 Okay, so we'll now have the third lecture by Professor Tim Dritzer, so please go ahead. Thank you very much so let's see. The last lecture I spoke at the end of the lecture a bit about split and non split group extension so this is pure group theory and in particular I mentioned just to give you one example sorry I think I lost my annotation tools just give me a second spotlight. Yep, and I gave you an example that there are four extensions of C2 by C4 to split to non split or if you want to where C2x on C4 with a trivial action and to when it acts by inversion so all focus abilities are there you get really four different groups here which all sit in the same sort of exact sequence. And this is closely this problem of whether the extension split or not turns out to be closely related to what we've been talking about, which is an embedded problem. So, mainly, when you try to embed an extension is Galois q. So think of q being C2 into a larger Galois extension which has this Galois group. So remember when you have one Galois extension and another that means that one smaller group is a quotient bigger group. Then we've seen that sometimes this is possible. This is not possible. So recall that for instance, we've looked at which theorems that not every C2 extension. Let's say of q can be embedded in into a C4 extension, let alone C8 extension, there are abstractions to doing that. And similarly, you cannot always embed a quadratic extension into Q8 extension, because if your extension is QD, then from that which theorem, this D has to be represented by a form, which is some of three squares. So numbers minus one or seven, which are not something three squares. The corresponding quadratic extensions can never occur inside a Q8 extension. So we say the embedding problem is abstracted. And now it turns out that it's not a coincidence that it's these two groups here, C8 and Q8 in this list, for which they're embedding problem is abstracted and it turns out that when you have a split extension. So in this case, when you're looking at these two groups, so split extension of a group by an Abelian group, then the embedding problem always has a solution. So you can always lift a Q extension to a G extension. So this is what I'd like to talk about. So I think this theorem was discovered a few times, and it appears in different sources, which don't quite quote one another, but I think the first one who noticed this was Saltman, and he proved this result here that if you've got G which is a semi direct product of a split extension, a semi direct Q, where A is Abelian, then you can solve an embedding problem and lift any Q extension to a G extension, or let's say, if you've got a regular, that's the case. We, he was also interested in extension was Galvan group Q, let's say over Q, T1, TN can be always embedded into regular G extension over the same base. And I don't know a very explicit version of this theorem, unfortunately, that would be quite helpful and I put it as a research problem. But I do know an explicit version when the Abelian group A by which you extend is cyclic. So let me give it to you, because it's very similar to the theorem that we had last time about how to construct cyclic groups in the first place. And if you recall, the way we constructed cyclic groups using kumar's theory, we sort of said, well, if you want to construct a family of CN extensions over the ground field, then you're joined and through to unity, then you take an arbitrary element q z to n, we call the G, and then you look at its n throats and then you hook up some sort of combination with roots of unity, and these n throats of these guys in such a way as to get us to let the right Galvan group. So this is the same thing here. So this is the generalization if you want to this construction where you have an extra q extension at the bottom, because remember we're not trying to construct a CN extension, we're trying to construct a CN semi direct q extension. In other words, we've got already this quotient we've got a family here, and we try to extend it to a family here. In other words, we want to, we already have this field k, and we want to put a cyclic group CN on top, such that it has exactly the right action that we want of this group q. We've got a finite group q, that's our quotient group here, we have n greater or two, which gives us the cyclic group CN, and we are given a homomorphism, which completely determines in this context if you recall, the split extension. It's simple morphism from q to automorphisms of CN. So it's some homomorphism. And Galois extension, let's say if q q of t, it doesn't really matter for what I'm going to say, let's say if q and then we put them later in a family over the rationals with Galois group q, which is disjoint from q z. So the idea here that we kind of do exactly the same, we join n through to unity to k. So again, now remember what we're trying to do, we're trying to construct a cyclic extension of k with certain property. So what we're going to do we're going to join n through to unity to k. Now this condition that k is disjoint from q z Diane, which is kind of generically going to be always to anyway, in any family, many regular family. So the guarantee is that when you are joined, did I am to to k you get again is extension of the same degree the same Galois group that's not and that cross, and we can identify it with that one and that cross exactly as before. So we identify Galois group of kids that are over K, with Z mod n that cross by saying that they can integer J, take a residue class in Z mod n that cross. And then the corresponding automorphism, which sense and through to unity fixed primitive and community to its J power gives an isomorphism like this. And it will basically the same polynomials as we had before for CN, except that there's a one little extra summation here, which goes over our automorphism group on the bottom. So it goes over Q. So again, this is going to be a polynomial and so it's going to have an roots, it's going to give the CN extension here. So the product of X minus something and terms. And this something is some combination of answer to unity and and through of the following things. So again, I take an element. If you want a generic element now in K, Z, so it's this field where I joined into unity. So on this field, I have two Galois actions. First of all, there is an action of this Galois group here, like I had before, which sense and through to unity, you know, to the J powers. But also, because my K comes with automorphisms, it comes with an action of this Q here. On elements in K or if you want in case it again by by this group Q. So there are two different actions, one by elements, if you want to set what ends at cross and one by elements of Q. And so we take and through to various things of Sigma J of Q of G for varying Q and very in J. And then there are also possible ends routes we could take of various conjugates of this element G, and we put them sort of cleverly together into one sum. And then it turns out that we have exactly the same thing as before. We get a polynomial with rational coefficients, and then for most choices of G, it is going to be irreducible. And if you construct a field by joining roots of G to K. So you get an extension, which is a Galois, which is Galois, is Galois group CN, and it is set up in such a way that it's also Galois over Q. And it's Galois group over Q is exactly as we wanted it to be CN seminar. It's just not very hard to check by just, you know, acting on its roots by various groups and seeing what what this action does to them. So this is how it works again it's very explicit because you can do these computations very easily with the complex numbers route to the nearest integers and all that. But this has also been implemented in the package as a function, which is called extension which I'll illustrate probably in a second, which takes a family of Galois groups. With Galois group Q, it takes the group G and gets a normal it gives it give it a normal subgroup which is sickly for the end, and then it lifts this family to a G family. As usual, I'll try to watch chat in case you have questions. So let me give you a few examples just to see kind of what sort of groups you can construct using these constructions. I just thought that you can get quite far with. So, let me start with small examples such as symmetric group and three lettuce or dihedral groups. So, let's start with S3. And this is the one baby case of an extension problem, because S3, you can call G this is my group, it has a C to quotient, and this turns out to be this, it turns out to be a split extension. And this is C2L, which is sickly group of orders. So remember again, if you S3 is a group of permutations on three letters, and even permutations they form a subgroup, which is this one, and the side of a permutation gives a whole morphism to C2. And now, first of all, it's a split extension, which is a general result. And this follows from this homology nonsense that we talked about last time, there's some group H2, which classifies these extensions H2 of Q, A, and if Q and A are co-prime, it turns out that this group is killed both by the order of Q and order of A and it has to be trivial. So there are no extensions other than the trivial one, the semi-direct product, and this is called the Schultz-Rosenhaus theorem. It says that when you look at extensions of group A by group Q, and the order of Q is co-prime to the order of A, then every extension is split. There's one extension for every action. In this case, this is therefore a split extension. And it's a special case of what we now did, because this is split extension with a cyclic kernel, and therefore we know by this theorem of Schultzman that the embedding problem is always soluble, so every C2 extension, you can embed an S3 extension. And that's a neat result, which is, well, not completely obvious that every quadratic extension is contained in some S3 extension. In other words, you can always, for every integer D, cook up a cubic polynomial, which has discriminant D up to squares. The corresponding splitting field contains a quadratic extension QD. And if you, you know, use this function here, and you see what comes out, you say, well, let's look at the family, the obvious family of C2 extensions, just QA join root A, so it's a family given by the polynomial x squared minus A, and you ask for extension of this family to S3 over C3. And then what comes out then is this family here, so it's a family of S3 extensions, such as the discriminant of this polynomial up to squares is exactly A. So you can use it to construct an S3 extension with any given quadratic field inside. And it's automatically regular from these properties, that's not hard to see. And then you can do for arbitrary dihedral groups, which S3 is a special case. So for example, the hydro group Dn, it's also a semi direct product of a cyclic group by cyclic group CN by C2. And that means that, first of all, you can realize it regularly. And really again, you can embed any quadratic extension QA join root A in a dihedral extension, such that the quadratic cutout by CN is exactly the quadratic field that you want. And again, you can construct it the same way, even though recall, it's the same problem as we had already with cyclic groups, when you construct them even without any Q, the coefficients of these families the size grows quite, quite badly. So these are reasonably unwieldy families, if you take N to be large, for example D11 is pretty horrible. On the other hand, in the book, for example, Mali and Matsatom and Vizgala theory, they didn't even have a D11 family at all, because this was one of the groups that this was just too difficult to construct computationally. And, you know, now you can do it, even though the results maybe are not very pleasant to look at. So that's two examples. So now let's take a little bit of a deep breath and see what is it that we've done. So the last two lectures, we constructed regular families for cyclic groups, and now from there it's quite easy to get a billion groups as well, because generally it's easy to construct direct products of two groups, if you can construct regular families for both of them. For example, there is an exercise to do this carefully for C2 cross C2, but the same idea works for two groups. Basically, if you take a family for C2, such as x squared minus a, and take another regular family for C2 in different variable, x squared minus b, and just taking a product of these two polynomials. It's very easy to see that it gives you a regular family for C2 squared over QAB. This you can do for any direct product, and then Hilbert specialization here and tells you that specializing these things, one of these variables to, I don't know, specializing b to a plus one or something probably will give you a regular family in one variable if that's what you're often. We can do cyclic groups, we can do a billion groups. We can do direct products and we have this new construction that says that if you can do a group Q, then you can do a semi direct products by Q independently of N and independently of the action here as long as this extension is. So if you look at small non abelian groups, you discover some pleasure I guess that we've covered all of them, except for one, which is the cotonium group, which I'm going to do next. Because if you look at small non abelian groups in this case these are non abelian groups of order less than 16. All of them, except two, except q eight and a four, turn out to be a semi direct products like this cyclic by by a million, as you can see here in the list. So the only two groups missing a four and q eight and a four we already know there's Hilbert's result on a and. You can also apply even though again it's not maybe as explicit salzman's theorem because a four is actually the special case of it. It's, it has a normal subgroup C2 squared which is a billion and the quotient by which is cyclic of order three. And again, the two orders are called prime here so that's automatically split extension. So, someone will also tell you, and so does Hilbert that you can construct it for regularly. So the only missing group here is q eight. And it turns out that well this is this group is not a semi direct product of anything by anything. It's quite easy to see because this group has subgroup plus minus one to the elements in the q eight plus minus one plus minus i plus minus j plus minus k. So minus one has a property that it says group subgroup, which is contained in every other subgroup of q eight. So, all the other subgroups generated by IGNK they all contain it, and such a group can never be a semi direct products because for semi direct product you need two subgroups sort of a copy of your a and copy of your q one normal one not, which is a trivial intersection, and you can't do it, we can't do a trivial intersection is every to every subgroups contain the same elements, plus minus one in this case. So this is not a semi direct product. But nevertheless, it turns out, sort of, interestingly, that you can still construct using exactly the same construction. Next, I think there's a question in order to apply Hilbert's visibility theorem. No, the family does not to does not need to be regular so Hilbert's visibility theorem just applies in general it doesn't need the word regular of course then what you get is a family, which is not regular. But if you're just interested in the inverse galore problem of a q that's certainly absolutely fine. It turns out that q eight is not in itself a semi direct product, but it's a quotient of a semi direct product. So it's a quotient of a group which is C four semi direct C four of order 16. But to which the theorem does apply, which is sort of quite interesting, because many often when you do this kind of inductive constructions. You say, well, you know, suppose I already constructed something smaller groups, let me now do it for larger groups. In this case, if you want to construct q eight using this theorem, you first need to construct a larger group, and then take its quotient. So the class of groups that we're looking at, which are these kind of semi direct products of things by CN, it's not closed under quotients. There are groups like this, which are semi direct products like this, whose quotients do not have it do not have this property, which is nice for them because again we can construct q eight using C four semi direct C four. So let me do it's very explicitly hopefully you don't object and show you you know how do you do this kind of practice. So first of all, if you look, let me just do it on the group names database for for q eight. So this is the quaternion q eight of order eight. Here's a lattice of subgroups which I alluded to. It has a center plus minus one and then three subgroups generated by ij and k, and then there is a whole quaternion. So this is a quaternion by generate some relations. And what I would like to look at here, I would like to look at what is it a quotient off. And if I look what is a quotient off, it gives us the list of groups of which q eight is a maximum quotient. And in this list, there is this group of what a 16, which is C four semi direct C four. I don't have to look at it really, except the fact that I have to know how to produce it in magma and magma. This is a small group of 16 comma four, because that's what that's what this is. This refers to the small group database. It's a larger group. It's a semi direct product again with C four by C four it's a group of what is 16. And you can look at it's you can look at it's generators and relations, which tell you exactly, you know what it is a and B are generators of the two C fours. And this tells you how one of them acts on another. It acts like a like a like in a dihedral fashion by inverting it. And And what I was going to do, oh yes, you can see now that this group has a subgroup normal subgroup, which is cyclical for the two, such a quotient by it is this key. And that's the thing that I'm interested in. So what I'm going to do, I'm going to construct a family for this group here, and then take a quotient and construct a family for Q. So let me try to do it kind of on the fly, even though we should have advised against doing this life but let me try to do it anyway. So here is our group. There's a small group of 16 comma four, which, but my think will probably refer to the same name C four by C four, it's some group of what is 16. To construct a family I need to find the abelian subgroup in there, which I'll call a which is C four such quotient by it is also C four. So first let me compute its normal subgroups. It's convoluted with magma unless you have a macro doing this which of course everyone does. So here is a list of its normal subgroups and I'm going to select one from this list. Here is a list S of all normal subgroups of C four, I think there's five of them or I forgot that's quite a few of them if I remember correctly, maybe even 11. But I'm interested in a specific one I'm interested in, in one of order four. I want the order of this group to be able to form. I want to be cyclic. I want the quotient by to be cyclic as well. We didn't give an error message so I guess my search worked. So now I've got a subgroup a, which is some group, whatever, of order four, which sits inside my group G, and which has a property that it's C four and the quotient by it is also C four. So now, as I mentioned before, there is this extension function, which lets me take a C four family and extend to a family for this group. So we're C four now refers to the quotient. So let me take a family for C four, for C four. There it is. This is a built in one built in family of C four extensions. This is the first transitive group of order four. So here's my family. And I'm going to do I'm going to extend this now using this semi direct product construction to my G. So again, this is my group C four by C four for the 16. This is my normal subgroup C four which I found here. And this is the family for the quotient group. And this is now hopefully going to be the family for C four by C four. So it's given by some polynomial in X of whatever degree it found necessary was coefficients, which are now polynomials in a and this is a regular family. And now, as I mentioned before, my group has a Q eight quotient, you can still see it here on the slide. So therefore, inside the splitting field of this extension somewhere, there is a Q eight extension. So let me, let me find it by the same way so let me redefine my G to be now the group of this family. And this is one when constructed it, it computed some gala groups and probably redefine it to be a different permutation group, and the one that I started with. So let me redefine my G it's still the same group as an absolute group. Again, look at it subgroups, which are normal. And now what I would like to do, I would like to find a normal subgroup of order two, such a quotient by C eight. So here, let me, at the same assertion, there exists a normal subgroup list, such that its order is equal to two, and the quotient is Q eight. Well, I'll just go by group name by quotient of G by M. Again, it exists. So this subfield here, if you now go by gala theory. So this group here gives you a gala extension of degree 16 was this gala group. And this subgroup that I found of order two, it cuts out a quaternion extension. So what I would like to do I would like to ask for subfield inside my family, which is cut out by this end here. So this is a polynomial of degree eight, such that if I ask for a family by by this polynomial, I find a family of quaternion extensions. It's not the best family there is. So it's given by a polynomial of degree eight. And this is best possible because if I mentioned I think the first or second lecture that quaternions are one of those groups, which although it was order eight. There are eight points in the regular fact matter and and on nothing else. So the smallest polynomial you can have for this group is of degree eight. But it has quite large coefficients where large is measured by the degree eight. If you look for the best possible family rather I know for the quaternion group. It's a transitive group of eight, five correctly. It's again a polynomial eight, but it's coefficients you see a small one, which is of course nice because if you experiment with this feminist for example, you want to test some conjecture class numbers or whatever for quaternion extensions by plugging in a equals 1234 and so on, generating these coefficients and then testing your conjecture. Then of course, you would like these coefficients to be as small as possible. And for that they have to have smallest degree in a as possible. So normally you're looking for families of small degree, which this particular construction is not particularly good at. But nevertheless, what I just wanted to show you is that it is possible to construct you eight, essentially with the same construction, even though it is not a semi direct product. So, and I'll leave it as an exercise I just put one to do it for q 16 on the homepage is that the same approach actually works, and you can construct a q eight q 16 q 32 and so on regularly with someone's theorem, or if you want explicitly for any given sort of power here like I did here for q eight with a similar approach. This goes back to one of the questions I think I was asked when I was talking about this embedding problem is that if you want to do it let's say for q eight q 16 q 32. Can you formulate this embedding problem explicitly. Now, I don't know how to do this explicitly sort of for q 16 q 32 and so on because this embedding problems they, they do become well reasonably unwieldy. But interestingly, the approach which we now have it avoids somehow it magically avoids this embeddings probably completely, because, as we've seen, as you've seen in the construction of q eight. We never had to do any sort of obstructions, we just said let's take C for extended to C for semi direct C for we can always do it. Of course, C for I mean the quadratic fields for example inside it that restricted and so on so somewhere inside there, there are some obstruction problems hiding, but we never had to look at them in order to construct this group or even these groups here. So this is a very powerful approach. And you can ask yourself, how far can we get with this, what is the class of groups that we can construct by using, you know, the technology that we develop so far. So this class is called semi abelian groups and it covers a lot and a lot and a lot of soluble groups. So I think it's a very interesting one to mention here. So let me do it now. So what have we done so far so about from SN and AM is Hilbert's construction which let's take separately and let's not look at it. So one, we prove that we can realize regularly over Q of t cyclic groups, what we've just did, except maybe not always explicitly but it doesn't matter as you will see in a moment if you just interested in theoretical results split extensions a semi direct q when it's an abelian group and, you know, q already has a regularization, so we can lift from q to a semi direct q. And it's also not very hard, even though I didn't talk about it yet to realize direct products and what I mentioned very briefly with products maybe I'll talk a little bit about them in the last lecture, maybe not I don't know yet. But I already mentioned that these things exist. So direct products is when you take two different gala extensions which are these joined, and you look at the compositing. And with products is what happens when you put one on top of one another, you start with a gala extensions group, age, and then you put on top of it gala extension is going to be then generically the group that you get is what's called the product g with age, and it is not very hard to prove that if you can realize g and age regularly, then you can also realize g times h and g with age. And it is interesting because this is one of those examples where you do need regular and you do need to do it over q of t. And there is one sort of cute examples where we don't know how to realize a direct product if you just do it over q. There was for a long time, the smallest simple group which was not known to be realizable over q was a group SL to F 16. There's a paper by boss man from just here who really proves how to realize it constructs an explicit polynomial with that gala group. And, you know, I remember, he just proves using complicated theory of gala representations and trust field theory that you can construct explicitly one example. So we have one example of SL to F 16 field. But that means for example if you want to construct an extension with gala group SL to F 16 cross SL to F 16 over q that we don't know how to do it, because you need to disjoint once in a while. So it's a bit kind of embarrassing, but, well, that's sort of how it is. And similarly, just because you can construct over q, it doesn't mean you can construct it over a number of you, for example, over the splitting field of the extension that he's constructed, we don't know how to construct as a 16 extension. That's all fine, because basically by shifting x, if necessary, in such a way as to make the ramification law side of the extension disjoint. You can always cook it, you can always set it up generically it will always be the case that from two families with a given gala group, the same group or not same family or not, just by shifting the variable, you can easily construct direct products and these products. So these are the things that we can construct safely groups or more generally split extensions by a billion groups. And if you wish, with products and and direct products. And these constructions if you ask what they generate, they generate a class of groups which are called semi ability, which are important in in risk of our problem for, you know, for this obvious reason. I don't know if anyone uses the terminology, apart from people who work in in risk of our problem. But, on the other hand, I don't think anyone else uses the word semi ability anyway. So I think it's been reserved in that context. So, a few people looked at it and the terminology was coined by matzat and then dancer and because stolen acting on his earliest papers proved some equivalent conditions for for this clause. So let me state it as a serum by just putting all these together by matzat dancer and stole the following conditions on a financial group G are equivalent. So the first one is really what we're asking now, what can we obtain in finally many steps in the way as we obtain the quaternion group q8 by taking split extensions of a billion groups and then take equations. So you start with a group, which is just a trivial group see one. So after that you say okay well let's take an extension split extension of this group by an abelian group. In other words, let's take any abelian group G one. If you want to take its quotient, then extended by another billion group a to take a quotient extend by another billion group and so on, and proceed in this way. So you can finally many steps, you can, you can get your group G, then this is called this group is called semi abelian. And by salt months theorem, which remember proves that once you can realize q, you can also regularly over qft or qt one T and whatever you can realize a split extension by an abelian group. So it's immediately that the inverse gamma problem let's say over qft, which I call the script IG and qft is true for semi abelian. It just follows immediately from salt months theorem, and this generous and this construction. And then conditions towards three are just for the current ways of doing this. So, second one says that G is generated by abelian groups, some groups a one up to am so far it doesn't say very much because any group is generated by cyclic groups you just take every element, you know, generates a cyclic group. That by itself doesn't say very much, but the strong condition that you want here is that you can order them in such a way that a J normalizes a I for all I less than or equal to J. And another characterization, which is in fact due to make a stall is that G is a quotient of iterated with product of a cyclic group cm with itself. It's a very neat compact way of writing such a class, in particular, proving this theorem by avoiding salt money if you wish, because that's what he also did. It's quite easy to, again, realize cyclic group cm and then you can prove that you can realize with products and by doing it repeatedly and taking quotients, you can construct at least in theory, all groups like this, even though this approach, you know, this particular definition is not always practical. And, well, if you want to realize specific group, you certainly don't want to do it in this way. Because a group satisfying these equivalent conditions is called semi billion, and for these groups who do know the inverse Galois problem. Okay, so here is a bunch of properties for for semi billion groups. So first of all, if you look at any of these equivalent definitions, it's very easy to see that all semi billion groups are soluble. Non soluble groups that are definitely not covered by this construction. But otherwise it's quite a nice class because you see that it's closed on the quotients closed on the direct products it's closed on the research products, basically from, from definition, and I didn't mention it here. But it's also closed by this construction of taking extensions of a billion groups or every split extension of an abelian group by semi a billion group is against a billion. And there's a slightly more general version here, if you have a group G, which is generated by a normal abelian subgroup a and semi a billion group you then G itself is semi a billion. And in fact the converse is to that if G semi a billion, then this generated by a and you were used strictly less than G, and a and G is a billion, and this allows you to do sort of this gives you kind of inductive construction for a semi a billion groups, and also a way to check if a group is semi a billion. And then there are people who worked the inverse color problem. People whom I mentioned before like told son and dancer, they proved a variety of theoretical results that certain classes of groups are semi a million. So first of all, if you've got a new boat and group what's called class two. If you don't know what a class of a new boat and group it doesn't matter here, because this just means. That the derived subgroup of G is contained in its center. Then such a group is semi a billion. And if you have a soluble group, all of who still have subgroups are billion the G semi a billion, and then to show that if you look at P groups that for small powers of P all those groups are semi a billion. So in particular inverse gamma problem is true over Q of T for all groups here for all groups of order P to the n, where n is at most for and groups of all two to the fight. So the cotone and group for example q eight and q 16, they are in this category here. So we know that they're semi a billion. Therefore we could have predicted that at least for these two groups, you can construct them with this, even though they're themselves not semi direct products of anything. They can be constructed using this repeated construction split extensions by a billion groups. So we found out that semi a billion groups, they cover a lot of ground. So if you, for example, look at groups of order lesson 64 there is 319 of them that is quite a large list, and only five of them are not semi a billion. And I listed them here. There is a five or 60, which is not soluble and it's the only non soluble group in this list here. And remember, non soluble groups are certainly not semi a billion. And the other five groups. 123456 Oh, I said five I think there should be six here. Unless I copied group twice, which I probably didn't. So the other five groups that you see on this list, one of order 24 and four of order 48. They are closely related to the group SL two of three. So they all contain SL two of three is a sub quotient. So this is some sense it's a smallest example for which this technology fails in a sense that it's a smallest example of the group, which you cannot prove it's similar over QFT by this construction, you can take with product of slightly groups with themselves as long as you want, take quotients, take direct products take with products, and all these constructions will never give you a group SL two three. So this is a very interesting example, and in fact, quite fundamental for this list, because all the others are related to it. And a five we know how to construct anyway, using Gilbert. So what I want to do in the last, I think, eight minutes, if I know, if I'm careful with my time is to talk about just this group SL two three. This is a very interesting group for a variety of reasons, not because of inverse color problem, but because it just occurs everywhere all the time. In fact, from what I've seen, like the site group names, which I mentioned here where I use C for semi direct C for this is the group, which is, I think gets the most hits on that website, because most people are interested in this particular group for some reason or other. Now, I know it mostly as coming from elliptic curves. It's a large automorphism group of an elliptic curve. And because of this also it's the largest possible inertia group that you can have in a gala representation of an elliptic curve. And it occurs in many other contexts, and it also happens to occur in this universe gala problem as being a difficult group. I observed that this group is difficult. So this paper by Smith from 1999, where he tries to construct families of gala groups over QFT for all transitive groups of degree at most 20. Sorry, transitive group of degree at most eight. See there's a question, what does equal 20 means. Oh, I'm sorry, you mean this one. That's not a 20, even though it looks like 20, it's a 20. It's people in crystallography. They have a standard notation for symmetry groups of platonic solids and rotations of platonic solids. And this particular specific group is called 20 in their terminology. Thank you. So, so he tried to look at transitive group of degree at most eight, and this particular one. He did not know how to construct a regular family. And he comments on this, there is quite a lot of research that several papers at that time, written on this specific group. And, and despite that, no explicit polynomials giving a geometric extension of QFT seems to be no, which is kind of interesting because if you look for example an element TV, and if you and if you look at number fields which have color group SL 23. There's quite a lot of them there. And it's just a question you know how to fit them into a family, but that seems to be quite difficult. So it's a very interesting group. So let us, like, to be honest, in the same year, I think the book of my land matzat came out and they do have a family for SL 23, even though they don't say where they got it from. Whether they just, you know, fit from a database or where they had sort of more conceptual construction, but I know one construction which doesn't work and I would like to explain it here probably can't do it in five minutes but at least I'll start it. And the reason it's interesting is that if you can do the same sort of thing for general soluble groups, I think you could get quite far in getting soluble groups over QFT. Okay, so the way most people and the wheels going to think of this group is a semi direct product of Q8 by C3. So it's again a semi direct product of a group by group, but in this case, this group here is not a billion, and that makes all the difference. So it's a semi direct part of Q8 by C3 and nevertheless what we're going to prove that in this case you can solve the embedded problem and lift any C3 extension to an SL 23 extension and along the way construct a regular SL 23 hammock. So you see this. So the Quaternion group, let me just know for one's list of its elements, it's a group of one and eight. The its elements are plus minus one plus minus I plus minus J plus minus K, where the defining relations are this IJ and K, they have four, they square to minus one, minus one is central commutes with everything. And you have the relations IJ equals K, KI equals J and JK equals Y, which I'm sure we've seen before. And if you've seen this before, you must have observed that this group has an obvious C3 symmetry that you can permute IJ and K cyclically. You can't do an arbitrary S3 permutations because you cannot swap IJ with one another that ruins this relation here, but a cyclic permutation preserves the group structure. In other words, this group is what's called as an auto-automorphism of order three. So there is an obvious automorphism action of C3 on Q8. Or if you wish, there's an obvious homomorphism from C3 to automorphisms of Q8, where a generator of C3, let's say, permutes IJ and K cyclically. So there is a map, phi, let me call it phi, from C3 to automorphisms of Q8, and that means that we could construct a semi-direct product of Q8 by C3 using this action. This is not how you define a cell two or three, you can see here from the name, it's defined as a group of matrices, it's a group of two by two matrices over F3 of determinant one. So there's three to the four matrices over F3, 48 of them form a group G or two or three of determinant plus minus one, and half of them 24 have determinant one. So that's how this group S or two or three is defined, but it has this strategy. So inside it has a quaternion group of Q8 inside it has three subgroups are generated by I generate by J generate by K inside it it has plus minus one, which is in fact central for the whole group and inside it you have identity. So again in Q8, these three are index two, they are normal, they are not, you know, preserved by conjugation, but in SL two or three, where you have this extra element of order three which permutes, which acts on Q8 by conjugation, this IJ and K are permuted. So these elements here are permuted by conjugation by element of order three. And so you just end here by Galois correspondence, this is now what we're after. So let's take this picture and put it upside down, because that's what Galois correspondence does. Maybe I should have put this picture upside down then I would preserve the inclusions but I didn't. So this is the tower fields that we're looking for this, for some reason has disappeared should be a little K. So this is the base field over which we're going to work. Then we're going to take an arbitrary C3 extension of little K called capital K. And then on top of it, we're going to try to construct the quaternion extension, using which result and what we know about quaternions and some extra input that will need in such a way that the three that will have along the way will be permuted by this C3 action. And this will guarantee that the total extension that we're going to get is going to be an SL2 of three extension. So it will require a little bit of work, but in the end, it will, we will end up with an SL2 of three family. Okay, I think I just ran out of time so it's probably a good time for me to stop here. Okay, thank you for the great lecture. Please. Are there more questions? I'll watch the chat or otherwise if you wish you can ask them live. Is there a sense in which most solvable groups are not semi-abelian? No, I'm not sure about this. It's quite possible in fact that most soluble groups are semi-abelian, because if I remember correctly, this null potency class two, it's quite possible that, no, almost all groups are null potency class two, just because, you know, they're all two groups and in two groups. Oh yeah, yeah, I think you might be right. It kind of depends a little bit how you count, because I think saying that most finite groups on your bones in class two, so it can prove something for them, we're almost there. This is completely misleading. So, so I'm not, I'm not sure how to quantify this. Yeah, are these all subgroups of SL2 of three? No, because they're set because this is a center. And there's definitely a cyclic group of three in there, which by the way, I can't even see here this. So in this, so, okay, let me just show off where is the website. So, here is SL2 of three. So if you want a slates of subgroups, you will see that it's almost a whole list of subgroups. So this is a list of subgroups which I've drawn. These are three conjugate now subgroups of four to four, Q8 and SL2 of three. So the only thing missing in this list is that there is a cyclic group of three, which is three silver. It's not normal certainly otherwise this whole group would be a direct product created by C3. There are four conjugates C3s, and you can also take them and enjoy the center plus minus one, and then you get cyclic group of four to six. And incidentally, these are all possible inertia groups you can have at P equals two if you look at the elliptic curves. Yeah. Because again, this group, that's that's how this group occurs. Okay, I see there's an announcement. This is a crucial session today probably right after this lecture starting in 10 minutes. Zoom. It's very nice. And is there a magma code to do extensions by a billion groups as well. No, but there's a problem on my list of research problems which asks to do that. So, I don't know an explicit or the very explicit analog of the theorem that I've given for cyclic groups for a billion groups, only when the civilian group is an after vector space I have a version of this. And it's sort of alluded to that it should be possible by the people who prove this result, but I haven't been able to do that. So if anybody knows how to do that for general billion groups, I would very much want to hear that. Okay, is there a nice geometric interpretation of the reef product. The fact that you know if you have a cover was going to be G and on top of it a cover was going to be H and the whole group is expected to have that cover G reef H. So for example, D for that's your typical example for his product, you take a random quadratic extension of q you join a square root of I don't know I take a random correct extension top of it joining square three plus I hopefully that's not a square. And then you're going to go to D for extension like that. So we sport of C to buy C to, but I don't think I can say anything more than that. And maybe this is not the kind of geometric interpretation one but you, I mean it acts on the reef product acts on a tree. So, you can describe it that way. No. Yeah, I can. Anyway, you can talk more about that later I guess. All right, so let's see. Any other question. So, let's see there is. I guess that all that I think that's all the questions. I think that's all the questions. So all right so let's let's end here then and there's this I guess yeah so go ahead and join the social social meeting if you want. The next talk is in about an hour and 15 minutes from now. So, Christian Lutter will get her next talk. Okay. So, yeah, all right, so see you later.