So, yeah, so last time we talked about group operations, so that, so we have a group G acting on a set X, so this is a map from G times X to X, so G pair of G and X is mapped to something called G times X, which satisfies two actions, namely that if we kind of twice act with two elements in this way, we compose the action twice, it's the same as if we multiply the elements in the group and act by that, and then the identity element should act as the identity, so this is for all X and all G and H. And we had introduced two important concepts connected with group operations. One is the stabilizer and the other one is the orbit. So, the stabilizer of an element will just be the set of all elements in the group which map it to itself. So, this is GX, which is the set of all G and G, such that G times X is equal to X, and we also have the orbit, which is just all the elements of X. So, the orbit of an element X and X will be the set of all elements of X, which we can reach by acting by elements of G on it. So, this is G of X, which is the set of all G times X is G and G. So, this is a subgroup of G and this is a subset of X. And we had seen that if we have a group operation, then we get a decomposition of X into this joint. So, then the orbits, the different orbits form a decomposition of X into this joint sets because they are actually equivalence classes of an equivalence relation. And we finished by the orbit stabilizer theorem which said that where we have a group G acting on a set X and we take a system of representatives for the orbits. So, in other words, this is a set which contains one element out of each orbit. And then we find that the number, so I want GX on X and in this case the X is finite. Then the number of elements of X is equal, this is kind of trivial, sum over all these representatives of the number of elements in the orbit of X. This is just the statement that X is the union, the disjoint union of the orbits. And we can rewrite this as the sum over all X and R of the index of G in GX because one finds that the number of elements in the orbits is precisely the index of GX in G. So, this is actually a very simple result but it's useful and therefore it's a theorem. So, now we want to look at the special case of an action, namely the action of a group by itself by conjugation, group G on itself. So, first what does it mean, conjugation? So, we have G a group and we see they say that G and H in G are conjugated if there is an element A in G such that say H is equal to AGA to the minus 1. And you can easily see that this is an equivalence relation. Okay, this is a trivial exercise. And so we can look at the equivalence classes which are called conjugacy classes of elements are called conjugacy classes of G. So, okay, we want to say something about these conjugacy classes and we want to see that this actually is an operation of G on itself. So, let's first define something more. So, for an element X in the group, the centralizer of this element consists of all elements commuting with it. So, let X in G. So, the centralizer of X is just, so I write it with Z of X to be the set of all G and G such that GX is equal to XG. So, these are all elements which commute with our given element X. Have some simple remark. First, the centralizer of any element X is a subgroup of G. That's very easy. Then, I can look at the center. In the exercise, I use the different notations C of G, but I want it now with the Z. So, the center of G is, I mean, we know it's a normal subgroup of G and this will always be is a subgroup of the centralizer of any element X. But that's obvious by definition. I mean, the center consists of all elements in the group which commute with all elements in the group. So, in particular, that's a subset of all those which commute with a given element X. And it's also obviously a subgroup. And finally, we have an element X in G is in the center if its centralizer is the whole of G. Again, this is trivial because the elements in the center are the elements in the group which commute with all elements in the group. And that's precisely what's being said here. So, this is really a remark. There's nothing to prove. Everything is directly from the definition. So, it turns out that the conjugacy classes are somehow very important invariant of the group. And if one wants to understand the group, one has to understand the conjugacy class. We will not really do that. Maybe in the next semester, you will also hear something about representation theory of groups. And you find, for instance, there's a close relation to the representations of a group and the conjugacy classes of the group. So, but now we want to describe this and we want to describe the conjugacy action of a group on itself so that the conjugacy classes will actually be the orbits. So, the definition. The conjugation or action of G on itself is just the obvious thing. It's just defined by G times H is equal to, say, G H G to the minus 1. So, the element G acts on the element H by this conjugation. So, by definition, the orbits are precisely the conjugacy classes. These equivalence classes are precisely the conjugacy classes. So, the conjugacy class of an element G is the same as the orbit because, you know, after all, this is precisely how it was defined. And we note also that the stabilizer of an element, say, X in G with respect to this conjugation action is precisely the centralize of X, you know, because by definition, the centralize of X is equal to the set of all G and G such that G X G to the minus 1 is equal to X and this is obviously equal by multiplying by both sides by G and multiplying on this side by G to the set of all G and G such that G X is equal to X G, which is... So, now I said it precisely wrong. So, this is the stabilizer is, by definition, this and this is the centralizer of X. So, this statement is also obvious. So, now we just basically, we just want to reformulate the orbit stabilizer theorem for this conjugation action and then it gets a new name. It's called the class equation. So, it's just another way to account, in some sense, the elements in the group. But it's actually just a special case of the orbit stabilizer theorem, but still it's a useful result. Therefore, it gets its own name. So, theorem of class equation. So, we take G, a final group. And so, we take R to be a set. So, a subset of G, which contains precisely one element from every controversy class. So, such that every element in G, minus the centralizer of G, every element in G, which does not lie in the center of G, is conjugated to precisely one element in R. Precisely one element in R. Okay, then the number of elements in G is equal to the number of elements in the center of G, plus the sum over all elements in R of the index of the centralizer of X in G. So, in some sense, note that, in some sense, I have... Here, it's formulated in some sense in an unnecessarily complicated way, which you will see in a moment. But this is usually the way one wants to use it. So, that's why we write it like this. So, note we can say it also as follows. So, maybe I can prove it and prove it, and then we can see. So, we take a system of representatives of the orbits of the conjugation action of G on itself. Then, maybe I'll read this. Then we can apply the orbit stabilizer theorem and that says that... So, by the orbit stabilizer theorem, we have that the number of elements in G is the sum over all X in R The index... So, R prime was our system of representatives. The index of the stabilizer, with respect to the conjugation action, of this X in G. And we remember that we had just seen that this stabilizer was actually the centralizer of X. So, we get this formula. Still with a prime. And now we have to... We see that if our element X lies in the center of G, then it commutes with all other elements. So, its centralizer is the whole of G. So, this index is equal to 1. And so, we put just... So, it means that on other words, we also have that if it lies in the center of G, then it's also clear that the orbit G of X is just equal to X, because, after all, by conjugation, it commutes with every element G in G. So, if I make G X G to the minus 1, then this is equal to X. So, it means that the center... So, therefore, we can just put minus the center of G. So, if you have an element in the center, then it is automatically equal to its own conjugacy class. So, in R prime, all the elements of the center are represented. And we just remove these. And in that case, if we have an element in the center, then this index is equal to 1, because these two are equal. And so, this formula is just equal to the number of elements in the center of G plus the sum over all X in R, the index of G C X. So, we see that this is just a reformulation of the orbit stabilizer theorem, where we also kind of have singled out the elements in the center, because there the orbit contains only one element. Okay. So, now we want to use this to show that any group which has a number of elements, the square of a prime number is a b. So, we know... So, let p be a prime number. We know, we proved it some time, that if we take... So, if the number of elements in a group is equal to p, then p is cyclic. In particular, it is what? I mean, it wouldn't make much sense otherwise. Then G is cyclic. So, now we want to see what happens for the next case when we have a square. So, now we want to show that if the number of elements in G is p squared, then G is a billion or commutative. And for this, we want to use this result. So, the first statement is... So, maybe... Do I give the definition? Yeah. So, first I want to give the definition. So, if I have a group... So, p is still a prime number. A group p, G, is called a p-group. If the number of elements in the group is a power of p, if there exists a number n bigger than 0, such that the number of elements in G is equal to p to the n. We'll see that... When we do the Seelow theorems, that they also have something to do with p-groups contained in given groups. So, now the first statement we want to make that the center of a p-group has at least... The number of elements in the center of a p-group is at least p. For position, let G be a p-group. So, the number of elements in the group is some power of p. Then, the number of elements in the center of G is at least p. Okay. So, we want to prove that. So, we assume that the number of elements in G is equal to p to the n for some n bigger than 0. And we know that the center... I mean, we know the center of G is a subgroup of G. So, we know that the number of elements in the subgroup always will divide the number of elements in the group for finite groups. So, we know that the number of elements in C of G divides the number of elements in G. But obviously, the divisors of p to the n are just the powers of p. So, 1 p p squared under p to the n minus 1. Thus, the number of elements in G is a power of p. So, it would be p to the m where m is bigger equal to 0. But it could also be that m is equal to 0 so that the center has only one element, in which case we would not have proven our proposition. So, we only have to see... Therefore, it is enough to see that the center of G is not equal to 1. That there is at least one non-chival element because we know that the number of elements in the center is a power of p. So, it must be positive power of p. So, it's at least p. So, we assume by indirect argument that we assume that the center of G is equal to 1. So, in particular, the center of G has precisely one element. So, then we apply the class equation. So, what does it say? It says that the number of elements in G, which we happen to know is p to the n, is 1... 1 is the number of elements in the center plus the sum over all x and r of the index of the centralizer of x in G, where r was a system of representatives for the non-chival conjugacy classes. So, it means that... So, x are a system of representatives for the conjugacy classes with more than one element. Or, in other words, for the conjugacy classes of elements which do not lie in the center of x. So, here, the sum is precisely over those elements for which the centralizer is not equal to the whole of G. The centralizer is a subgroup. So, maybe I can also write this. So, this is equal to the set of all x. So, representatives for... No, anyway. So, I'll just say it again. So, if I have an element here, we take precisely the representatives of conjugacy classes for which the centralizer, for which the conjugacy class is more than one element, so where the element does not lie in the center, which is equivalent to the fact that the centralizer of x is not equal to G. So, for x and r, we have that the centralizer of x is not equal to G. But the centralizer of x is a subgroup. So, but z of x, in this case, is a subgroup of G. So, we know that the number of elements in a subgroup is a power of p. And so, we take p to the n, divide by a smaller power of p. So, the number of elements here will be... This index is a positive power of p. So, we find that for all x and r, we have that this index is divisible by p. So, to say it again, we know that this is a subgroup of G, which is not equal to G. So, its number of elements is a power of p, which is smaller than this n. And so, the quotient is a positive power of p. This is divisible by p. But then, we see our contradiction. We have that p to the n is 1 plus something divisible by p. That contradiction, p to the n is equal to 1 plus a number divisible by p. Okay? And so, with this, we find that indeed it's not true that the center consists only of one element. And so, therefore, it is a positive power of p and therefore it's bigger or equal to p. And now, we want to prove this statement, which I announced here, that if I have a group such that the number of elements is p squared for prime number, then it's a bn. So, it's a proposition. Every group of order p squared, where p is a prime, is a bn. You know, for instance, that all groups of order 4 are bn. On the other hand, we know that the symmetric group in three letters, which has six elements, is non-a bn. So, anyway. So, we take such a group. So, we have to show... So, we want to show the group is commutative. So, we can reformulate this in a different way. So, we have to... So, g is a bn. If, for all elements, x in g, we have that the centralizer of x is equal to the whole of g. This is just by definition. The centralizer of x is the set of all elements in g which commute with x. And so, if for every element in x, the set of all elements which commute with x is everything, it means every element commutes with every element. So, let's see. So, we want to show that. So, obviously, if x... So, now, if we have such a... And so, let's take an element in x in g. So, if x is an element in the center of g, then obviously the centralizer of x is equal to g, essentially by definition. On the other hand, if x does not lie in the center of g, that actually will finally not be able to happen if g is a bn, but if we assume we have an element x which does not lie in the center of g, then it means that if I take the centralizer of x, this certainly contains x because obviously x commutes with itself, and it also contains the center of g because the elements in the center of g commute with every element. But x does not lie in the center of g. So, this contains p elements. So, it means it contains these elements. All the elements in the center plus also x itself. But we know the number of elements of the center of g is bigger equal to p, and so the number of elements in the centralizer of x is strictly bigger than the number of elements in the center of g because we have also the element x. So, that means the number of elements in z of x is bigger equal to p plus 1. But the centralizer of x is a subgroup of g, and so g has p squared elements, and we have a subgroup which has at least p plus 1 elements. As the number of elements of a subgroup must be divisor, there is no divisor which is bigger equal to p plus 1 of p squared except for p squared itself. So, it follows that as number of elements of c of x is a divisor of p squared, so number of elements of g equal to p squared, it follows that z of x is actually equal to g. So, we have found that for every element x in g, the center is equal to g. So, it means g is commutative. I mean, if you think this to the end, obviously we find that this case actually does not occur because the group is commutative. We have proven it's commutative, so it x is always in the center of g because all elements are, but we are making an indirect argument, so that's okay. Okay, so this was this result. So, now I want to, are there any questions about this? Comments? Time? So, now I want to look at something a bit more special. I want to look at the symmetric group again, and I want in particular to understand the conjugacy classes of the symmetric group. So, I will, we have talked about the symmetric group in n letters before, and now I want to look at it slightly more carefully. So, talk about the symmetric group S n. So, we want to describe this group more precisely. We will introduce new notations for elements in the symmetric group in terms of so-called cycles, and then we want to see that the conjugacy classes of the symmetric group can be described in terms of these cycles, the length of the cycles. So, we call that S n, this was the set of all bijections from the set 1 n to itself with the, as a group operation, the composition. So, sigma tau was equal to sigma composed with tau. And there's a, okay. We have introduced the notation. So, for sigma, so we write sigma as some kind of a matrix where we write on the top the numbers 1, 2 and so on until n, and in the bottom we write the images. So, sigma of 1, and so this is certainly a way how one can describe all permutations. Now we want to introduce a new notation in terms of so-called cycles, which many of you might also know. So, you can just say it definition. So, say let we take some elements a1 to ar, the distinct elements of our set 1 to n. So, we take, we pick out r numbers from 1 to n, different numbers. And then the cycle a1 to ar is the following permutation. Well, it's what one might think if one understands the word cycle. It means that it should be something which maps a1 to a2, a2 to a3 and so on, ar minus 1 to ar, and ar back to a1. And all other elements which are not among a1 to ar should stay fixed. So, this is the map. Permutation sigma in Sn, which is given by sigma of ai is equal to ai plus 1, if 4i equals 1 to ar minus 1. And sigma of ar should be, well maybe just say that, whatever. And sigma of ar, so a1 for i equals r. So, sigma of ar is equal to a1. So, you really have like, if you would kind of put them all on a circle, you kind of turn it once around. And obviously, and sigma of b is equal to b for all b which do not belong to the set a1 to ar. So, this is the cycle. So, we call r, the number r, is called the length of the cycle, of the cycle a1 to ar. And we say that if I have a cycle of length r, I also call it r cycle. And in particular, the two cycles are somewhat more important than they are called transpositions, two cycle transposition. So, if you have a two cycle, it just means you have two elements a1, a2, and the corresponding permutation kind of exchanges them at least all other ones fixed. And you may be called, so we call a1 to ar, the support of the cycle. And we call some cycles disjoint if their supports are disjoint. So, cycles a1 to ar, b1 to bs are called disjoint. Well, if the corresponding sets of elements in one n are disjoint, so if and only if I take the set a1 to ar, then this does not intersect set b1 to bs. We don't have any common elements. So, we can first, we have an obvious remark, cycles are after all, these are permutations. And it is, I claim it is obvious to see that disjoint cycles commute. So, that means if a1 to ar and b1 to bs are disjoint cycles, so this set of these elements does not intersect, there's no common elements here are disjoint cycles. Then a1 to ar times b1 to bs is equal to the other way round. Now, that's kind of obvious because what these things do to elements has nothing to do with each other. So, if I apply this to any element k, then if k is among b1 to bs, then I will apply this part to it and this will not do anything. And this will do the same here and the same if it is one of the a's. And if it doesn't belong to any of these, it will anyway be the identity. So, in any case, they do the same thing. Now, we want to show that every element in the symmetric group can be decomposed, I mean written as a product of disjoint cycles, essentially in a unique way. Every element, say sigma in Sn is the product sigma equal to sigma1 times sigma whatever, s of disjoint cycles. And so if we require that, so disjoint means pairwise disjoint. So, no two of them have any element in common. No, whatever. So, if we require that if I take the union of the supports of the cycles, that this is equal to the whole of the set 1n, then this decomposition is unique. If we require that the union i equals 1 to s, the support. That the union of the supports of the cycles is 1n, then this decomposition is unique up to obviously reordering. Obviously, as the disjoint cycles commute, I can also order them, but I claim that the number of cycles and precisely which cycles they are is unique. It's determined by sigma. So, this second part I will not actually prove. So, I kind of make this an exercise, although it essentially follows from what I prove. But, and so this decomposition, so in this, so under this assumption, so the, maybe I don't need that now, it's coming later. Where is it? Yeah. So, we will call, so, definition. So, if we have such a decomposition into pair voice disjoint cycles, such that the union of the supports is everything, we call this the cycle decomposition of sigma. So, in case, so, composition sigma equal to sigma 1 to sigma n into disjoint cycles with whose union is 1n. So, such that the union of the supports is called the cycle decomposition. So, by the buff, it is unique. So, we can look at an example if I have it somewhere, well, whatever, so we can look at, for instance, let's look at, so, cycle decomposition of say the element 1, 6. What is it? 2, 2, 3, 4, 5, 6, 5, 3, 4, 1. So, how do we do this? So, we can say, we start with the element 1, it is sent to 6, element 6 is sent to 1. So, we have a cycle, so we have, yes, one cycle is 1, 6. So, if you take the element 2, it's mapped to itself. So, that's also a cycle. Now, we take the element 3, it's mapped to 5, 5 is mapped to 4, and 4 is mapped to 3, so back to 3. And that would be our cycle decomposition. So, one thing that one should obviously notice is that if you have a 1 cycle, it's just the map which sends one element to itself, and all other elements to itself, it's just the identity. So, 1 cycle is just the identity. And so, this is why I had to assume that the support is everything in order to get uniqueness. Otherwise, I would have had to assume that it doesn't contain any 1 cycles. That would also be possible. And then, you can also see that this notation, the notation is not precisely unique. So, if you have this element, for instance, 3, 4, 5, so, or whatever, if you have the element 1, 2, 3, this is the map which sends 1 to 2, and 2 to 3, and 3 to 1. But I can also write it as 2, 3, 1, because it's the same thing. 1 is map to 2, 2 is map to 3, and 3 to 1. So, the decomposition into cycles is unique, but the notation is not unique. You can always kind of rotate them and get the same element. Okay, so now I want to prove this theorem. Maybe I can first prove the theorem. So, as I said, I wanted to only prove the existence, but the uniqueness is actually quite easy. So, prove. So, well, we will find that... So, how does one do this? What are these cycles? So, what you actually find is that, you know, if you take the element sigma, it has the property that it sends an element in the cycle to the next element in the cycle that permutes them. So, you find that under the operation of the group generated by sigma, the supports of the cycle will be precisely the orbits. And so that's, therefore, the way how we want to do it. So, let H equal to the cyclic subgroup generated by sigma. So, if I take the map... So, we have this map of multiplication of elements in one N by elements in H. So, we have the multiplication from H times one N to one N, which is just... We take an element in the subgroup generated by sigma and we apply it to this. So, tau i is mapped to tau of i. So, this is obviously an operation of H on one N. Basically, it's just we have the operation of the symmetric group on the set one N by just applying, you know, the element in the symmetric group to the corresponding set. And now we restrict it to the subgroup H. So, we have an operation. And so, we can look at the orbits. So, we choose some elements, B1 to B. Do you want that? S such that they... such that if I apply H to B1 and so on to H applied to Bs, these are the different orbits of this action. So, we know that... So, this B1... the set B1 to Bs is just a system of representatives for the orbits for this action. And we know that the set one N is equal to the disjoint union of the set. So, now, I mean, I've called here H is just all the powers of this one element sigma. So, let's see. We want to write this. Now, let me see. So, for each element, I equals 1 to S. We choose the minimal power of sigma such that if I apply this power to Bi, I will get Bi back. So, let Mi be equal to the minimum over all k positive integers such that if I take sigma and apply it k times, so sigma to the k of Bi is equal to Bi. This element sigma is an element in the finite group. So, there's some positive power of it, which is the identity element. And so, certainly, there is some minimal positive power which is the identity on this one element Bi. So, then, it is easy to see that if I take this element Bi, sigma of Bi and so on, sigma of Mi minus 1, so the Mi power of sigma applied to Bi, these are distinct elements. They are pairwise distinct. It's like you're doing some exercise. If you would have that, you know, for some smaller number, so if two of these would be equal, then you could, you know, multiply with the corresponding power of, so, sigma to the L of Bi then it follows that if I take, so forth, let's say 0 is more than equal, then you see that you can just multiply with sigma k to the minus k and you would get that Bi is equal to sigma of L minus k to the Bi. And so, we have found a smaller power which is equal to Bi. That's, so we can, so these will be pairwise distinct and how does sigma act on them? So, they are pairwise distinct and we also see if I take the orbit of our element Bi, I claim this is just this element. It's just these elements Bi, sigma of Bi, sigma Mi minus 1 of Bi because, in fact, what you find is that if you take sigma, so in fact, by definition, if you apply sigma to any element here, it gives you the next element and if you take this one and you apply sigma to it, you get this. So, in fact, if I look at this set here, Bi sigma of Bi until sigma Mi minus 1 of Bi, this is a cycle because we see precisely, by definition, if I apply, so let's see, maybe I can explain it slightly better. So, if I, so I get anyway, so, if I, obviously, this is by definition, if I write it like this, it is a cycle, I can write down any distinct elements, I get a cycle, but what I have is that if I take sigma I of, so if, so if X is an element in the set, Bi sigma of Bi sigma Mi minus 1 of Bi, then if I take sigma of X, this will be, as I said, the next one here, cyclically, so this is the same as sigma I of X if sigma I is the cycle because the sigma precisely acts by rotating this by one, putting everyone to the next and the last one to the first and this is precisely what sigma I does by being a cycle. So, thus we get the sigma I for equal to this thing, Bi sigma of Bi until sigma Mi minus 1 of Bi for I equals 1 to S, disjoint cycles, I mean they are disjoint because the elements here are the elements in the orbit and we know the orbits are disjoint and the union of the orbit is the whole of X, so who's such that and the union of the orbits, union of their supports is 1n and now if we take any element in 1n, I claim that it acts by the product of these cycles, so we want to claim that sigma is equal to sigma 1 times sigma S, so this is the cycle decomposition and so we take an element in 1n and so we have to show that these two permutations do the same thing to it, so obviously if you apply sigma to it you get sigma of X and if you apply this we get something else, so let's see what we do get, so let X, so we know that the union of these sets here is the whole of 1n, so it means that X lies in one of them, so then there exists and there is a unique I in, so then there is an I in the set 1 to S such that X is an element in the set, sigma of Pi, sigma Mi minus 1 of Pi, well in fact if we take all these elements the union of all these elements is X so we precisely have that X in fact X is equal to Pi to the power J for some J with J so not J, so now you should protest if I write nonsense, so sigma to the J of Pi for some J which lies between 0 and Mi minus 1, because the elements here are precisely this, Di and then what does sigma do to it so then sigma applied to X obviously is equal to sigma to the J plus 1 of Pi which will either be equal to Pi if this was Mi if this J was equal to Mi minus 1 or otherwise it's just the next one so the index is like this so that's what the sigma does and what does sigma, if I take sigma I of Pi of X so if I take sigma J of X for K for J different from I then this is equal to X because the this cycle only permutes the elements which lie in the elements which are here so if it is if our X is none of these elements it is not moved by this so the sigma I sends Pi to sigma of Pi and so on but if I have an element which is none of these here it will just stay fixed and on the other hand if I take sigma I of X this is equal to sigma J of Pi and you know this is just we apply the cycle to it and this precisely says we go one step further in the cycle so this is Pi if J is equal to Mi minus 1 and sigma J plus 1 of Pi if J is different on Mi minus 1 and so we see that and so if I take sigma 1 times sigma S of X then these things commute so X is not so the sigma K which are different from the sigma I will not do anything so this will just be equal to sigma I of X and this is which is as we have seen the same as sigma of X so it's kind of I mean I made it maybe a bit more complicated than it is somehow these different cycles have nothing to do with each other and so our element X will lie in one of these cycles and then only the corresponding cycle X on it the other ones don't do anything and therefore you get the same result so this will prove the result so we see in particular that actually we in the cycle decomposition we find that the supports of the cycles are actually the orbits of the elements under the powers of the thing we want to decompose so as a corollary we find that every element in the symmetric group can be written as a product of transpositions you see where this plus every element sigma NSN is a product of transpositions so we have for this we just have to see that every cycle is a product of transpositions so assume we have because every element NSN is a product of cycles so if every cycle is a product of transposition we are done so we write down any cycle A1 to AR is a cycle and I claim I can write it as a product of transpositions in fact I can just write down what it is so I say I take A1 A2 A1 A3 and it goes on until A1 A2 so I claim this cycle is in this way a product of transpositions so transposition after all was a two cycle well and you know you can just that's kind of clear so here you want to say that this thing maps A1 to A2 so if you take A1 A1 is mapped to A2 then A2 never occurs again so A1 is mapped to A2 now A2 is mapped to A1 here and A1 is A3 but then A3 never occurs again so A2 is mapped to A3 and so on so it's easy to see so this is easy to check so this shows that is a product of transpositions and as every element in the symmetric group is a product of cycles we are done so how much time are basically none let me see so I maybe just give a preview so now we want to use this to describe the conjugacy classes I mean as I said I will only say so we will find that if I have two cycles of the same length then they are conjugated to each other so if you have a if an element in the symmetric group has a cycle decomposition such that the lengths of the cycles are the same for both if there are two elements which have cycle decompositions with the same lengths then they are conjugated to each other and vice versa so that means the conjugacy class of the element will be determined by precisely by the lengths of the cycle in the cycle decomposition of the element and if you look here so such a cycle decomposition is a decomposition into the wet set one you know in particular if you take a support of the cycles this is a decomposition of the set one n into the disjoint subsets so the number of elements in each cycle if you take the tuple of the number of elements in each cycle this is a set of numbers which add up to n and you do this up to the ordering so you find that the conjugacy classes are in one one correspondence to the partitions of the number n so of the ways how you can write the number n as a sum of numbers and so this we will explore the next time and the other time the other thing we will explore the next time is that we also look at the sign of a permutation so if you have a permutation then I had told you here that it is a permutation as a product of transpositions and this writing this permutation as part of permutations of transpositions is not unique there are many ways but you find out that you can if you take minus one to the power the number of transpositions you need to write this thing this is something which only depends on the permutation itself and this is called the sign of the permutation we will prove this and okay that's maybe the beginning and then after that we will start preparing ourselves for proving the Silo theorem


Today's topic is Marshallian demand function and what are the properties of the Marshallian demand function? Marshallian demand mostly tells us the relationship between the quantity demanded and its price and it tells that there is always a negative relationship between the price and the quantity demanded and here the consumer derives his utility through the consumption or not directly through the consumption rather through the expenditure that he incurs on the consumption. So he conceals the other fact and he prefers to take the utility through the indirect form. Usually when we say that there is a property it means that these things should be the particular part of that either parameters or the policies. So when we say the properties it means that they are the very necessary things. So whatever we are going to conclude related to the Marshallian demand function if any of this property will be violated that general conclusion will not be followed. The first property that we are going to discuss it is the adding up. What we mean by the adding up? Generally the property of monotonicity it tells us that the consumer is having a desire to grab more and more of any commodity and he is desirous to get more of the quantity that he has already preferred. But when we add up all the things and when we add up all his desire but there should be a limit on this and what will be the limit? So that limit tells us that whatever the desire of the commodity to a consumer is there when we add up all these things it should at least be equal to the budget line or it should be less than the budget line it should not be more than that. So when we include the properties and we take the multi good case and we say that the consumer is having the demand function for commodity X1 or the commodity X2 or the commodity X3 up to commodity XN and then we multiply that commodity with their respective prices it means now the price of the commodity X1 multiplied by the X1 it will having the expenditure that the consumer is incurring on the purchase of that commodity. So if we say that if the consumer is buying all the properties of one, two, three, so when he gives the value of every share to him then when we multiply the value of his commodity with quantity then he comes in the form of a complete expenditure on that share and when we add up all the assets in his consumption bundle then he comes in the form of a total expenditure so it means that will be the total expenditure so here the expenditure incurred by the consumer for the purchase of all his desired commodities it should be within the budget limit and it should not exceed the budget limit. Second property that we are going to deal is the property of homogeneity and what we mean by the property of homogeneity is that if we are going to increase the units of the demand function means the prices price of good X1 price of good X2 and any other and if we are going to make them double by all the total function is multiplied by the same unit so if we are going to double the prices and at the same time if we are going to double the income then the output level of the consumption that the consumer is going to demand will be the same as like the previous so the demand function if multiplied by any unit t it will yield the consumer at the same demand level because income and the prices they both will be increased or decreased by the same unit so whatever will be the change that will be offset by the incremental part of the expenditure through the incremental part of the income and now coming to the example of the Marshallian demand function if we take the function of a utility where the two commodities X and Y they are utilized and now they are in the form that X have the elasticity of 0.3 and Y is having the elasticity of 0.7 and that this demand function if we are going to derive through the quantities demanded of X and quantity demanded of Y through any Lagrange function then we come up up to this point and now here we see that whatever is the shape of the consumption function or the utility if we just multiply the whole if I am going to put here this demand function and rather this demand function but if I am going to multiply the respective subject constraint that we are having here in the form that income is equal to price of X into X plus price of Y into Y so if I am going to multiply this by 2 or if I am going to multiply by this 3 whatever will be the form the resultant demanded quantities they will remain the same and the same relationship or the same behavior it can be exhibited by any other form of the demand Marshallian demand function that can have any form of the CES or the Cobb Douglas so even in this next equation we see that now the utility function or the demand function is something different but their budget line or the subject or the constraint is again this price of X into commodity and price of Y into Y and this budget if multiplied by the double or if multiplied by 2.5 or if multiplied by the 3 whatever but because this change will be having the same level of the change on income and the same level of the change on the their respective prices so the resultant quantity demand is will remain the same now coming to the third property that is called the negativity what we mean by the negativity is that if there is any budget constraint that enables the consumer to purchase a quantity of a bundle that is X1 and at the same time consumer is having the bundle of X1 and X2 or the Q1 or the Q2 but that budget makes him possible to purchase the commodities quantity X1 it means now there will be no other bundle like this or no other budget constraint that will enable him to purchase the quantity of the X2 if we look at it then we talk directly about consumption on its expenditure that if any budget which we want to take and that budget gives our consumer this affordability that out of the bundle of X1 and X2 is purchasing the quantity of X1 so we should keep in mind that along with it its affordability and the limit of the budget is also that apart from that it has no other purchase it is not possible so if we look at it then the concept of this negativity and the fourth the main thing is that is the slutski equation we have already studied related to this the slutski identity and here through the marshalia demand function now we can see that it makes possible to the economist to derive the slutski equation as we are having a single objective that we have to maximize the utility and if we have to maximize the utility at the same time we have to check that what is the change in the quantity demanded by the change in any price so when we measure this thing we utilize the two ways one is the utility maximization and at the same time we can utilize the other form that is the cost minimization and this part is the Hixian and this curve that we are drawing on the right side that is we are having through the marshalian demand curve and this marshalian demand curve that was having the dx into price of x and price of y and income and here now this income was substituted by the expenditure function so this equality of Hixian and the marshalian demand curve makes possible for the economist to draw that what is the amount of the expenditure required to attain the same level of the utility so when we take this two equal this part of the marshalian demand function this is further decomposed in these two parts and this says that this change it is in the two parts change due to price of x and this is the change that is due to the change in the expenditure so marshalian demand function through its decomposition provides the facility to the economist to have the analysis of the change in quantity demanded through the price via substitution effect and via income effect so we see that in this slutsky equation this right hand side that we have noted through our marshalian demand curve it shows that the changes in the price of x that we were dealing with they affect the demand of the commodity not directly rather indirectly via changes in the expenditure function so here the change in commodity demanded will be due to change of his expenditure that the consumer has to make in response to price change means if price has increased he has to adjust for that by increasing the expenditure or if price has decreased then how he has to adjust so that change in expenditure and then what is the change in expenditure related to the price of x and mostly this part is just equal to the commodity the previous that he has purchased so this makes possible to the consumer to have this income effect through the marshalian demand function and this mathematical representation of the income effect we are having here so this marshalian demand function makes possible to give the decision or to give the reply of our question that what will be the change in the quantity demanded through the change in the price and this change will be decided by the indirect utility function


Hi everyone, Sean with JaspysCaseBreaks.com here doing 2017 Top's Chrome Monster four box break, two random teams each, number two. So, no veteran commonship in the break, 15 spots, everyone gets two random teams in the Major League Baseball, chasing Cody Bellinger, Aaron Judge, Alex Bregman, Matt Olson-Rookies, plus, so you get four cards per pack, 10 packs, plus there's two exclusive packs of Strictly, the Chrome X Factor cards, which are super nice. So, let's see how it goes. Uh, we've got our list of customers from JackBizBajoon, LastSpotMojo, all the way up to Guy Canashiro. So, let's copy this list because we need to double it up because everybody's getting two teams with each spot. So, you can see Jack there with LastSpotMojo, Guy with first spot purchased, Jack again, LastSpotMojo, and Guy. And the teams from the Diamondbacks down to the Washington Nationals. Dice roll to randomize both lists and pair them up three into two five times. Roger Ness down to Michael Craig, five times on the teams. From the Philadelphia Phillies down to the Washington Nationals.  So, let's sort this by teams and trade window open. Any trades? So, I'm going to turn on real quick. I'm going to flip over to the camera. Here's the four boxes we've got for this break. You can see they're all tagged with RT for random teams since it's all came from the fresh one case last night. So, this is random team two. So, we have two more sets of four. That'll be four random teams three and four. So, I'm going to turn on the Nick camera real quick and I'm going to open up the first box while the trade window is open. Trade window closes by the time this box is open. No chat for trade yet, but, well, Roger, at this point he would have been on the Phillies. So, yeah, actually, never mind. You have both teams. I only saw that you had the Marlins. But, yeah. I don't think he's, uh, you might have Sixto Sanchez adjacent. You'll have guys that were in there the first time he ever, you know, donned anything with the Phillies. Phillies must be regretting that, right? I mean, lose him for one year of J.C. Rialmuto. Cool. Trade window closed. No talk of trade. Got to print and rip. So, here we go. Right, Roger? He's not just good. He's real good. Uh, last spot Mojo got Jack, the Mets, and the Yankees. There you go. A little Subway series action there for Pajak. Justin Verlander, Refractor for the Tigers, Matt Gashira. Joe Musgroh, Rookie Green to 99. Houston Astros, Nick Stover. Too bad that's not an actual, like, auto of his AutoCard. That'd be cool. Corey Seeger, Refractor for the Dodgers. There you go. Cody Bellinger, Freshman Flash for the Dodgers. Matt McLaughlin. Robinson Cano, Hyper for the Mariners. Matt Olson, Bass, going to the A's. Sean Matik. And, wow. Rookie Refractor, Aaron Judge. There you go, Jack. Last spot, Mojo. Cindergarde as well. Going to the Jack. And how about a base Aaron Judge in the same box? Also for Jack. Nolan Aronato, Refractor for the Rangers. That's exactly what you're looking for. That'll ease Jack's pain of not having hockey right now.  I mean, you haven't even gone through the X-Fractors and where I've already seen three of the four names that were in the thing. I didn't do anything, Jack. I just picked the boxes. You bought the team or bought the spot. Can we go three for three? Can we get an X-Fractor as well? Josh Bell, Rookie X-Fractor. Manny Machado, NL MVP potential. Hopefully it's Freddie Freeman. I thought maybe that was Cody for a second. Hunter Dozier, Rookie X-Fractor. So that Hunter Dozier, Rookie X-Fractor goes to the Royals and Matthew Shearer. That Manny Machado, X-Fractor, the Orioles, Guy Canashiro, and that Josh Bell, Rookie X-Fractor for Darren Redding and the Pirates. I want to be that lucky, Eric. Box two. You know, Eric, it's that last spot mojo. 60% of the time, works every time. That's right. I don't think we did hit him. Let's find him. Robert, we didn't see him. You know what? I don't know why his name's not on that list either. He's a quality rookie. I'm gonna protest until Nick adds Ben and Tendi to that list. Benny Bicep. Actually, no. I think we did hit him. Somebody mentioned Benny Biceps last night. I'm pretty sure. I'm pretty sure. I think we did hit him. Glass Now, yes, Roger. Danesby Swanson. That guy had was the biggest prospect. Should have been the best. I just, I think I'm a little tainted on him. But yes, he's in here. We just hit, we had a base already. But Tyler Glass Now, absolutely. We hit him last night, I think. Yon Mankata. I like him a lot. White Sox next over. Wilson Contreras, Refractor for the Cubbies. And there's Benny Biceps. Andrew Ben and Tendi. Red Sox, Robert. Zach Ranky, Hyper. Chad Pinder, Negative Refractor for the Athletics. I've been really cool if it was a Matt Olson. A's, Sean Maddock. How about another one of them? Alex Bregman, our first sight of him. Rookie Refractor. Houston Astros, Nick Stober. These boxes are loaded. Josh Bell, Rookie. For the Pirates. Darren Redding. Wilson Contreras, Refractor going to the Cubbies. Carlos Correa, Refractor for the Astros. Francisco Lindor, Refractor for the Indians. It's Alfonso Ribeiro. Matt Strom. Andrew Toles. Come on guys. X factors need to deliver. I'm gonna leave this note on here. Okay. If you could just, when you roll it back there, if it could be like. Everything on top? Yeah, but if you could just put that out facing out. Yes. Yeah, I get you. We're on the same page. Another nice box. You need to find something those X fractors though. I mean, we've got two Rookie Refractors of two of the big names that we were looking for. Box three. In the boxes last night, I think we had what? We had two autos in one box, one of them a numbered green and then none in the other three. Robert confirms. So game shows. You know, they have like that assistant, like the Vanna White, but like of like lower end game shows. Do you think that they get these girls from like lower end acting classes or do you think they go straight like straight model, model casting? Wondering, do they like go actresses? Or do they go just go for a model? I don't I don't get that one Rex. Not catching what you're putting down on this one. Rex's humor is a little bit too high brow for me sometimes. No more Mazzara Refractor for the Rangers. Manny Machado Refractor for the Orioles. Dan B. Swanson for the Braves. Nick. Just a little a little jaded by him. Freddie Freeman. Here's my pick for NL MVP. Jose Abreu. AL MVP. Let's go. Come on. Justin Bohr Refractor for the Marlins. Did she really, Roger? How did I not know that? That's incredible. Starlin Marse, Hyper. It's a pretty cool picture of Mookie Beds. I mean, I know he's in the running for NL MVP and with the fact that his team won the World Series, does that play in? Hunter Renfro. He's now on the raise, but rookie Refractor there. Going to the Padres. Xander Bogart Refractor for the Red Sox. No more Mazzara Refractor. Going to the, was that Rangers? Mazzara? It was, right? Whoever it was, it's going to go there. Corey Seeger Refractor for the Dodgers. Oh, okay Rex, now I got you. You might be delayed a little bit on the, definitely what she said. That's crazy, Roger. I didn't, I don't know why I didn't know that she was on Deal or No Deal. There you go. Benny Biceps, Rookie X-Fractor. Boston Red Sox, Robert. Congrats. Cool. Last box. No autos yet. Are they waiting in here? Was she like a recurring role, Roger? Now this just adds to more questions. I want to know more about the Royal Family. How did, was it, is it Prince William that's married to her? Or Prince, what's his name? What's the other Prince? You forgot about Mookie Betts? In the running for the NLVP or in general? Oh, the fact that he's in there? Harry. Whenever I see the name Harry, I think of Dumb and Dumber makes me so happy. Oh, so she, she didn't get kicked off because of, oh, she was on Mad Men? What? This just, this gets deeper. No, I need to know more. I love Mad Men. Why don't I know what she played? Was she in a prime role? So many more, all these answers to my questions just add to more, more questions. Is that in her bio, Roger? You've never seen Mad Men? Robert. Just get out of this stream and go start watching Mad Men. Take the next week off of work. Tell them that I said it's okay. And tell them I said it's okay. I'll send you a, I'll send, send work a note. Just go watch it. Manny Margot, Refractor Rookie for the Padres. Garrett Cole Purple for the Pirates. I hear he's pretty good. Numbered to 299. Pirates, Darren Redding. Hunter Renfro Rookie going to the Padres. Matt Carpenter, Refractor for the Cardinals. Wands in Suits. She was in Suits, not Mad Men. Okay. Well then, that's not as exciting. I've never seen Suits. Clayton Kershaw Hyper going to the Dodgers. This is more Matt Olsen. Les Ryan Healy. Another Bregman. Houston Astros. Nick Stober. Gary Sanchez, Silver, Refractor for the Yankees. Manny Margot for the Padres. Lindor for the Indians. And another Bregman. 30th Anniversary Refractor for the Astros. Nick Stober. Last chance. Come on, Big Rookie. No. But we did get a lot of them in the packs themselves. So there you go. Very nice. No randomizers, no nothing to do there. Guys, that was 2017 Tops Chrome Monster. Four box break. Random teams, two each. Number two. I'm Sean, jaspyscasebreaks.com. Number three will be in the store shortly. Let's see what else we can find. These boxes have been really hot. Obviously really nice. The rookie class. jaspyscasebreaks.com. Number three in the store. I'm Sean. Thanks for hanging out. See you next time.


Hi everyone Joe for jazzy's case breaks calm coming at you with 2022 panini legacy football 4 box pick your team number 9 And if you follow the instructions right here, you'll got it. You get a chance to win a bit of break credit No vet common only insert rookie cards number cards and the hits relics and autos will ship Let's get on with it. Just grab four boxes from the cabinet right here Another four box break in the store if you want to run this back We've got time for it tonight Brandon that last spot mojo got the Cardinals The last spot thanks to him. Thanks to everybody for getting in on it. Appreciate it Let's go. We were talking a little bit about Kevin Durant just off-camera before we started this recording in Tyler I mentioned yet drives me crazy because I was saying Kevin Durant like what What does he want? He made this big fuss about going to Brooklyn and now he's unhappy the going gets tough He wants to get traded but he want to be in broken in the first place And I was like, yeah drives me crazy when athletes sign these big contracts or whatever Then one year to the deal they want out or something Yeah, I don't understand that Or I think in from a football sense Like Kyler Murray who kind of made this big deal about wanting an extension But he was only what like a few years into his career. It really hasn't accomplished I mean he's accomplished a lot. He's a he's a he's a solid player, but Kind of made a big deal about that. I was like, what have you done? Right poor Blake Griffin. Yeah, right? They did Brandon's pointing out meanwhile Blake Griffin was the opposite sign of the contract and they traded him They're like, I don't know. It was epic But that the team saying we made a mistake. We shouldn't have had him sign this contract Let's get this off the books ASAP And Blake Griffin's one of those guys where it's like if only he could stay healthy. All right box one Let's go. That's an eight out of ten for the ages Alvin Camara lights Camara action that is For Jack and the Saints eight out of ten nice low number on that one. It's a cool-looking insert. I dig it and we've got Malik Willis Three color patch and autograph 20 out of 25 tight titans. Why am I blanking on Malik Willis? Yeah, Tennessee Titans Went to Liberty Liberty Liberty Liberty That is for EA and the Tennessee Titans Nice rookie patch and autograph Out of 25 too, is he gonna get some is he gonna get some time this season? Maybe are they gonna have some Malik Willis only plays maybe What's Ryan Tannehill's contract situation? All we're saying is autograph looks like an earthquake seismograph reading Now he's gonna make the earth shake for Tennessee. That's what he's indicated This Desmond Ritter Josh Allen to 299 Then we got a blue 16 out of 20 Javante Williams for my rivals the Broncos that's for Michael. Yeah, Malik Willis could go to could could use a Could use a weekend at the Jaspie's autograph cam It's definitely a legend right here. Bo Jackson to 150. That's gonna go to Jack who has my Raiders Under the lights Josh Allen Got a rookie to 199 Brian Robinson jr. For Patrick and the commanders And we got an orange Mini Andrew Booth jr. To 125. That's for Minnesota Aaron Miller on points So we'll randomize this to one person in the break winner take all on the point. That's how I usually do it Hopefully we won't say anymore, but Unfortunately takes the place of the of that second auto Juan Soto could he could he tie the game up with a solo shot? Todd are saying Raiders look good at the Hall of Fame game for what it's worth. I agree for what it's worth a little worried about that Raiders offensive line, but I suppose if With all the weapons there. I suppose they can kind of Suppose they could run like quick plays To get the ball out of that out of Derek Carr's hands a lot faster and They could they could find ways around it, but I'm a little nervous about the Nervous about the offensive line and I did the defense is still a big question mark. I know they You know, it's not like Jaguar starters were playing so the defense looked good in that respect But it's not like they were playing against starters. So Defense is still a question mark But I do like I do like the Raiders defensive coordinator And in his philosophy and approach Patrick Graham, I think he was on the Giants So I think he's really good. I do like the McDaniels higher. I think You know, I think a lot of people who are criticizing him about the Denver years and the the sort of bailing on the on the Colts I mean that Seems like yesterday, but it was a little while ago and I think he's definitely I think he's definitely You know kind of learned a lot from both of those experiences and And yeah, so I like it. He's saying all the right things. He I think he's I think the GM Also from New England is pretty solid They think the draft was really solid so hopefully sort of the You put the drama the Raiders drama of last season behind us and kind of move on is more of a professional organization Bears fan tough sledding for Tyler. I'm rooting for Justin Fields though He could be a really fun quarterback for the Bears if They kind of give him some weapons and you know, give him a chance to succeed There's Phil Sims to tonight a Brandon saying I think this is the first year excited to be a Dolphins fan Yeah, I mean that's got it. That's a fun team There's Josh Jacobs Playing with a chip on a shoulder. I like that 50 out of 50 50 option did not get picked up. So That's for Jack and the Raiders. Yeah, Tyree kill big move Jalen waddles fun to watch to was gonna have You know, Tua just has that x-factor where he can just seems like he finds ways to win Now that he has weapons he he doesn't have to shoulder the load too much so he can use his his weapons around him to To make some things happen. I think the Dolphins defense has been a little low-key underrated, right? 127 out of 150. There's John Mechie the third for Houston. That's gonna go to Aaron and Is and there's Vilas Jones jr. Speaking of the Bears and Is the AFC East super scary outside of the bills? I Mean the the Patriots looked mortal without Brady although they had a really good deal with Mack Jones, but You know Jets are still still figuring things out But with the extra the extra wild card Wow, I didn't realize that John Mechie battling acute Promeolo Cidic leukemia, you know, he now said late July. I didn't realize that it's crazy He'll likely miss his rookie season fighting fighting that disease Seems like this is I mean, it's cancer. It's always scary, but but Seems a battleable. So hopefully There's a professional athlete. Hopefully he'll be able to battle it. I'm sure he'll fight as Alvin Camara to 299 Saints Another little mini right here my homies Patrick Mahomes Wow shots fire Jilo saying Mahomes has better left-hand accuracy The right-handed Mahomes better left-handed accuracy than to his left hand.  There's EA with the Titans It's true. Look it up. I mean Are these Mahomes what did it once in a game? That's a small sample size No wonder if Kenny Pickett will we'll see some time and all these moves and Padres They've lost again Yeah, four runs and four games with the Padres Oliver saying back-to-back shutouts Tough times. I feel like that the entire team's got to be pressing because they're like, okay, we got all these big names Ba-ba-ba I'm sure they're definitely pressing at the plate They're they're they're grinding there. They're grinding the bat. You know what I mean? They're making sawdust out of those bat handles See that's the thing. It's one thing to get all the big names on a team Then it's the another thing to actually execute The pressure is there No, did they go I was only half paying attention in this game They got hosed on a replay could have been should have been one-one Small sample size rebels saying but but rebels Royals doing better than the than the Superstar Padres It's like Duke Tyler saying when they had Zion Archie Barrett camera. It's just way too much star power. Yeah the pressure kind of gets You can get to a player. It's out of 150 Aaron Donald for the Rams. It's Jalen waddle mini six out of 75 And a nice rookie year Brandon can't believe that they that the Royals traded with Maryfield Yeah, I didn't I was a little surprised by that too There's a Robert Quinn to 299 for the Bears It'll be for EA more Desmond Ritter for the ATL. That's for Julian And what do we have back here? We've got Carson strong To 100 for the Fly Eagles fly that'll be for James James Carlson with Carson another picket for the Steelers that'll be for Michael and there's Alvin Camero and There's Romeo doves It's a cool looking card nice path. What bowl is that some sort of bowl patch uniform patch little baby autograph You might need to go to Jaspie's autograph camp 5 out of 10 nice low number on that too Blanking on where where he goes Had him last break to Packers. That's right. I feel like someone was saying earlier This goes to Aaron Miller. It's almost saying earlier that This might be Someone to kind of keep an eye on might be interesting this season. Yeah, that was kind of a weird way to play with Rebel thinks Whitmerfield wanted out that made a spectacle about the Novak seen thing for the Toronto road trip And now he's only getting back just because he's a team with a chance of the postseason Rebel saying that's a bit of a slap in the face a lot of Royals fans are pissed Yeah, there's Ed Reid to 299 and there's Aidan Hutchinson Autograph under the lights autograph. He's supposed to be I think he's a lion I want to say he's potentially Really get great game wrecker Yeah, your second overall pick for the Lions a Heisman Trophy runner-up for what it's worth in training camp. I've heard that he's already showing Leadership and poise and apparently Aidan Hutchinson's sister is a Miss Michigan apparently What a year for the Hutchinson family All right fourth and final box ladies and gentlemen Right Tyler Tyler's right. Yeah, probably he didn't want that with anyway. You got you got another You got you got the better way All right. Yeah last box Brandon has the Cardinals and last fall mode. He maybe wants to see a Tyler Murray auto That'd be nice So good luck Brandon. Good luck everybody. All right boys and girls good luck Final box two autographs There's Jerome Ford to 125 for the Browns. That's gonna be for Aaron We got a blue Tony Gonzalez. That's one out of 50. It's for the Falcons. That'll be for Julian And we got an orange Kenny pick it 74 out of 199 You might marinate behind Trebisky this year, but maybe you'll get a shot or two And we got Derek Stingley Jr. Who is a corner out of LSU place is with the Texans now Didn't have to go go too far And that will be for Aaron. There's avian Howard to 299 We've got Trey McBride to 150. Xavier and we'll go to the Dolphin. That's gonna be for Chris Trey McBride to 150 for Brandon Still looking for that second autograph But under the lights Justin Jefferson Nice one for the Vikings. That's gonna that mini card going to Aaron Miller and looks like the final autograph is gonna be eight out of ten Rookie Auto Jermaine Johnson the second for the New York Jets the J ETS Jets Jets Jets Chris Maxwell On the board with an autograph At the very very end any other Low numbered card here or something like that to close it out a bonus auto maybe Another Kenny Pickett under the lights insert We've got a Trevor Lawrence to 100 for the Jaguars that will be for Nestor Trevor Lawrence might have a really nice season And that's that my friends another four box break of legacy football in the books we've got Got some stuff to do here. We got to give away some money got to give away those 400 points So let's flip back to the list now everyone gets a shot at the points Let's gather everybody's names here. We'll use one dice roll for for both lists And now if you follow the instructions here, you got to buy at least an even number of teams. I'll turn into entries So let's sort by your first names here Aaron bought four teams. That's two entries Chris bought Six teams that's three entries EA got two teams. That's an entry Jack got four teams. That's two entries James got two teams. That's an entry Michael two teams one entry Nestor two teams one entry Nick two teams one entry Patrick two teams one entry and Rick with two teams one entry. That's total of 14 entries So two out of the 14 are gonna get a gonna get break credit 125 dollars each. So I got Aaron Chris EA double check this got Jack James Michael Nestor Nick Patrick and Rick All right one dice roll for both lists. It's gonna be two and a five seven times This is for the point they montable get four hundred points after seven one two three four five six and Seventh and final time Chris Maxwell after seven 400 points Right here going your way Two and a five seven times. This is for the break credit top two after seven one two three four five six and Seventh and final time. Good luck. It's Chris and Jack. Congrats to the both of you break credit 125 bucks each going your way Thanks for watching everybody. I'm Joe for jazby's case breaks calm Pick your team nine legacy football in the books I'll see you next time for the next one


Welcome back. I'm Jay Fidel. This is catching up with Kakaako, especially Kakaako Makai. And the reason is there's a bill in the legislature that would effectively repeal the prohibition on residential development in Kakaako Makai. And we want to talk to people who know something about the planning aspects of Kakaako Makai. And one of those people is Wayne Takamene. He's here with us today. And he's the chair of the Kakaako Makai Community Planning Advisory Council. Welcome to the show, Wayne. Hi, Jay. Thanks for having me on. Absolutely. So what is it to be the chair of the Kakaako Makai Community Planning Advisory Council? Talking about, you know, the history, your history and the history of the council. So tell us about that. Well, my history in Kakaako Makai began as a student at Roosevelt High School, and I started body surfing at Point Panics. Then later on, in 2005, we started talking about the AMB RFP to build 200-foot condominiums, just a few blocks away from Point Panic. So we started to get involved, and we created what's called the People's Prefer Plan, which was a rendering of a proposal that instead of condominiums, we would try to create a cultural experience in Kakaako Makai. And some of these included like a exhibition for water sports, Hall of Fame, and other things like farmer's markets. So we had a fish and farmer's market concept. And it began to get more popular as people started to realize that we would be losing a great recreational asset in Kakaako Makai, because it's really the only other sport that's undeveloped in Honolulu along the shoreline. And when we did our surveys as the Kakaako Makai Community Planning Advisory Council, we found that the number one thing people wanted to see was more park space in the Makai or shoreline area, an expansion of the shoreline promenade. And this promenade is part of a concept that was introduced by Governor Wahe'i called the Lay of Green, which would be a lay of parks setbacks from Aloha Tower to Waikiki. And this is where the whole thing started to come together. This was a very powerful concept, the Lay of Green. So later on in 2005, we defeated the proposal for condominiums and the legislature created laws with the prohibition of residential development by the HCDA in Kakaako Makai. They also created a law that prohibited sales of state land in Kakaako. So even if you were to develop a condominium, you wouldn't be able to sell the condominiums because it's prohibited by state law. So this already, what they were planning to do in the lingo administration was to develop these 200-foot condominiums and they had like a series of three of them that were right next to the Kuala Basin Harbor and then sell the land to A&B. So A&B, I think for $20 million, I think that was the number you were talking about, $20 million would get all the profits from selling condominiums. And you know, it's crazy. The real estate market in Kakaako, on the Makai side of Alamata Boulevard, I think this was like 10 years ago, they were saying penthouses for $36 million. So I don't know if that will fit into the definition of affordable housing that Ohio is talking about. Well, so during that 2005 session, they also made a resolution that required the HCDA, Boy Community Planning Development Authority, which has a jurisdiction in Kakaako Makai to create a working group that will help to implement the development and planning and implementation of a master plan for Kakaako Makai. So that became the Kakaako Makai Community Planning Advisory Council. And so by the legislature, we were allowed to get $500,000 to do this planning thing. And later on, it was revised to $750,000. And we completed the master plan in October of 2010. Oh. And what did it provide this master plan? Well, this master plan was created by several groups that were part of it. And we broke it down into focus groups. But excuse me, I'd like to read the vision for Kakaako Makai that was adopted by the HCDA. Kakaako Makai is a community's gathering place, a safe place that welcomes all people from Keiki to Kapuna with enriching cultural, recreational, and educational public uses. A special place that continues the shoreline lay of green with scenic beauty connects paramatic vistas, Malka and Makai, and encourages ecological integrity of land, air, and sea. Kakaako Makai honors, celebrates, and preserves the historic sense of place, Hawaiian cultural values, and our unique island lifestyle for present families and future generations. So this is what we proposed. And this is what the HCDA approved. I'm sorry to say, but the Hawaiian, the OHA concept of making residential in Kakaako Makai that will likely be sold to the highest bidder is really something that is contrary to our vision for Kakaako Makai. But the plan was never implemented? Well, in 2011, we started meeting with OHA. The venture was Colette Machado, a beautiful woman. And, you know, she said she was impressed with our Kakaako Makai master plan. And then it was then a concept where OHA would would be taking Kakaako Makai as compensation for their seeded land, seeded land, I guess they weren't paid. They just didn't feel that they were compensated well for that. So we thought, okay, fine, if they want to develop our plan, then it would be okay. But later on in the session, OHA introduced the bill, or it was introduced by the supporters of OHA, but the bill was actually to repeal our law that prohibits development of residential development in Kakaako Makai or planning by the HCDA. And through to the different organizations, and there are over 55 organizations that participated in the creation of the master plan for Kakaako Makai, including OHA. OHA had two representatives that were part of what we call the steering committee. They were ex-official on the steering committee, and they were active in the creation of the master plan. But that repeal of the residential laws in Kakaako Makai was defeated, and it had repeatedly been brought up by OHA. OHA's law to take the land in Kakaako Makai, I think it was supposed to be a $200 million settlement was approved, and they have, since that land was, I think it was designated as public trust lands, so they have acquired that. Several times they tried to repeal the law of residential in Kakaako Makai, and I believe in 2012, the Attorney General said that that proposal for OHA to repeal the law is actually unconstitutional, because it would be favoring a entity over the general public. So did OHA ever explain to you, Collette Machado, or anybody else over there, why they had changed their position from supporting your master plan to doing the reverse? Well, we had a, there was an area Senator for Kakaako at the time was Senator, former Senator Brickwood Galutaira. He is also right now one of the just newly elected at large OHA board members of trustee. He appeared to be favored the residential development in Kakaako Makai and repealing the law. It's really unfortunate, but then Governor in 2010 was Neil Abercrombie, and in 2005 when he was, I believe he was a House of Representatives for Congress, he actually came down and had an interview and he spoke against the A&B development, so at that time we believed that he would be, would support a view of keeping Kakaako Makai free and clear of residential, and then later on he changed his mind while he was governor and said, oh, they should give the Hawaiians the land and let them build residential. So unfortunately that didn't go well with the public and he became a one-term governor. Brickwood also supported residential in Kakaako Makai and he was very unpopular with his reelection hopes and he was soundly defeated by the current Senator Sharon Moriwaki. So, you know, it's very unpopular when we did our surveys as CPAC, it came out near the bottom. You know, nobody really wanted residential and maybe it's because the people who were participating just wanted to see something else besides residential. A lot of people feel that it would actually become a gated community, like something like how you know, Koalana is with the hotels there, you can't, I think there's two public parking spaces, you know, to go to the beaches there, we feel that it would become a gated community and that, of course, the land wouldn't be state land because if you're going to build a 400 foot condominium, developers are going to want some interest in that development and they want compensation which is going to probably be that they're going to own part of that condominium and sell the units for financial gains. So, you just did a panel in our previous. Other people who said the same thing. So, if you were to appear in front of some committee in the legislature today where they are considering this current version of the bill that would allow residential development in Kakaka Makai, what would you say to them? Probably the same thing I said in 2011, I think in 2012, last year, I think there were like about four or five times that Ohio has tried to repeal this law and failed. So, similar to what I've said before because there's nothing new. I mean, I don't think we see anything new with the land. I think a lot of people are just, you know, tired of Ohio coming back and then we have to go and seem to drill. Many times before, I'm sorry, I didn't count how many times they tried to repeal the law but it seemed like a whole bunch, you know, and so we're all getting a little tired of this but we're all pretty steadfast in our beliefs that residential in Kakaka Makai is something that shouldn't be there. Yeah, one other thing I wanted to ask you is just suppose, hypothetically, just suppose that they're able to get it through and they are able to build a 40-story high-rise condo there. What would happen in your view of it and your, you know, your master plan community development view of it? What would happen to a Kakako Makai? What would it be like? Well, Jay, you have to remember that this land was a dump. They had incinerators there. They were burning all the opala and incinerators created ash. These ash is, there's documented in several phase two environmental studies that it's bad stuff. It's lead arsenic, PBCs, have reports that they should take cattle off the boats, burn them through these pools of DDT to kill all of the parasite and all of that stuff is still on the ground. And so if they're going to build a huge building, they're going to have to dig up underneath, put in infrastructure and when they're doing that, they're going to be hitting toxic waste. And another thing that comes up is that in Kakako Makai, when they were able to throw up these big buildings, it changes the flow of water, underground water, and then you'll see sinkholes popping up. So what happens if there's a sinkhole in Kakako Makai with all of these toxic waste? You have high levels of lead, you have arsenic, PBCs, benzene, things all, not very good stuff and you can't really go and dig that stuff up. If you disturb any of that stuff, then the equipment you're using to disturb it becomes contaminated and you have to decontaminate that. If you want to move some of that toxic waste out, then you have to get a permit to dump it somewhere and I don't think you'll find any place on Wahoo that's going to take that toxic waste as a disposal site. You're having a hard time just finding regular waste to dispose. So the land really does not work well with high buildings. It's just something that there's going to be a problem with and then for us to use the ocean for recreation, it's terrible. What are you going to do? You're going to put this stuff up and then you're going to end up saying to the public, can I go in the water? Can I use the ocean recreational places because there's toxic waste? There's also underground systems under Takakumakai. It was basically reef before. So what happens when you disturb all of this toxic material? It's going to come dislodged from the sediment and then it could flow out into the ocean and affect the reef and also surface out there in the water. You've identified a bunch of problems. I mean the toxic soil, the fact that it would be disturbed, it would leach off all around it. The water problems in general, the access problems, I guess that's a community issue, maybe a cultural issue for native Hawaiians. So this all sounds like it needs to be thoroughly investigated by WAVE and EIS. Has your group taken a position on whether EIS would be appropriate before the legislature acts? You know, I think the legislature should just shut it down. I mean this is something that President Densho and Takakumakai have come up previously so many times and this law was enacted in 2005. It's nearly 20 years. I think you should say it's not going to happen. But we would look at it and say, well, it's a free country and we'll put up our our point of view and which is also the point of view of the attorney generals saying that this allowing OHA to be given an exemption to this law is not something that is constitutional. So there are several fronts on this. And I think one of OHA, they want to develop, I guess, affordable housing. And I believe the legislature would swap land with them someplace else. So the issue of affordable housing is something that I don't believe is something that really holds water. That they want to build affordable housing in an area where penthouses are going for 35, 40 plus million dollars. It's not affordable for your Hawaiian people. I mean, there's opportunities other places that will have more land available for OHA to build their affordable housing for their culture. But I don't think Takakumakai is that place. And sounds like you're going to continue the effort to stop this, the repeal of this bill, to stop residential development.  Yes. And it's not really stopping them, but we support the law, existing law, the law that has been created in 2005 and continues to this day. So we support that. I mean, I know a lot of people who are Hawaiians and they have their beliefs in what they should do in Takakumakai. A lot of them have come up to me and said, Wayne, if this comes up, let us know. We don't want residential here. So there's the people that are Hawaiians that don't really believe that this is the correct thing for OHA to pursue. Wayne Takamini, the chair of the Kakaako Makai community planning advisory council, which has been following the action in Kakaako Makai for many, many years. And we really appreciate you coming on and talking with us, Wayne. Thank you so much. Thank you, Jay, for this opportunity to educate people on what the issues are. I think you've done a great job doing that, Wayne. Aloha. Aloha.


This way to Bear Country. You'll know when you're there. As soon as you enter, you'll feel like a bear. A great grizzly bear. A bear and a stain bear. We are the Bear and Stain Bears! Mama! Hey, Deep. Let's do something. Oh, no. Come on. You can't just do this all day. Let's play a board game. Okay. Can we play? Hurry. Sure.  I want you to know something. I love you. But I play games seriously. You're going to lose. Whatever.  Yeah. Oh, I'm a teacher. A lawyer. Who wants life insurance? I do. Who wants fire insurance? Count me in. Murray? Geez, Murray. Got enough kids? Take some breath. Land on an estate plan. All right. Who's getting an estate plan? I will. And I don't even know what that is. Murray? You got to do an estate plan.  Look, it can wipe out all that money. It can take all probate fees and court costs and... Listen, what happens if you're ever incapacitated? It'll wipe out a big chunk of that. Come on, Murray. Yeah. What do I know? I'm only a lawyer. Please don't do that. Get $50,000. He's your Murray. Who's your friend? I know it's a tough time, Dee, but Murray had a will. And he left his estate to you. What a sweetheart. What's an estate? Well, think of it this way. An estate is everything Murray owned. His cars, his houses, his money, his stock, everything he owned. So all of Murray's stuff is my stuff now? That's great, right? Yeah, it sounds great, but Murray didn't have an estate plan. That's sad. I'm only seven. I don't know what you're talking about. It's time for me to put my lawyer hat on. Okay, I'm ready. Look, you've got an inheritance. But since he didn't have an estate plan, it's going to affect what you receive. How so? Dee, when Murray didn't do an estate plan, he made a big mistake. Your inheritance is going to get a lot smaller, unfortunately. You're going to lose money because of probate. That's going to take a big chunk of it. Oh, my gosh. And then there's executor fees. And then don't forget Murray was incapacitated and he lost some money in a conservatorship. What? And then there's taxes. And heirs fighting the estate. Oh, Dee, it doesn't stop. You got me kidding. It could have been avoided. You know, hamsters always avoid doing an estate plan. Some people do, too. What are you waiting for?