 So we're having Tim and Dr. Chitra's second talk on the inverse Scala problem. So please go ahead. Oh, thanks, Bjorn. All right, so just to remind you where we were, was it only yesterday? We proved that, reviewed Hilbert's Revisibility Theorem and proved that as San is a Galois group of Q. And remember the proof was that he start by a field where you're just formally joined in determinants, just an independent transcendental variables. You let S and act on this field and then you compute the invariance. And because in this case, the invariant field just happens to be purely transcendental over Q, it automatically gives you a family of SN extensions over some rational function fields such as this. And then by Hilbert's Revisibility we can get a San as a Galois group of Q as well. Now, there's an obvious question, is that can you do it for other groups rather than SN? And so let me start with this, it's called a NERTHES method or NERTHES program because she basically asked this question and initiated quite a lot of work and invariant theory to look into this. So suppose now G is not a symmetric group but just an arbitrary group. Find it, of course, like all groups here. So it's a subgroup of some SD. In other words, G acts again as permutations on D things. So let me again take D independent transcendentals, D determinants and let G act on them by permutation. And now this is not a full SD action but nevertheless it does. And you can do the same thing. You can compute the, you can start with this field capital K which is Q a join alpha one alpha D, rational function field in D variables, G acts on it linearly preserving Q and you can ask yourself what's the field of invariance and in particular if this field a purely transcendental extension of Q. And then if the answer is yes, then G is a, we've just realized G is a Galois group over this field and by Hilbert's visibility, we also realize it over Q then exactly as we did for SN. In fact, what you did, what you do need is a little bit less. I said, let G be a transitive subgroup of SD so it acts on D letters. Actually you need a little bit less. You need a rational representation of G. So it's enough for G to act on Q vector space and then we can do this whole yoga but it doesn't really matter. So what NASA did is that she approved for all subgroups of S4. This is indeed the case and this field that you get here is purely transcendental in particular. You can prove like this that all the subgroups are realizable as Galois groups over Q. So this is a very encouraging beginning but it turned out that in general, these fields, this rational function fields, so these fields of invariance, they can be quite complicated and the answer in general is no, they are not in general purely transcendental extensions of Q. So the first example was by Swan in 1969 and then there was a for cyclical group of 47 and then there was quite a detailed analysis for a billion groups by another Henry Glenstra who proved in particular that C8 is the smallest example of a cyclical group for which the answer is no and this field here is not purely transcendental. And I should also say that as far as I know, this is also not really a practical way to realize Galois group over Q. This fields are quite complicated, they can be complicated to compute and to decide whether they're purely transcendental extensions of Q or not. But anyway, this sort of, this was on the, I guess still is an approach to the inverse Galois problem and just one remark on it, you can reformulate it in an algebraic variety terms. So what is this field capital K that we started with? The field Q join alpha one, alpha n. So this is the field of rational functions on an affine space A n over Q. And when we compute its invariance, that's the same as today the quotient of A n by the action of G. So again, G acts linearly on this vector space in our simplest case, by just permuting the coordinates. So this quotient here is some n dimensional variety, which is to be affine. And notice question is simply equivalent to the fact to the question, whether this quotient to be is a rational variety. So we call rational variety is the one which is bi-rational to P n or A n. And that's a question about its field of rational functions. So it's a question, whether it's field of rational functions is purely transcendental of Q. Now, of course, you can say, well, asking whether it's rational variety, even if the answer is no, well, that's a little bit too much to ask, is it? You don't need it to be a rational variety. For example, even if it has at least a rational point, at least let's say that this map here, you have a natural projection map from A n to A n over G and assume it's underemified over that point, then the fiber of that point gives you a G extension of Q. And therefore you just realize G is a Galois group of a Q. So if you want to solve the inverse Galois problem, really you don't just need this to be a rational variety. That's between overkill. You just need to be able to prove that it has or it has enough rational points. Or for example, if it has rational curves or rational varieties of high dimension, then you can realize this group G over some function fields, possibly with less variables than M. But in any case, you've solved the inverse Galois problem over Q of t as well. So this is just nice way to think about it. This is just a geometric interpretation of what Hilbert has done for SN and what you could try to do for other groups. Okay, so as far as I know, this approach kind of doesn't get you that far in the inverse Galois problem. So let me talk a little bit about the approaches that do and focus at least this lecture, mostly on soluble groups. So let me start by one very easy observation that if you realize group G, then automatically you've realized every quotient to G. So in other words, let's say it was little field K, maybe Q, if you try to realize Galois groups over Q or Q of t or Q t, Q t 1 t n. And suppose you found an extension of little K, capital K, which has Galois group G, then and Q is a quotient of G. So suppose you have a normal subgroup and in G, such as quotient is Q. Then of course, just from this tower and from Galois theory, you see that this quotient is, it is also, if you find a Galois extension, it's Galois group Q. See, there was a question in the chat. It's okay, I just, let's go ahead and take care of it. So in this story, when you take a quotient of f i in space, maybe you get a finite map. So the quotient variety is always n dimensional, but it's not necessarily a n or a rational. All right, so again, if you've realized the group, you can realize a quotient of it. And by our definition of regular, if you realize G regulate, let's say over Q t 1 t n, so no subfields are constant. Or if you want, there are no algebraic elements in capital K over Q, then of course, every such subfield is going to be regular as well. So every quotient group you've realized as well. So in other words, it's completely elementary that if you've realized the group, you've realized a quotient notation here. As I know, not a little unfortunate, but I don't know a good notation for quotient groups. So this Q over Q is really a quotient group of the rationals. I try to distinguish between what we're writing. And similarly for the inverse Galois problem over Q of t. And by the way, if you're playing with this package of computing Galois groups, it has a function which is called a subfield, which does let you compute subfields of families and therefore construct quotient groups like this explicitly. And now a natural question is, what about the converse? So can we always embed, can we always go the other way and embed a Q extension, where Q is a quotient of G into a G extension. So in other words, suppose you have a group, it's a quotient of a larger group. You've realized a smaller group as a Galois group. You want to realize a larger group as a Galois group is that always possible? Because if yes, there is a large class of groups, soluble groups on your photon groups, which you can build step by step with very simple steps, just simply groups. And then so for instance, every two extension of Q, you can build by repeated quadratic extensions. So then you can ask yourself, can I realize a soluble group? For example, Q over Q of t by this kind of repeated construction. And for that, you need to be able to solve this question. Well, this question goes under the name extension problem or actually embedding problem. I think maybe I should have said this is the one which normally use. And the answer turned out to be no. So in general, it is not possible to extend just an arbitrary Q extension to a G extension. So let me just look at one example of that and to see that there are local obstructions for doing that. So for this example, I am essentially talking about an embedding problem when you try to embed a quadratic field into a C4 extension. So let's suppose first that you've constructed an extension of Galois group C4. So let K over Q have Galois group cyclic of order four. So in other words, so this is C4 here and it has a unique intermediate field which I'll call L. So the top field is K, the bottom field is Q and there's a quadratic extension in between which is called L. And let us look what sort of restrictions this imposes on a field L to be contained in a C4 extension and you can do it by considering places above infinity. So Q has an infinite place and you can see how it splits in L and how it splits in K. So how do you do that? You have to look at the, what's called decomposition group, the infinity which is a subgroup of C4. So again, C4 is my Galois group. I have a unique place infinity here. I take an infinite place above it in K and I look at the set of elements in the group which leave that place in there yet. Now this decomposition group because I'm dealing in the infinite place, it's a Galois group R over R trivial or C over R of order two. So this is the group of order one or two which is contained in C4. So it's really two possibilities. It could be trivial or it could be C2 contained in C4. And now if you decode what it means. So in this case, there are four places above infinity in K because places above infinity in K are in one-to-one correspondence with left cosets of the Galois group by the decomposition group. So there's four of them in this case. And in the second case, when the infinity C2, there are two places above infinity in K because these are cosets of C4 by C2. And if you want to compute places in an intermediate field, there is a general result which is for which you don't need this group to cyclic or Galois or anything. If you know the decomposition group and you want to know how your prime decomposes not in the top field, but in any intermediate field, then the only thing you need to know, you need to know what is a subgroup that cuts out this intermediate field. So in this case, C2 in C4. And you have to look at double cosets of that group and the decomposition group. So in this case, the places above infinity in L are double cosets C2 backslash C4 forward slash D infinity. And if you compute what it means in these two cases, you see that, well, this is either C1 or C2 and you see immediately that there is a one, sorry, there are always two of these places above infinity. So in other words, which has proved that independently of what this decomposition group is, there's always two places above infinity in L. So the two possibilities again are that there can be four places above infinity in K or two, but in any case in the quadratic field in between, there's always two infinite places. And that means that this field is real quadratic. So K doesn't have to be real, it can be real or complex, but L has to be real quadratic field. So what we just proved is that if you have a C4 extension, then the unique quadratic inside it is a real quadratic. And in other words, what it means is that if you try to solve an embedding problem, you start with the quadratic extension and you say, well, I have a surjection from C4, which is this Galois group to C2 to this Galois group and I have a quadratic extension, can I embed it into a C4 extension? And the answer is no. So for example, QI or Q join, well, any negative square root so any dimension of the quadratic field cannot be embedded into a C4 extension of Q. So this embedding problem is quite, it's not trivial, already for simplest groups you can think of such as C4 and C2. And it's really Witt who studied this in great detail for at least small groups and he has complete results in particular for this case that I'm talking about C4 to C2 and the Quaternion group created, he was also interested in. So let me state his results and I outsourced the first theorem as an exercise, Q1 on the whole page. There's mini course. So with theorem is the very pretty statement. It says suppose K is a field of characteristic not two so little K and we're interested again in a question whether you can embed a quadratic extension over K into a C4 extension. So let's take a quadratic extension, non-trivial, K join square root of E and then it can be embedded into a C4 extension of little K if and only if this number E, which is in K is a sum of two squares. It's a very nice, very pretty result. So if you do it over Q, then for example, five is a sum of two rational squares. It's two squared plus one squared. And therefore, you know that Q square root of five, you can embed into a C4 extension. And in fact, there are many such C4 extensions and one of them is Q join Z divide. So this cyclotomic field here has got a lot of that much for that and it contains Q of five. But on the other hand, if you take QI or Q would three or any other rational number, which is not a sum of two squares, then such quadratic extensions cannot be embedded into a C4 extension. And you see therefore that if you try to, well, do what Schultz-Reinhardt and Schubert-H have done to prove, for example, that every soluble group is realized over Q by some sort of inductive proof, which is essentially what they did, then you do run into this an obstruction problem because on various steps in your construction, if you are not careful, these extensions will not even exist. Just see if there are any questions which are unanswered, no, not yet, okay. All right, so and another result, which is also very nice and there have been generalizations of it by Sarah and others by looking at these quadratic forms, which play a role here for other groups, but this is the first result like this proved by Witt in 1936 for Q8. It's the same sort of problem, but for the Quaternion group Q8. So if you try to construct the Quaternion group Q8, it kind of goes like this. Q8 has center plus minus one. So this is a group which has eight elements, plus minus one, plus minus i, plus minus j, plus minus k. It has center plus minus one. And if you divide by the center, you get C2 cross C2, generated by classes of i and j in the quotient. So the way you can construct Q8 extension, you can start with a C2 cross C2 extension. That's very easy, right? It's just by quadratic, take arbitrary numbers A and B in your field, as long as they're independent quadratically and you do get a big quadratic extension, then, well, this is a C2 cross C2 extension. And then you ask yourself a question, when can I extend to a Quaternion extension? Is it possible? And if not, what's the necessary and sufficient conditions? And this sufficient necessary and sufficient conditions were found by Witt. He proves that a big quadratic extension like this can be embedded in the Quaternion extension. So in other words, there is a quadratic extension on top, which makes the whole thing Quaternion, if and only if the following two quadratic forms are equivalent over, sorry, it's supposed to be a little k over the ground field. So this is just a standard form, x squared plus y squared plus z squared. And this form has coefficients A, B and A times B. So A times B defines a third quadratic, which you don't see here inside this big Quaternion extension, k a joint square root of A. So again, this is a very neat, necessary and sufficient condition and you can make it completely explicit in a sense that if you have a witness to the fact that these two quadratic forms are equivalent, then you can actually construct this Quaternion extension explicitly. So the second part of the theorem says that if P is a three by three matrix, which let's say was determined one over A, B, normalize it like this, such that it transforms this quadratic form into that quadratic form. So P transpose times diagonal matrix with A, B and B on the diagonal is the identity matrix. Then you can classify all Quaternion extensions containing a little k, A, B, root A, root B. There's essentially only one line of them. So you can a joint square root of this expression here, where this P11, P22 and P33 are the diagonal entries in this matrix capital P and root A root B and root A, B are in your big quadratic. So this gives you a Quaternion extension and every other Quaternion extension which contains this big quadratic is really the same except before taking a square root you multiply it by an element of a little k. I see that my little k became capital K and fortunately somewhere here, but it's the same people. So this again is a very nice result. But again, you see that this is, well, it's a non-trivial condition. For example, you see if you stare at this that if you want this quadratic forms to be equivalent they represent the same numbers and the numbers represented by the form on the right are sums of three squares. So anything represented by this form, for example, A, if you plug in y equals zero and z equals zero, x equals one, you see that A is represented by this form and similarly B and AB, they're all sums of three squares. So in other words, if you take a quadratic extension K and join a root A and you ask yourself can I embed it into a Quaternion extension? And then the answer is no unless A is a sum of three squares and that's more or less if and only if in fact there's a very few exceptions to that. But again, it means for example that if you work over the rationals, then well, sums of three squares are positive. So things like QI, for example, again, you can not embed it to a Quaternion extension of Q. How hard is it to find such a matrix P? Yeah, that's a good question. And as far as I know, there are computer algebra systems like Magma which have nowadays equivalence of quadratic forms over general number fields. But from what I understand, it's quite a difficult problem to do it. And in sort of, I don't really know very much about it, but it seems not very easy over a general number field as far as I know. For the second question, do we have a result where the extension of degree to the K can be embedded in a Quaternion to the K plus one extension? I don't know actually. So this is, this is a type of as I mentioned, this is a type of problem which has been considered quite a lot because in the example that you ask or the example that I do here, this is what's called a central extension. So the little C2 which you want to put on top is contained in the center of the group. And this is exactly the type of extensions which have been considered a lot by Sarah, Master and others, because they looked especially at double covers of alternating groups, so-called AM twiddle, which are like this where you construct AM and then you want to extend it to an extra quadratic which is a central quadratic extension. And so in that case, there is a result which essentially involves a trace form of the field in question, which kind of explains where this form comes from and which quadratic forms you need to be equivalent. So I think there is a result like this. I don't know if anyone has formulated explicitly for Quaternion extensions and well, how practical it is, I guess, depends how efficient your algorithms are for proving that quadratic forms are equivalent or rather giving a matrix which does that. Okay, so right. So as I just mentioned, especially for double covers of alternating groups, people have studied this a lot. And in particular, it has been proved that this double covers of AM, so the smallest interesting one is a group SL25, can be realized over Q of T using these methods. But as I would like to mention here, these proofs are not sufficiently explicit in that they do not give a family. So as far as I know, for these groups, even again, for SL25, when MS5, it's a double cover of N5, there are no families known over Q of T and I stated this as a research problem P1 to find such thing. And again, as I mentioned before, Schultz-Reichhardt and Schubert, they do construct soluble groups. For example, the soluble groups over Q inductively and they need to deal with this problem. It's essentially sort of by ramifying a lot of primes along the way and occasionally changing groups in order to prove that these instructions do not exist in their context. All right, so what I would like to do next is to talk about cyclic groups because as you've seen from this extension problem business, even for a cyclic group of order four, which is not like a very large group, it's not entirely clear how to construct C4 extensions of, let's say Q of T because, well, if you do it very naively by starting with a quadratic and trying to extend it, this is not entirely trivial. So the question is how we do it for general cyclic groups and generally what are the, and after that I'll talk a little bit about one case where this abstraction problem does not exist that gives a large class of groups that we can construct called semi-abelian groups. All right, so let's look at cyclic groups next and construct a regular CN extensions of Q of T for arbitrary M. So this is the problem I'd like to talk about to start with Q of T. How do we construct a family over Q of T of regular CN extensions for arbitrary M? Now on the one hand, you can say, well, from algebraic number theory, we know everything about CN extensions because this is just part of class field theory. We know everything about abelian extensions of Q and every CN extension is contained in some cyclotomic field, Q zeta M. We know the Galois groups. We know what these fields look like. So, you know, how hard can it be? But if you think about it, if you try to do this, you find that class field theory is a little bit kind of orthogonal to this families over Q of T business. And while, for instance, you can just construct lots of C4 extensions in the class field theory, take, for example, the prime one, what four, five, 13 and so on. Then in Q zeta M, there will be a C4 extension. It's very difficult then to put them into a family over Q of T. So, and there is rather easy solution to this, how to construct this CN extension, which I think has been rediscovered a few times. And I think first goes back to Smith and then Dan Sar and Schnapps have refined it a little bit and it uses Kuhner theory. So, let me explain how it works. And this is the one which is also implemented in the package. So, suppose you've already constructed an extension, let's say over Q, with Galois group CN, C4, if you wish. Then, no, no, how do we understand CN extensions without class fields theory? Well, there's something that's called Kuhner theory, which is in some sense much more explicit and much more amenable to what we are trying to do. For example, every quadratic extension of Q is just obtained by joining square root. And similarly, every CN extension can be obtained by joining nth root of something, but it's only true over fields which contain nth root of Q. So, in other words, if you start with a CN extension, so Q, so degree n extension Galois with Galois group CN, then it's not true that it's obtained by joining nth root of rational number. That's completely untrue, because if you join nth root of rational number, you don't even get a Galois extension. You need nth root of unity to make it Galois. But if you join nth root of unity, then this becomes true. So if you join zeta n, then this extension here, K zeta n or Q zeta n, which of course usually will also be a CN extension unless you've been very unlucky and you've chosen zeta n to overlap with K. So this will be again normally CN extension. And now by Kuhner theory, this is obtained by joining nth root. Of course, this nth root is not of an element Q, it's of an element in Q zeta n. So this K zeta n can be obtained by joining nth root of something to Q zeta n and this is something I'm going to call G. So G is going to be some element of Q zeta n cross. So in other words, every extension that we care about, every CN extension of Q, it's contained in this extension here where you take Q zeta n and then you join nth root of everything or if you want nth root of G for varying G. So let's look at this fields a little bit. Now, if you look at this field here, it's a Galois closure, it's actually reasonably large and this is what it looks like. It has the Galois group over Q, which is what's called the wreath product CN by Z mod n Z cross. So where does this come from? So first of all, this Z mod n Z cross, this is just a Galois group of Q zeta n over Q. So if you recall, the Galois group of Q zeta n over Q is Z mod n Z cross and this isomorphism is completely explicit. If you take an element J in this group here, so it's just an integer core prime to n, that there's a unique automorphism in Q zeta n over Q which sends a fixed primitive n through the unity to its J's power. And this gives an isomorphism like a completely canonical isomorphism between these two groups. Yes, so that's the wreath product CN with Z mod n Z cross. I think that's what I, I don't know my sign of wreath product maybe is not very well visible, but this is what it is. So generally, so this is the group here, Z mod n Z cross. So what I'm doing here, I'm taking a Galois Z mod n Z cross extension by joining n through the unity and then I take a CN extension of top on top. Now normally when you do this, so when you take an extension with Galois group G and then on top of it an extension with Galois group H, the Galois group you get for the full extension over Q is this wreath product of G by H. So the wreath and which is quite a large group. So it's, it's the way it's defined, you take this number. So that's what ends at cross. So all the five function of M copies of CN. So you take CN to the power five of N and then you let that one ends at cross act on it by permuting the, by permuting the copies in a regular action. And the reason you get the group which is so large if you think about it is the following. So when you are joined to Q z to N, N through to G you get a CN extension. But of course this group here is that what ends at cross it has a lot of elements. So there are a lot of automorphisms of Q z to N over Q and if you apply such an automorphism to G so to the element of which I take the N through then I get a different element in Q z to N and taking N through to it gives you a completely different degree of extension. So if you want like this to be small so this whole gala group to be small then this field here has somehow had to be invariant under the action of z mod N z cross and then generally it won't be. There's a completely independent CN extensions which you get by joining N through to G and through to sigma two of G and sigma N minus one of G where the sigma J's I denote this automorphism which sends zeta N to zeta N to G. So there's lots of CN extensions of top on top and if you compute the whole thing make it gala by taking the gala closure you get this very large group here. But nevertheless it turns out that if you play with this and cleverly choose a combination of these sigma and J of G's then inside this group you can construct a CN quotient. So here is a theorem which I think goes back to Smith in some form and as I said to then search and snaps in more generality but essentially it's a very explicit results which says that here is the CN extension for you. So this is how it goes. So in the notation as above let G be an arbitrary element of Q zeta N and let's identify the gala group of Q zeta N over Q with z mod N z cross as above calling the elements sigma J where sigma J sends zeta to zeta to the J. Then you just define the following polynomial. So it looks a bit large but essentially what happens here is that you take this N through of this different sigma J of G's. So again G's an arbitrary element in here you hit it by sigma by various elements in the gala group here and then you take N throughs you pick a choice of an N through an arbitrary choice for each of them. They don't have to be compatible in any way with some complex number or number in this field if you want where you compute it, it's just a complex number. And then you take these guys here and raise them to some powers throw in some root of unity for a good measure and consider these expressions as roots of a polynomial of degree N and then it turns out that this polynomial has the right Gala invariance properties. So this polynomial f sub G it depends on this auxiliary parameter G that you've chosen q zeta N is a polynomial with rational coefficients. And moreover from the computational perspective it has integer coefficients if you start with G which is integral so which is an maximal order zeta N. And of course polynomials with integer coefficients are very nice to compute over the complex numbers. You just do a complex computation. You take these guys here compute them as complex numbers to some precision multiply everything together round to the nearest integer as long as you control your precision you'll find these polynomials and very, very quickly. And for most choices of G again from the risky density point if you want this polynomial will be irreducible and it's Gala group is going to be C N over q. And moreover, if you think about it this construction again every C N extension of q is contained in one of these fields. So in fact, you can prove that in some under some assumptions I think the assumption is that N is called prime to five M or something like that then you do get an arbitrary C N extension of q like that. So let me just give you an example for N equals three and four where these cyclotomic fields q, z, or q they're very small they're just quadratic extensions of q, q, z, as three so q root minus three if you want or qi respectively. So you can just do it in full generality this whole construction. So you take G to be an arbitrary element in this field q, z to N. So A plus B z to N if you want where A and B just you can think of them as independent transcendentals. So it's an arbitrary element of q, z to N and then you plug them through the formula above it's quite easy to do by hand we don't even need a computer algebra system for this and then you find two polynomials which I wrote here for N equals three and N equals four f sub G of x which is this cubic and f sub G of x which is this quartic and these are families which turn out to be generic so they cover every single C three and C four extension of q families over q, a, b those two parameters a and b and this is how you construct C three and C four extensions and this works beautifully for a general N. So if you take G to be just make it one parameter now A plus z to N and then this is the choice that I implemented with the packages family of cycling group of N. So for any N if N becomes large then it's a wait for a little bit longer because these polynomials grow quite quickly and they have large degree in A but it will give you such thing hopefully eventually and these families are not very useful for large N because the so their degree in A so in other words the degree in these variables here is quite large for large N is N times five N and as far as I know for even for cyclic groups it is not known what are sort of the best possible families for cyclic groups in other words for lowest degree in A. I think it's a very interesting question and I sort of made it as problem p five on the homepage but in the case of this air construction and it does allow you to construct regular families over q of t and sorry and so at least for cyclic groups we know that they can be realized as gamma groups. Now to answer a question in the chat is it well understood which g work and why some fail? Well it's kind of the usual thing if you take g which is by accident happens to be an N power in q zeta N and it's not entirely clear how do you explicitly formulate this in terms of coefficients then of course it fails and you get a trivial extension and similarly if it has some sort of bad in various property so you just take it with zero I don't know you could get this polynomial to be inseparable but it's generally a little bit messy so I think the best thing you can really say is that most choices of g will work I don't think you can do much more than that really. Okay so now that we've constructed cyclic groups and then I didn't mention it before but if you can construct regular families for two groups it's very easy to construct regular family for the product of two groups just basically you're multiplying them together then once you know how to construct cyclic groups you can construct a billion groups and the next very important case which kind of brings you close to soluble groups is this one which is called split extensions by a billion groups so this is the one I'd like to talk about in detail. So recall we've already observed that there are these embedding problems or extension problems which are in general not soluble so you cannot for example if you have a group G and a quotient Q you cannot always extend a family or just field if you want with Galois group Q to embed it to a field with Galois group G and the two examples where it's not possible is C4 surjecting onto C2 and Q8 surjecting onto C2 cross C2 where you can't always lift C2 extension to a C4 extension and similarly here. So the problem here is that these extensions are what's called non-split extensions and for split extensions with a billion kernel it turns out that you can always solve this embedding problem. So let me say a few words about this. So suppose A and Q are groups. A for the moment is arbitrary but in the moment I'll take it to be a billion which is why I call it A. So just maybe to answer the question when I say that most G work I really mean that's the risky sense that if you go back and for example here you think of A and B are two independent transcendental so there is a whole sort of FI playing worth of choices then there will be as a risky close subset of those EMBs which do not work or at least the risky open which will work even though I think what I said is also true. So it's a risky dense and not a risky open, that's the same. Okay, so let me look at the split extensions with a billion kernel again. So A is going to be a group in a moment it's going to be a billion for the moment arbitrary and Q is also going to be a group. So this will be a normal subgroup and this will be a quotient. First of all, just recall that an extension of Q by A it's an exact sequence of groups like this. One goes to A goes to G goes to Q goes to one. This just encodes the fact if you want that A is a normal subgroup of G and Q is a quotient. So in other words, whenever you have a group which is a normal subgroup in a quotient group it's an extension of Q by A. So it's an exact sequence like this. And this extension is called split if this exact sequence splits. So in other words, if there's a group homomorphism going back from Q to G, let's say phi such that if you compose by taking phi and then pi again you get the identity map, it's the usual definition. So if you think about it, the image of Q under phi will be a subgroup of G, which does not meet A and which projects subjectively onto the quotient. And such subgroup is called the complement to A and G. So it generates G together with A and it meets A trivially. But it's not necessarily a normal subgroup. Otherwise this will be a direct product. So if such a thing exists then you call it a split extension and if it doesn't you call it a non-split extension and group theories very often use notation like this with a colon or a dot, the right G equals A colon Q if G is a split extension of Q by A and A dot Q if not. Now, every finite group can be built from finite simple groups repeatedly deductively if you want to use any extensions. So if you start just with simple groups and then you allow extensions of simple groups by simple groups, extensions of what you've got by simple groups and the other way around and so on and eventually you'll get to all groups. And therefore this notion of extension is very important group theory because it allows all kinds of inductive constructions as long as whatever it is that you're trying to do works for simple groups and it works for extensions. And in particular soluble groups can be always built from cyclic groups or even cyclic groups prime order like this. And now that we know from the inverse Galois theory point if you everything about cyclic groups we know that they can be constructed regularly if you want to get from there to soluble groups it's really trying to understand extensions. So now suppose from now on A is a billion then what happens is that, well we have an action of G on A by conjugation, right? When you have a normal subgroup A of G then conjugation preserves A that's precisely definition of a normal subgroup. So every element G acts on A by conjugation but if A is a billion then A itself acts trivially on itself by conjugation and therefore this action descends to the quotient. So it's really Q that acts on A. So the action of conjugation of G to odd A has A in the kernel. So we get a map alpha, well let me call it alpha from Q to automorphisms of A by taking an element of Q, lifting it to G and this action does not depend on this lift and then acting by conjugation on A, okay? All right, so now we have three pieces of data. Whenever you have an extension it has a normal subgroup A it has a quotient Q and it has an action which is also canonically associated to this picture of Q by A. And it turns out that in the case of split extensions which I should have said here or I should correct my notes. So in the case of a split extension this data A, Q and this homomorphism here from Q to odd A, they determine G uniquely. So you can construct G from this data from A, Q and such homomorphism and you get a unique split extension of Q by A which has this action here. So in other words, if you want to understand split extensions that's very easy. You just need to decide what your A is, what your Q is and you pick a homomorphism from Q to odd A and for any such homomorphism there's a unique split extension like this. So there's a formula which I wrote down here how to construct this split extension which is called a semi-direct product. In this case, it looks almost like a direct product except not quite in the first component because it uses this alpha somewhere along the way from Q by of Q from Q, A and alpha. Okay, so split extensions we understand completely and if you want to understand all extensions of Q by A with a given action alpha then they're classified by some homology group. So there's a homology group H2 of Q, A. So this is the group homology where you have a group acting on a Nabilian group and the action is given by that alpha there and it's this homology group which classifies all extension up to an obvious notion of an isomorphism of extensions where the zero element of this group corresponds to the semi-direct product to a split extension. Okay, so maybe let me just end with an example here. So let us look at extensions of C2 by C4. So in other words, I'm looking at groups G which fit into an exact sequence like this. They have C4 up in the kernel and C2 as a quotient. So G divided by C4 equals C2. So in this case, what you need to know the quotient is C2, the normal subgroup is C4. So what could be the possible actions here? What are the actions of Q1A? In other words, what are the possible group homomorphisms from Q to odd A? Now odd C4 is just Z cross, so it's just plus minus one. So there are only two automorphisms, it's identity on C4 and there's inversion on C4 which sends X to X inverse. So there are two possible actions, alpha one, let's say which sends C2 to odd C4 by letting it generate X identity. It's a trivial action and alpha inverse, alpha minus one which sends C2 to automorphism of C4 by sending a generator to the inversion X goes to X inverse. And now in both cases, it turns out that this homology group which I alluded to, it's cyclic of order two. There is a state homology theory for finite groups which says that if this group here Q cyclic which it is in this case, then you can compute this homology groups very explicitly by taking in this case just invariance of A under the Q action and dividing by the image of the endomorphism of A which is where you take the sum of all elements in Q and then you let it act on A. Now if you do this computation in this case for both this homomorphisms, you find that the quotient is Z mod two Z. So there are two extensions for each alpha one and alpha minus one. So for each of the choice of this action of Q and A there are two extensions like this, the trivial one, the semi-direct product and the non-trivial one which is non-split. So let me just list them here. So in case of alpha one, the trivial action the split extension of C2 by C4 is C4 cross C2. So this is a mableon group. This is a normal subgroup. This is the quotient and the quotient acts trivially by conjugation and the non-split extension, the unique one is cyclic group of order eight. Again, this is a normal subgroup in here, quotient is C2, the action is trivial but this is a non-split extension. There is no way to lift this to a homomorphism into here such as the composition is going to be identity. And similarly, in the case of alpha inverse if you want to construct extensions of C4 by C2 by C4 you find that there are two, one split extension which is by Hidl group of order eight D4 and one, which is non-split extension quaternary group of order eight. In both cases there is a cyclic subgroup of order four in the case of D4 these are rotations. In this case, the reflection gives you a splitting of this exact sequence. And in this case, you think for example subgroup change by I, but there is no splitting. So in this case, this is a non-split extension. And I will start next time by explaining what it has to do with solving the embedding problem. Okay, I think it's been 15 minutes so let me just stop here. Okay, thank you. So let's please join me in thanking Tim for this lecture.