I have a few more applications than are in the notes I might maybe give you some updated version of the notes if I find the time to I mean when I find the time to put it in the notes so I want to give some first applications so first as a preparation for that I want to briefly talk about the direct product of groups I think I very briefly introduced it in the beginning now I have to say a little bit more so definition let H and K be groups so the product H times K is you know as a group is of H and K is as a set just the product so H times K is just the set of all H K such that H is in H and K is in K okay so the set of all pairs let's clear the multiplication is component wise so multiplication so this is a group so the multiplication so if I have H1 K1 times H2 K2 obviously this is done component wise so this is H1 H2 K1 H2 K2 and the neutral element is just having the identity on both sides so so the neutral element is one one and the inverse of an element H K will just be component wise the inverse and obviously you can immediately check that this works also it's really standard so now we want to characterize you know find the criterion for a group to be isomorphic to the product of two subgroups and call this theorem it's maybe exaggerated so that G be a group H and K subgroups of G and we assume three properties first that G is equal to a kind of the product of them so what I mean is if I take the set of all products of an element of H and an element of K then I get in this way all elements in G okay so I can reach every element in G as a product of one element in the subgroup H and one element in the subgroup K second statement is that these are actually not just subgroups but they are normal subgroup and the third one is that they don't intersect if I take H intersected K obviously as they are H and K are both subgroups they have to contain the neutral element of G but the claim they assume that the only contain the only common element is the neutral element so they kind of products of these elements give everything and they intersect as little as possible and then the claim is then G is isomorphic to H times K okay so this is not an application of the CELO theorem this is something we want to use for the applications it's just something which one yeah I would think so why did I mean is there a misprint yeah you're right I mean there's a misprint obviously I don't know how did I do that yeah obviously it has to be component I said it's component wise so it is correct in the sense that what I meant is correct but it's not correct what I wrote okay so component means wise means obviously the same component is multiplied but anyway this the rest is correct yeah no so they are always you know I'm I will always make some misprints obviously I can always claim I just do this to check whether you pay attention okay so let's try to prove that so question is yeah maybe I would try to so there you know there's an obvious map from H times K to G just by sending an element H in H and K in K to the product so we have a map maybe I call it phi or whatever theta from H times K to G which sends element H and K to H times K okay so there's obviously such a map um you know here just the product in the group G so we want to show this is an isomorphism so it's we want to show it's both bijective and it's a group homomorphism so first one has to look at the first one here the first statement just says this map is surjective you know that's what it says you know it says that every element in G can be written as a product of an element H times K which means precisely it is an image of the map so one says theta is surjective that's certainly a good start now we want to show it is also yeah let's let me see yeah we want to show maybe it's a homomorphism so let us take so if you have an element H is it this one yeah so let then we take an element H in H and an element K in K and we want to we look at this expression H to the minus one K to the minus one H K want to know what that what yeah well that's what I want to see yeah yeah that is that is the idea but I have to explain this also to the others so we can write this as H to the minus one K to the minus one H K now H was a normal subgroup so if I do this this is also an element in H because H is normal so I multiply with an element in H so this is an element in H and now obviously you can do it the other way around as he also observed so we have H to the minus this is also equal to H to the minus one K to the minus one H K and by the same argument this is you know the conjugation of this element with an element in H so this is also an element in K and so the whole thing is an element in K so we find that this element as he said is an element in H and in K so it means H to the minus one K to the minus one H K is an element in H intersected K but by the statement number three this says it is the identity element now what does it tell us we can now bring these to the other side so we multiply by H we multiply by K so it follows so on this side we multiply by H and on this side we might and then afterwards by K so this means that H K is equal to K H so we have found that every element of H commutes with every element of K which is certainly a good thing to know and that immediately shows us what yeah that this is a home office so I'm yeah but again I will write it down so if we now we take K one H one H two in H and K one K two so this was H in K and then we can ask ourselves so if we take phi oh no it was theta no theta of um say who do you want it yeah same which we wrote is it better yeah maybe do like this H one K one times theta of H two K two okay we have to see that this is the same as theta of the product so what is it so this is here this is H one K one times H two K two but now we know that every element of H commutes with and every element of K so therefore this is the same as H one H two K one K two but according to our definition of theta this is just equal to theta of H one H two K one K two so which precisely means that this is a home office and finally in order to see that's an isomorphism we have to see that kernel is what okay so that that that is indeed true but you know okay it's true that if you have a bijective if you have a subjective map from a finite set or onto a finite set with the same number of elements then they are yeah yeah but did we I mean well I'm not quite sure whether we have completely seen that the number of elements in G you know one at least it's not completely evident that the number of elements in G is the product no that is the num that is the product I mean you have to see that all these elements are different it's maybe kind of obvious but you know still we might want to give an argument so so then finally to see that theta is an isomorphism we have to see that the kernel of theta is equal to the neutral element so it's equal to one one well so let H K be an element in the kernel of theta so that means H times K is equal to one so in other words I can multiply by H to the minus one so we have K is equal to H to the minus one so now that means K is an element in K but it's also equal to H to the minus one which is an element in H so this is an element in H intersected K equal to one so that means K is equal to one and as it is equal to H to the minus one it's also H is also equal to one one and so the map is injecting okay so now we want to briefly apply this to cyclic groups so classify and say something about that so the first just recall so recall a group is cyclic so a finite group whatever if g is equal to a for so maybe a quarter group H for some a in H and so so if H is cyclic of order say m we know we have seen that H is isomorphic to just the group for instance z mod m z no so the integers from you can either view this as integers from zero to m minus one where the addition would be given by the by adding them and taking the rest by division by m anyway so corollary so we assume assume the numbers n and m are some positive integers and n and m are relatively prime so that means they have no common divisor so the only common only integer which divide positive integer which divides both of them is one then we have that if I take the group z mod n m z this is isomorphic to z n z mod n z times z mod m z okay so I mean recall that this as a set this is just a set of classes zero one until m minus one with the addition so so if we take the subgroup generated by the number n so so the element n has order m in z mod n m z and m has order n in z mod n m no because it's just you add them and then until you add until you get to n times mz so we have therefore that we have two subgroups we have we have a subgroup so z mod n m z has a subgroup of order n and the subgroup of order m namely the ones generated by these elements and the intersection of these two subgroups consists just of the element zero in this group no because you can just see but anyway it's so and so the intersection is zero which is the neutral element and obviously they are normal subgroups because this group is commutative so by this theorem I just wiped out it follows that z mod n m z is isomorphic to z mod n z times z mod m z so we see in this case if these two numbers are relatively prime you can such a product is always isomorphic to the product what we want to do now we also want to see how is it if instead we take yeah so now as a first step we want to kind of classify the groups whose order is the square of a prime number they are somehow of this form so this is now now we are using this helo theorem so corollary let p be a prime and g a group of order p squared time is it then either g is isomorphic to z mod p squared z or g is isomorphic to z mod p z times z mod p z note that the previous theorem doesn't apply here because obviously p and p is not relatively prime so these are actually turned out to be not isomorphic okay so so the this kind of the first statement is that if we have a group such that the number of elements is the square of a prime number then there are only these two possibilities so maybe i can you maybe remember that we proved that a group whose number of elements is a square of a prime is commutative so we showed okay so we already know quite a lot so now we use this corollary to the so which was Cauchy's theorem which said that if so g is a group whose number of elements is the square of a so is a is a power of a prime number so in particular the number of elements is divisible by this power of this prime number so then g contains an element of order p so by Cauchy's theorem g contains an element how do i want to call it a of order p so so we let say h be the subgroup generated by this element a so this is a subgroup with p elements now we take any element p which does which lies in g but not in h and we call h prime the subgroup generated by b well there are two possibilities either the subgroup is the whole of g or it isn't no minus h so this is not i mean at least for me the notation of one set minus another is like this you could sometimes it's also used as dividing by an action from the other side but i never use this notation so for me it's this i mean but if you're more familiar with it you can say this but this also i find confusing so whatever you write is confusing it just means it's an element of g which is not an element of h and this for me is my standard notation so either we have that h prime is equal to g and then obviously g is isomorphic to the p square because it's cyclic p squared z because you know we know that cyclic groups are just of this you know there or you know h prime is not is i mean it's a strong subset i mean a strict subset of of g so well the number of elements in h prime is a divisor of p so if we so if a h number of elements if h prime is different from g then it follows that the number of elements in h prime is equal to p because it must be divisor of p it cannot be p squared because otherwise it would be the whole of g and it cannot be one because you know this is an element which does the element b lie does not lie in in h and therefore it's not equal to one okay and obviously h prime is cyclic because it's well it was defined as a cyclic group so h intersected h prime is a subgroup of h positive h yeah of h and it's and again it's therefore its number of elements is a divisor of p so the number of h intersected h prime is a divisor of the number of elements in h or in h prime which is p and as i hear from behind you know it cannot be that h intersected h prime is equal it has p elements because otherwise h would have to be equal to h prime which is impossible by our assumption so and h intersected h prime is different from h or from h prime because otherwise you would have that h is equal to h prime which contradicts our assumption that this is generated by this element b which does not lie in h so therefore it follows that the number of elements so it must be a strict subset so the number of elements in this intersection is smaller than p so it can only be one and so if the subgroup has only one element it means that this is equal to the neutral element in the group so we find that again we have two uh subgroups so our group is commutative so h and h prime which prime are normal subgroups and we can easily see that h times h prime is equal to our group g maybe i can leave that to you but it follows quite easily and the intersection h intersected h prime is equal to the neutral element so it follows that g is equal to the product and obviously both h and h prime were cyclic groups with p elements so this is just isomorphic to z mod pz times z mod pz okay how much time yeah now we want to come to a slightly more difficult result so remember that i had introduced the dihedral group as an example let me briefly recall it so i had described it as the set of symmetries by reflections and rotations of an n-gon so if i have d to n so here we go d to n to dn and what i had described is it's a group so d to n dn is a group generated by two elements by elements a and b with the property that which one was it that a to the n is equal to one b squared is equal to one in the group and bab or if you want bab to the minus one is equal to a to the minus one and we also want that such that so with two n elements so i mean obviously we could have a group with one element which has this property a and b is equal to one and one but we have this so and we had seen that this exists as these rotations and reflections of an n-gon and we in fact can say what the elements are the elements are so as a set d to the n was equal to the set of all a to the i b to the j such that zero is smaller equal to i is smaller than n and zero is smaller equal to j is smaller than two you can see that these are precisely two n elements and what yeah yeah i said it i think but i didn't write it um so you can see that these are precisely two n elements and you can see that if you think of it a little bit that any product in the a's and b's you can bring into this form by applying these two relations whenever you have a power of a bigger than n you can reduce it and whenever you have a power of b bigger than one you can make it at most one and you can commute the a's you if you have an a and b's on two sides you can bring it on the other side by using this relation and so in the end you can bring it into this form i mean there's you can imagine that so this is this group and now um i want to claim i want to classify a large number of finite groups this is again an application of silo theorems so theorem yeah i'm not quite sure doesn't make sense to start with the proof i would have to see what i do so let p be a prime and let g be a group of order um two p so twice a prime then either g is cyclic so isomorphic to z modulo two pz or g is isomorphic to the heli group dp okay so there are only two possibilities for the isomorphism type of a group with two p elements so it doesn't make sense to me to try to to give the proof i will do it next time maybe we can just see for instance that the things we already know tell us quite a lot about groups of small order so i can make a little table so if we look at the number of elements of g and on the other hand we look at the isomorphism classes of the group g so what up to isomorphism how many what are the different groups so if the group g has one element it's just one so that's not very exciting if the number of elements in a group is a prime number we know it's cyclic so it's isomorphic to z mod in this case 2z so it's same here so isomorphism class so i abbreviated last so if four so four is a square of prime number so we know there are two possibilities it can be z mod 4z or z mod 2z times z mod 2z and these are actually also not isomorphic as you can easily see because this contains the element of order four and this doesn't then five is again a prime number now six is twice the prime number so there are two possibilities it can either be equal to the cyclic group or it can be equal to the group d3 so note that we also know some other groups of order six so one group of order six that we know is the symmetric group in in three letters but now we know the symmetric group in three letters is not is not commutative and this one is so we know that this is actually isomorphic to s3 and then group seven elements yeah okay if i define d3 as a sub as a subgroup the way i defined s3 as a d3 as a subgroup of so if i define dn a subgroup of s and they are equal yeah if i look at this abstract thing here obviously they are isomorphic but whatever and then seven it's z mod 7z we have eight well yeah we know some we have z mod 8z we have z mod 4z times z mod 2z and we have z mod 2z three times but you know we don't have any theorem about this i mean we don't have a theorem which classifies the third for third power of a prime number what the numbers is so there might be more and they know so we don't know so there might be more and there are more and we don't know them certainly there there are some non commutative groups but we haven't understood them and then nine and ten we still know so just to up to ten so for nine we have z mod 9z and z mod 3z squared so times itself and for ten we have z mod 10z and we have d5 so which are in each case the complete classification so for instance we see that with a few things we have proven so far except that we didn't yet prove this and we already have classified except for the number eight all groups of order up to ten and you know it goes a bit further 11 well 12 is not so nice and then anyway okay so maybe that's that's enough now so we meet again on Friday I think