 So now we will discuss the first Grigori group. I have sketched the definition here on this blackboard, and now we discuss it more in detail. So this is a group acting on a root of 3, and the action of the generator's ABC is defined recursively. That is, A just permits the branches, and the action of BCD is defined recursively in the way presented on these pictures. Maybe I mentioned also some equivalent language. So these are groups acting on root of 3. And initial language of the original Grigori-Chuk paper was using intervaling changes. So you can see the interval from 0 to 1. Let's say 0, not included 1 included. Then A permits 2 halves of this interval. B acts in the following way. So we have 0, 1 half, 3, 4, 7, 8, and so on. We can see the points of the form 1 minus 2 to some integer power. And we say B acts in the following way. On the first interval that we have drawn, B acts like A, meaning permits the halves of the interval. On the second interval from 1 half to 3 fourth, B permits x like A, that is, permits the halves of this interval. Then on the third one from 3 fourth to 7 eighth, it acts 3 value, identity on this one. And then again, with period 3, we have again like A from 7 eighth until 15, 16th, like A, like A, and so on. So we have a sequence. In this notation, one can say a B acts like A, A, identity, A, A, identity, and so on with period 3. Now C acts like A, meaning my drug, you can picture, so. On the first interval from 0 to 1 half, it acts like A permitting the halves from 1 half to 3 fourth, it acts 3 value. Then from 3 fourth to 7 eighth, again permitting the halves of the intervals and so on. So C acts like A, identity A, identity A, and so on. Again, with period 3. And D, the action of D is a negligence, or only we start with identity, that is meaning from 0 to 1 half, A, D acts like identity. And then from 1 half to 3 fourth, permitting the halves of the interval, from 3 fourth to 7 eighth permitting the halves of the intervals, and so on. Identity A, A, and with period 3, identity A, A, and so on. This is just the same definition I explain drawing trees. The halves of intervals correspond to the branches of the tree. And though what is nice with BCDs, it's convenient to use both notation, so it's easy on these pictures to draw explicitly the action of BCDs. What was good that BCDs didn't permute the first level, and the only interesting action is occurring along this ray. So something really interesting in the language of the interval occurs only close to 1, and that's made possible to make these drawings. In general, for some more complicated groups, usually it's more convenient to use nice and relevant language of trees. But I wanted to mention the languages of intervals just to mention that some orientation not related to what we discussed today, just pointing out these groups defined as groups exchanging subintervals, but it is important in this definition there is an infinite number of intervals involved in the definition. But just to mention that even if we consider very different or priori class of groups of interval exchanges, when we allow only finite number of intervals, many group theoretical questions seem to be open. So here there are people who know, who are really specialists in this subject, who can say much more than I can do. But apparently, if you can see, for example, a group with finite number of intervals, real group of interval exchanges, it's not known whether the growth is always exponential or at least, you can confirm, right? Exponential or polynomial, but this group somehow fits in a more general context where infinite number of intervals is considered. So this is our group. So by definition, it has four generated by four elements, a, b, c, d. But actually the first observation that three of them suffice. So the generating system will be considered, and which is really an important generating system, a, b, c, d. But in principle, this group is generated by three elements, not only by four. Just because there are some straightforward relations in this group, I firmly that there is a remark. First of all, a square is identity, and the same for b square is identity, c square is identity, d square is identity. And d is equal to the product of b and c. And actually, so actually, b, c, d generate a finite abelian group for elements. So c is equal to product of b and d. And b is product of c and d. How do we see that? First of all, a square is just in volution. So we see from the definition a square is identity, right? Now what about b? On each interval, b acts either like a or like identity. So a square is identity, and identity square is obviously identity. So just on each interval, we see that the restriction of b square to these intervals, we consider that the definition is identity. So b square is identity. And the same for b, c, d. Just on each interval, we had only a, or identity. The square is identity. Now, y, d is a product of b and c. What happens if you take the product of b and c? So on the first interval, so let's write the action of b and c, right? On the first interval from 0 to 1 half, b acted like a and c acted as a as well. So we have a square, meaning we have identity on this first interval, right? On the second interval from 1 half to 3 fourth, a acted like a, I mean b acted like a, and c acted like identity, like identity. So we had a times e, we had a. On the third one from 3 fourth to 7 eighth, a acted trivially, and c acted non-trivially as a. So we have identity times a. We see, and then we know with period three we have the same action, right? So on the first interval in our definition, we have identity. On the second, we have a. On the third, we have a, and so on. So we see the product of b and c is the d, and analogously, we have these relations, which will be important for us. And so we can reformulate this statement saying that if we have an element, take some element in our group. This group is called first Grigorych group. This is the first Grigorych group. Take some element in our first Grigorych group. Suppose it has some length. So with respect to s of g is equal to some n. Then there is a nice standard form to write, to express g in terms of a, b, c, d. We can say then, then g can be written in the following way. We write star for b, c, d's. The first star could be present or not. Then we have a, when we have some among b, c, d's. Then we have a, and so on, the last a, and maybe at the end some star present or not. So each star denotes b, c, d's, but not necessarily the same one. So why do we have that? Just since a squared is equal to identity, no need to write a squared, so shortest word for g cannot have a squared. And since the product of two generators among b, c, d's is the third one, so there is no sense to write a product of b, c, or c, d. So if we have a shortest, actually not only that there exists an expression like this, actually any shortest word for g will have this form in the generating set s. And this will be important for all what we will prove about this group. And now we want to show that the growth of this group is sub-exponential. So this was a crucial discovery of Grigorychuk. We want to show some main lemma, so it's a lemma, but it's actually a theorem rather than a lemma. We will show that there exists alpha smaller than 1, such that the growth function of the first Grigorychuk group is smaller than x, some constant, and to the alpha. We'll write a constant if you don't specify which alpha we take, we can forget. To prove this, we will use a certain self-similarity property of this group. This group is self-similar, and actually it's self-similar in several possible meanings. So for us, we will need the following. Actually, what we really will show, which is stronger than the previous claim, we will show the following. That there exists h, a finite index subgroup in G, and a map phi, there exists phi, from G to the direct product of several copies of G, such that the following holds. So s is the standard generative set. s will be as always a, b, c, d. From h, we will need an application from this finite index subgroup, h to Gk. Such that the following holds, we can see the, so we can see that for any h in h, we can see that its length, ls of h. And we can see that the length of, so in Gk, we can see that the length function in Gk will be just the sum of ls of h1 plus ls of g2 plus ls of Gk. And so we can see that the length of the image of h, meaning, so we can see the l of, let's say, h1 plus l of h. We want that this is much smaller than the length of the element we started with. We want to show that this is smaller than some constant gamma. So we want to say that there exists gamma smaller than 1, such that for any h, this is smaller gamma times lh of h plus some constant. Just for, yes, these are where phi of h is equal to h1 hk. So this is some very strong actually contracting property. For example, even if you would require that gamma is strictly smaller than 1, it's already a quite strong property. But for sub-exponential growth, we needed that it's strictly smaller than 1. Just before we explain why this, we have this for the first Grigatu group, maybe I make some general comments on important groups. The existence of such map is a kind of multidimensional analog of a dilation in important groups. If you have an important group and if you have some map of an important group, let's say n, important or virtually important group. A priori, we can start with any group, but only virtually important groups would admit the dilation that we define right now. So we can see the map, we can see the endomorphous psi from n to n. And it's called to be dilation. If for any n, the length is multiplied by some constant greater than 1 such that for any n in n, the length, we fix some generating set. So we can see the sum group n and s is generating set. And so ls, we want that ls of any element is greater than 1 plus something, some constant, times of the image psi of n, of psi of n, is greater or equal 1 plus c, the length of our n. Some very easy example. If n is equal to zd, if n is equal to zd, if you can see the psi, just multiplication by some number, for example, by 2, if you can see the psi, that sends z1, zd to, for example, z1 to zd. It's clearly a group of homomorphism, right? And yes, I didn't finish the definition. And so I defined one property here. So this property to multiply by some constant greater than 1 is clearly satisfied for this endomorphism of zd. And I forgot to say, we also ask that the image psi of n is a finite index of nn. So here, for example, it's clear for zd, we have this map that multiplies all coordinates with some number. So clearly, the length is multiplied by something. And it's clear the index is finite. So the index here is 2 to the power of d, for example. Not all important groups admit dilation, but many do admit. So another example, if you consider, for example, a group like this, of upper triangular matrices. So in one of the exercises, you had the group of upper triangular matrices, right? So with one on diagonal and some coefficients, so some a, b, c, n, z. And if you multiply now, if we send it to something, we multiply by some large number. Actually, 2 is already not bad. But if we multiply by, let's say, if we send it p, a, p, b, we can't multiply c by p. It wouldn't be a group homomorphism. But if you multiply it by p squared, then this is very good. This is a group homomorphism. From our matrix to some finite index subgroup. So here the index will be p times p times p squared. It will be the subgroup in this p squared in our group, right? And one can check that this is a dilation. So I guess it's already dilation with the definition I gave. But it's even easier if we take some iteration of psi. Actually, there are several versions of possible definition of dilation one uses. Sometimes one requires the property I have explained not for psi, but rather for some power of psi. But anyway, so if one understands just how we prove it, so the length of this element is approximately a plus b plus square root of c up to some constant. But having this in mind, so yes, in a quality way. Here it's smaller, so the image is smaller. And here the image is bigger. We wanted the image, yes, here the image, I have not yet explained. Yes, we want it to be, no, the lemma stated is correct. So h is a finite index subgroup that I stated. I explained why we have inequality in different directions for the Grigory group and here. So this map for the first Grigory group that we will discuss now, and that's what we need for the proof of the upper bound, is not an analog of dilation. It's an analog of the inverse map to the dilation. So just for important groups, for a billion groups, in a natural way, usually one looks dilation as a map from g to a finite index subgroup, but one can reformulate. It looks artificial in the case of a billion groups, but becomes not artificial, but really essential for the first Grigory group. If we have a dilation, just reformulation. We can see that just the image of our dilation, right? Let's phi be the inverse to our upside. Phi defined not everywhere, but phi is so. Phi is defined on a finite index, right? Phi is defined on psi of n, and this is a finite, by definition, it's a finite index subgroup, n. Finite index subgroup. It should be the identity map, sending everything to the identity and you would have a map that's satisfied. I just injected. Yes, sorry, I didn't say this is injective map. You will do injective. Thank you. Yes, phi, injective. Actually, that is, it's not so important. In our example for the first Grigory group, it will be a morphism. But the dilemma is true even if you would not assume that. We will construct what we will construct will be a group homomorphism. But just for this general conclusion about the growth, it's important we have just some map. It's enough. So just this analogy, to finish this analogy with a billion on an important case, so if you consider the inverse to the dilation, then the inverse divides the length by some constant, right? So the longer than for this inverse, we have phi of anything, of any n, is smaller than the length of its length, will be smaller than some constant gamma, smaller than 1. We have some constant gamma smaller than 1. And the length is smaller than gamma length of our element. This is our analogy. So now, I have mentioned so many groups admit dilations. For example, groups of upper triangle matrices, no matter the size, m times m's upper triangle matrices admits a dilation. It's not difficult to see. Just looking on homomorphism of this form. For this group, actually, I don't know if it was explained good in the formulation exercise, but there is a natural generating set which consists of you put of elementary matrices. You put one here. This is the first generator that corresponds to our coefficient A. And you put here one. This is another generator. Actually, these two generators are enough. We could consider them a generating set, or we can add also the matrices. It's not important. We could add the matrices like this. But this matrix is a commutator of these two. The statement is true for both of them, because the length is like this. Actually, it doesn't help at all. So you had some indication for this in one of the exercises. So actually, it's not in any way up to constant. Let's say up to constant that it's clearly up to some constant. But to understand that it's really dilation, if you don't know the constant a priori, we don't see it. Why it's really dilation? If it would be really equal like this, it would be straightforward. We have dilation. But that's why it's a bit more convenient, though not really necessary, to take a power of our phi. Then we see no matter whatever. Even if you don't control good a priori, this constant, it doesn't matter. We see its power certainly multiplies by some number greater than 1. So this is our analogy. As I said, for example, all Unitreaga-Marx group do have a dilation. And this is one of the reasons why nilpotent groups have polynomial growth. There are many ways to see why nilpotent groups have polynomial growth. Even those that don't admit a dilation do have polynomial growth. But for our analogy, it's helpful to know just something which is called Frank's lemma, that if you have a group with dilation, a priori, we don't assume nilpotent just any finitely generated group with dilation, admitting a dilation, dilation, then V of n is polynomial is smaller than n to some power. Why? Because let's estimate the growth function. We have some element, G. It's with a finite distance of the image of our dilation. So there is some constant related to the index. So G exists. We know that it's constant C. And for any element G, we can choose some G prime in the image of our dilation, which was phi psi, I think, right? Psi, it was psi, right? For any element, we can find an element which is close to it and which is in the image of our dilation psi. And since it's in the image of the dilation, so it's equal psi of something, psi of some other element H. And we know that the length of H is much shorter than the length of G. The length of H is smaller than some constant gamma times L of G. What does it show us? We see that V of n is smaller than some constant times V of n divided by gamma. Let's write it several times. V of n is smaller than, or not divided by gamma, but smaller than one, multiplied by gamma. We can write this is smaller than this. Then applying it again, this is smaller than C square V of n gamma square, smaller than C cubed V n gamma cubed, and so on. So we want to take log n times. If we log n times, divide n by some, so multiply by some number smaller than one, but divide by some number greater than one, we will obtain some constant. So we can deduce that V of n is smaller than C to the power some constant log n. Is it OK a lot? So if we take x, so it's exponent of constant log n, so yes, it's smaller than n to some power. So this front lemma gives you a very easy explanation why if you have a dilation, the growth is at least polynomial. Now our main lemma formulated above will be this multidimensional analog. We don't have a dilation, because we have a map not to the group itself, but to a product of several copies of the group. So we certainly cannot guarantee it's not true at all for aggregature groups that the growth is polynomial. But there is an analogous, also not very difficult, analytical lemma of the same client that will show if we have this situation with the contraction, then the growth is at most x n to some power, smaller than one. And the main idea behind this construction is really to check why we have this contraction, why we have this map that contracts from a finite index subgroup to a product of several copies of our group. So maybe the main thing is to explain why we have the contraction. What is our map? What is the letter used for aggregature group phi? What is our map phi? And what is our finite index subgroup? And how many copies of the group we'll have, which we did not by K? Actually what is possible is our H. We could consider a stabilizer of the first level of our tree or in the language of the intervals, meaning elements that don't, that all points of the first half sent to the first half and all points of the second half sent to the second half. But it's a little bit easier if you don't care too much about our coefficient alpha, only if you want to claim that it's smaller than one. The easiest way is to go to the third level. So we consider the stabilizer of the third level in our tree or on the interval, meaning we divide by 1, 2 to the power 3. We consider the subintervals of the length 1, 8, 2, 8, so on. And we consider only the elements that don't interchange points of some intervals with the points of another interval. This is clearly a finite index subgroup in our group, right? So what is left? So any element making some permutation on the corresponding level, we can multiply it. And so we get some element in the stabilizer, meaning the index is not larger than the permutation group on 8 elements, right? Actually, it is smaller, but for us, we just need finite index. So the index is not more than, index is not more than cardinality of symmetric group on 8 elements. So this group is good for our setting. And we want to consider a map. So we have this group h will be this stabilizer. We want to consider phi from h to 8 to g to the 8. And what is this map? This map, simply on each branch of the tree, or clearly on each small interval of the size 1 over 8, writes the corresponding element. So h, we can write h1, h8. Just in our definition for our generators, the restrictions to the branches of the first level, that we can repeat and take whatever level we like, right? So restriction of the generators of the group to some level also expressed in terms of the generators if we reinterpret smaller branches as if we really identify them with the whole tree, right? So this is the map we are considering. Since it's true for generators, it's true also for any element in the group. So actually, any element in the group, we can write this way. So we have this third level, third level, I think. And then just we look what we have on the first branch, what we have on the second branch, and so on. We have 8 elements like this. And our main goal is to check that this map is contrasting in the sense explained on the other blackboard so that the sum of all the lengths of h i will be much smaller than the length of our element. Why is it so? Here we use this normal form that is explained here. So any element we write in this form, we take some element h in our subgroup h. We write it as stated. So something, a, then sum of bcds, a, and so on. And we want to see what happens if we consider its restriction to some level. But first of all, before we go into the third level, let's consider just restriction to the first level. So we write suppose that h was in the stabilizer on the third level, in particular the stabilizer of the first level. Meaning we can write h as, it's not the same h, maybe because h was on the, let's write it as g1, g2. I don't use h1, h2 because they, in my notation, corresponded to the third level. So we write h like this, right? So on the three meaning h is like this, g1, g2. And big branches are not interchanged. Oh, in the language of interval, they have g1, g2. This is our h. And we want to show two things. First of all, we want to show that even for distinction on the first level, the length of g1 plus the length of g2 is not greater than the length of h. I don't say yet that we multiply by constant smaller than 1. But it's already helpful to know at least it's not greater than the length of h plus constant. And then we want to say another thing. We look on this normal form here. Our stars are BCDs, right? And we look how many among the stars let n be the number of B's among our stars in our normal form, numbers of B's. And analogously, we consider nc, nd, which is the number of C's and the numbers of D's. Then the number, just because we have this form, the number of A's. I don't make a notation for the number of A's. Because the number of A's, we know that it's nb plus nc plus nd plus minus 1 or plus 0, right? So I say up to constant 1, which is not important for our argument, the number of A's is the same as the total number of BCDs just because they're alternated, right? And now, a second claim that we want to say that if there were many D's among BCDs, if we assume that nd, the proportion of nd among nb plus nc plus among BCDs, BCDs was at least 1 third, then we want to say that then L of G1 plus L of G2 would be greater or equal to 5 over 6 L of H plus constant. So if you would have to work only with words that have important proportion of D's, it wouldn't be necessary for us to go to third level. We would see already for this restriction on the stabilizer of the first level, we have gained this constant smaller than 1. So why so? Just let us see how we can estimate the length of G1 and how we can estimate the length of G2. Again, so we have written our word with A's and BCDs. And to write, we'll find now some words, not necessarily minimal ones, but some words representing G1 and G2. And to do this, we do the following. Just each time we multiply by something, we look what happens on the branches. So before we multiply it by A, here we had some BC or D. We just need to see what B gives us on each branch. So just by definition of the action, we knew that B acted like AC, C acted like AD, D acted like identity and B. We see that B gives a contribution A on one branch and contribution C on the other branch. C similarly gives contribution A on one branch, D on the other branch, and D gives trivial contribution on one branch and contribution B on the other branch. And when I say one branch, it depends on how many multiplications by A we had until some position we are considering right now. If we had an even number of multiplications, for example, if we had at the very beginning of our word BC or D, we really looked like written on the first branch, it's A, on the second branch it's C. But if we had some not even number of A's, for example, if we have some odd number of A's, for example, 1A, if you look on this star in our normal form, then meaning we had that the halves of our interval or the big branches of our tree are interchanged, meaning that if we're controlling the contribution of this element, if it was B on the first branch, it acts like C and on the second like A. But what is important for us is just the total number of contribution our element BCD can give for G1 and G2. So we see the total contribution of each B is 2 because 2 generates A and C is 2, of C is 2, of D is just 1. This shows that the length of G1 plus length of G2 is not greater than 2 times NB plus 2 times NC plus ND. Plus, yes, if it was, we suppose this element in the stabilizer of the first level, that's it. Because we just observed that we don't see contribution of A's at all. A only tells us whether intervals corresponding to the I interchanged or not when we look on their contribution. So it's really BCD who defined these contributions. And A's only shows if the big branches I interchanged at the moment we control it or if they're on the initial places. Then in the length and the length of H, we knew that it's more or less the length of H. It was 2 times NB plus NC plus ND plus minus 1. So we see if the proportion of ND is at least 1 third, we obtained our claim, right? So our second claim on the middle blackboard. So we have proved this claim. Now what if not? First of all, it's not so important, but we won't use it, that the proportion was 1 third. Actually, any positive proportion would give us some contraction, the worst one, but some. But we don't use it. We just observe that at least for one element BC and D's, the proportion at least 1 third, right? So obviously, for some either as we used in this claim, ND divided by NB plus NC plus ND by the total number of BCD's is at least 1 third. Or this happens for maybe for C and or NC divided by NC plus ND is at least 1 third, or maybe the maximum proportion is the number of BCD's is NB plus ND is at least 1 third. If the proportion of D is good, we already proven our statement because on the first level already, we have this contraction with coefficient of 5 over 6. And then we go to second and third, and by the first part of the claim, the total sum doesn't increase. That is up to a constant only. And so if there are many D's in our normal form passing to the first level, we have the contraction and we can pass to the third level as we claim, then the contraction is not worse than on the first level. Now suppose this doesn't happen. Consider the case, then there is a big number of C's. Use this rewriting procedure we have used. So instead of our H, we have this element G1 and G2. Then each C becomes ND, right? So for total contribution of C to G1 and G2, all C's become D. So we keep this G1 and G2 as we have obtained it. It's important we don't rewrite it as a shortest word. We just use this rewriting system and look on this G1 and G2. And we see that we can apply the second part of the claim just for these elements, right? So if the proportion of C's was large, then we can think then passing to the first level the proportion of D's now is large and then passing to one more level, we have a contraction. And similarly, so actually if at least one of the first cases holds, we wouldn't need the third level. It would be enough to go to the second level. But sometimes we need the third level because it may happen so that the proportion of B is large but maybe of C's is not large. Then we have to go to the third level. Then again, for the first passing, B's become C's. For there's one more passing, C's becomes D's. And we can always claim that the total sum, so finally considering in all these cases, right? In any case, so in any of these three cases, in any case, we obtain that the length of H1 plus the length of H2 plus the length of H8 is not greater than 5 over 6 L of H plus constant 3 or something. Maybe if constant is not important. Now this is the main observation behind the observation of Grigorychuk that the growth is sub-exponential. So the only thing which is left to prove, I have four or five minutes, right? So the only thing which is left to show that is this lemma, did I give the number? I did not give the number. The second lemma for the proof of sub-exponential bound for the growth function of the first Grigorychuk group, it's enough to prove the lemma written on the highest blackboard saying that if we have this contraction then the growth is at most xn to some alpha. And this lemma is really analogous to the Frank's lemma and we can formulate it just a simple lemma about function. So consider an increasing function. So let's f of n be increasing function, positive edit increasing function, such that the following holds suppose that for any sufficient large n, f of n is smaller than the sum, some constant, and then the sum of vn1 times vn2 times vnk. Yes, the sum of the products like this. Where the sum is taken over n1 nk, such that n1 plus n2 plus nk is smaller than gamma times n. Gamma times n. If we have a function satisfying this inequality, then we say there exists alpha, which depends in principle on our coefficient, contraction coefficient here, gamma. It's f, sorry, it's f. We will apply later for the growth function v, but I wanted to formulate just an abstract lemma about functions. Nothing to do with group theory. It's just a fundamental lemma about functions analogous to Frank's lemma. So suppose we have this. Then we say there exists some alpha smaller than 1, fn2 times fnk. So we have some k, so we have some k. A product. No, it's a product, just product. So we have some function, right? Defined on integer numbers. And we take a product, fn1, fn2, and so on, where we take all possible products. We take all possible n1 and so on, nk, with the total sum smaller than gamma times n. And gamma was a constant, so gamma was some constant smaller than 1. And k was some integer number. We are interested, it's true even for k equal to 1, but the claim is not interesting for k equal to 1, because much better claim of Frank's lemma. We have already proven it's valid. We are interested in k, at least, or 2. We say that there exists alpha, depending on gamma and k, such that f of n is smaller than x, some constant n to the alpha. And this would be enough to finish our proof. Because we do the same thing as we have done in the Frank's lemma, so far we have this analytic lemma. If we have proven lemma free, just we bound the growth. So we have something we use again. So for the first Greek integer group, we have our subgroup h, which is a finite index. So any element is at some finite distance from some element of h. So it will give some possible sum constant. It's enough to control the intersection of the ball of some radius. As we discussed with our subgroup h. And in the subgroup h, just we see how many choices we have for element of h. If the length of h is some l, then h, the image, this phi of h, is h1 hk. And the total sum, just by definition of contraction, l of h1 plus l of hk is smaller than gamma times l. It's not important here. We can also allow this constant. We can start with some constant and write the sum is greater. Like this. So just considering this intersection of the balls with our subgroup h and applying this analytical lemma, we did use the growth. Indeed, it will be xn to the alpha. So the only thing which is left to prove is this analytical lemma. And again, so it's done directly like front lemma here. So we just use the fact that proof of lemma 3, we want to prove that log f of n is smaller than some constant a, let's say, n to some alpha, with some alpha smaller than 1. We assume that this holds for any n smaller equal to some, strictly smaller, let's say, than some m. And we prove this claim for, we will show that if we choose a large enough and if we choose alpha close enough to 1, that we will be able to prove the claim of lemma 3 just by induction on our n. So we assume that for smaller numbers, we have this conclusion. Now, we look on what we know about f of n. For f of n, we have this inequality that we will use. Observe that if n is large enough, each n i is smaller than, because the total sum is smaller than gamma n plus c. So in particular, each n i is smaller than gamma n plus c. And this is strictly smaller than n if n is large enough. I mean, if we wouldn't have a constant here, it's true for all n. But if you want to formulate here with a constant here for any n large enough, we can see that the elements that figure in this inequality are smaller. And we can apply induction claim. So we can see each term here, so each term. We have some sum, but each term satisfies the following. So we take log, let's say log of each term. We take log, we have log of n1 plus log of n2 plus log of nk. And we just use the fact that n to the alpha is a concave function, right? So just here we use induction hypothesis. So it's a smaller by induction claim by n1 to the alpha plus nk to the alpha. And we will see that this is smaller that, well, it's certainly smaller than n to the alpha. But it's not enough for induction because we don't have one term. We have many terms. So we want to claim a little bit better. But still, using the same concavity, we can claim that this is smaller than 1 minus epsilon, the same constant a, and n to the alpha. Time to give it a state, but this is really elementary. You have something like gamma n divided the k to the alpha times k. And it is sufficient to choose alpha such that gamma n divided by k to the alpha times k is smaller than 1. It will give us some epsilon, 3k smaller than 1. Why we can do it? But by our assumption, gamma is strictly smaller than, sorry, this is gamma n, yes, I want gamma k. Yes, we want to choose some alpha such that gamma divided by k to the alpha times k is strictly smaller than 1 to have this conclusion. But we know if alpha would take alpha equal to 1, we would get just alpha divided by k times k gamma, which is strictly smaller than 1. And since it's strictly smaller one, if we can take some alpha a little bit smaller, and we still have a stricter in the quality, so we can always choose appropriate alpha. And so we have this claim. And now just we count how many terms we have here. The number of terms is at most, so each term is at most gamma times n. And the number of terms is at most the number of terms is not greater than gamma n times k. So this is just some polynomial. And here for each term, we have exponential So since we have an extra epsilon here, this epsilon will kill the polynomial term. So even taking the sums of all terms, we can conclude that the value of n is smaller than xn to the alpha. And this finishes the proof. And I have to stop here.