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Let a be an element from a group with finite order Iol ged(n.k) = 1. Prove that n. and let k be a integer with the centralizer of a. denoted by Cla) {T € G eq...

Question

Let a be an element from a group with finite order Iol ged(n.k) = 1. Prove that n. and let k be a integer with the centralizer of a. denoted by Cla) {T € G equal to the ceutralizer of ak . C(a ) . IU ar}. is

Let a be an element from a group with finite order Iol ged(n.k) = 1. Prove that n. and let k be a integer with the centralizer of a. denoted by Cla) {T € G equal to the ceutralizer of ak . C(a ) . IU ar}. is



Answers

Let V be the vector space of n-square matrices over a field K. Show that W is a subspace of V if W consists
of all matrices A = [aij] that are

Question given us when it took problematic particular election that one Q plus two cube plus three Q two and Q is equals to one plus two plus three. Dylan all square. Not approved by mathematical election. We just solid for let's say for and equals to one. And in that case our left hand side is one Q. Which is equal to the right hand side. Since that is one square and that is also one. Now when you look for any calls to K now we assume that it is true for N equals two. K. Assumed it is true and therefore we get one Q plus two cube plus kick you is equals to one plus two plus. Okay hold square. And now we need to prove for any close to K plus one. So when we write this year we get left hand side as once Q two Q bless KQ plus K plus one. Thank you. And our left hand side. Sorry, 19 side is one plus two plus K plus k plus one whole square. Now let's start with the left hand side. So here this entire term can be replaced by this value which is one plus two. Bless less clear square. And we lived with K plus one. Okay. So now we have two dumps and we need to simplifies us that we get this. If both are equal now to do that we can just neglect this step. And what we can use is summation formula. So here this is one plus two plus three T. K plus one cube. And this is some of Uh huh. Where are cube? Sorry? Where are they starting from? One till it's a K plus one. And we know some mention of our cube is if they are going to keep this one. So this escape plus one square. Okay. Plus one. And here we have K plus one plus one since N plus one whole square by four. And on simplifying get K plus one square. Okay. Plus two. Great. 44 So this is our left hand side. Now let's solve the right hand side part. So here we can apply submission formula inside the square down. So this is one plus two plus three plus K. K plus one. So this can be written as some of our our starting from one tree. K plus one whole square summation of our is given by formula N. But here an escape plus one. K plus one. Okay. Plus one plus one by two whole square. And this gives us K plus one square. This becomes K plus two squared over. And that is the same as this and total by induction. Thank you.

Okay, so we're trying to decide for which values of an to to the end is greater than in squared. So there's a couple of ways of going about this, probably the easiest way of going about. It is just trial and error. So we could try playing in a couple of images. Now you don't want to stop it one. Um because if you stop that one, you can actually get the wrong conclusion. So you plug in and is able to one on the left hand side, you get to on the right hand side, we get one and to act two is greater than one. Okay, So that's good, but if you plug in and is equal to two, you get to to the power of two. On the left hand side, on the right hand side, you get to the power of two. Well, these are equal, so this is no longer true. So let's keep trying if we do N is equal to three, we get to to the power of three on the left hand side, on the right hand side, we get three to the power of to back. In this case it's the opposite right. The left hand side is eight, the right hand side is nine. And we do four. We get to to the power of four. We get forward to the power of to in this case they're both 16, so they're equal. So for the first four cases, it's ambiguous as to what values of and make this statement true. If we do five, we get to to the power of five on the left hand side, on the right hand side, we get five to the power of to now this is 32 on the left hand side and 25 on the right. And so this is great of them. And then we can keep going. And what we see is that everything after five, including five, is going to have this relationship. And so our claim is going to be um to to the end is greater than and squared so long as and is greater than equal to five. And we want to prove this by induction. Okay? So to prove this by induction, this is our new claim. We're going to establish our base case. Our base case is going to be show that it's true and it is equal to five. Well, when I was able to five, we have two to the power of five on the left hand side, which is 32 we have five to the power of two on the right hand side, which is 25. That's in fact through the 32 is greater than 25. So we'll say chuck mark, basis case checks out. Okay. And then we'll say our inductive hypothesis is that we're going to say suppose that for some fixed imager K, the statement is true to the power of K is greater than Okay squared. Okay. Now, in order to do our inductive hypothesis, we're gonna actually have to establish an algebra so I'm going to do this in a different color off to the side. Just that we don't confuse it with our actual proof. There's an interesting thing here and you'll see why it's relevant just one second. But it turns out that for integers, if K is greater than or equal to three, then two K squared is greater than K squared plus two K plus one. And I promise this will be relevant to something and you don't have to do it in this order. You would have seen why this would be important later on if you continue with the problem. Now to see this, we can just find where these two things are equal by setting them equal to each other. Okay, rearranging really quick gives us k squared minus two K minus one. Whose roots are if you were to plug in the quadratic formula, the roots would be one plus or minus route to which as decimals are approximately 2.4 something. It doesn't really matter. And the other one is minus point 06 something. Okay, but the point is that any time K is greater than 24? We actually have that two K squared is greater than K square post to keep this one. And if you were to see this, if you were to put both of these things, two k square, it looks something like this. Okay. Whereas K plus one squared is like shifted over but it's less narrow. Okay, I'm not exactly right? Yeah, clean that up a bit. So let's narrow a little bit wider. And so any time we're above this number? Right here. Okay, this one right here, which is the two K squared one is going to be above the K plus one squared one. Okay. Mhm. Now, why is that relevant? Because when we start to go to our inductive are inductive step, we're going to look at two K +12 to the K plus one, which is the same thing as writing two times two to the K. But by our inductive hypothesis, we know that this has to be greater than two times k squared, right? Because our inductive hypothesis is that to the power of K is greater than K squared. So we just insert that for two to the power of K here. But we just showed and read that two K squared is greater than this is strictly greater than K squared plus two. K plus one. Which factored is just K plus one squared. And so in other words we have the two to the K plus one is greater than K plus one squared. Okay. And so we're ready to make our conclusion. Our conclusion is that we have shown that to the power or two to the power of five I suppose, which on the basis case two to the power of five is greater than five power of two. And that if for some Okay, But in particular some K greater than equal to five to the power of K is greater than okay squared. Then it's also true that two to the power of K plus one is greater than cables, one squared. And so it must be true for all in all right. Hopefully this was clear seeing another

So let's start with the question. The question says prove that. Are said 10 elements has any two in minus one and two. And managed to buy six subsets containing exactly three elements whenever and is an integer greater than or equal to three. Okay, so what we can do is we can use the mic to prove this thing. Okay, we're probably using induction. So let's start by writing model things we have been given. Uh so it has given that N is an integer bit uh where N is greater than or equal to three. Okay. And to prove uh yes, I set with elements has and into n minus one into n minus two by six subjects containing exactly three elements. This is what you have to proof. Okay, so we are going to prove it by induction. Okay, Okay. So let us assume the first step let PN Sorry, uh let B n B a set bet. Okay. Elements. This is what was the question? Okay, and elements uh has N into n minus one in two in minus two by six subsets uh containing. Okay, Okay. Uh exactly three elements. Okay, first thing we have to write it in your question from you. Okay you have a new give the solution. You have to write this thing as it is. Uh Now let's do the basic step. Okay, so normally be a substitute uh be of one, but here it is given that N is less than equal to three. So basically you'll find B of uh three. That is N equals three. Okay, so was set with three elements. Uh Mintz is of uh Okay. Abc Okay. Uh uh The only subset um I just with three elements. Uh Sorry about the spelling mistake. Uh Three elements is one. That's a little Okay. Uh Okay, so there is only one subset with three elements. Okay, now let's see uh substitute in the formula and Legacy. Okay, so it was in in two in minus one in 20 minutes to buy six and 33 into three minus one is equal to 22 three minutes to within one. And this thing is six by six is nothing but one. Okay, so pay of three is true. Okay, now let's go to the induction step where we assume that P K is true. Okay, so if PK is true then uh what if a set with K elements has? How many elements came to came? I just one sorry, not elements. Then became this one came in this two by six subsets with exactly three elements. Three elements. Okay, we assume this to be true. Okay, I'm gonna take it as a question when this thing is true, PK is true. Okay, so what we need to do is we need to approve P k plus one is also true. Okay we prove this then we can say this induction is right. Okay so let's stand. Okay so you are set with K plus one element. Okay so how it will be? It will be in the form of heaven to uh Kay okay okay. Element and one more element will be there. Ok Plus one. Okay so uh you know the previous key element? Uh it's subset. Okay we know that the earlier subset. Okay uh even a two up to aK Okay it has uh K into k minus one and two K minus two by six uh subsets. Okay uh this from the previous thing we know this set has this many subset and it contains exactly uh three elements. Okay uh while uh okay in two k minus one by two subsets uh with exactly three elements, Three of which one will be but a k plus one. Okay. See then only it will be accommodated in this set. Okay? So what we can do is okay in two k minus one by minus two by six plus this subset. Okay, this number of subsets for this case came to k minus one by two. Okay, we will solve this. Okay uh Okay let's go to that. So what we can do is we can take came to k minus one comment from this. Okay, so this will be further, it is to k minus two by six plus one by two. Okay, so uh let's take a simple wheel. Okay minus one. Okay. Uh So it comes out to be uh okay minus two plus three. Okay, so this is came to k minus one and two K plus one by six and uh copy this expression to the next page. Uh Got a paste it. So how can I make it on the farm orphanages? Uh Let me try not in Formula. Just I have to bring the question in form of this. K plus one. Uh K plus one minus one. And K plus one minus to this farm. I have to bring Okay by six. So what I'll do is uh from the previous thing I had this thing. Okay I'll take this K plus one out. Okay, K plus one. I rearranged the terms. Okay And k minus one by six. Okay. Plus one is outside. Ok I can like this writing desk. A plus one minus one. Okay. Can like to take this? Okay. And the camera is one. I can like the desk A plus one minus two. Okay so I hope uh we got the terms that we required. Okay. Plus one plus one. Plus one. Okay, So from this, PK plus one is also true. So since PNS true, be cares true, B K plus one is also true. And the conclusion is by principle of mathematical induction, P N. Is true for all positive in teachers. And I hope you understand the problem. Uh, thank

To check this subspace. What we need to do is we need to verify three exams. The first axiom is zero, vector should be belonging to a subspace. W. I mean the set W If you won't, you too belongs to W. Then we should verify that you want to see. You too also belongs to W. What is the closure property? Third, if you belongs to W then land up. You should also belongs to W Where lambda. Is it number from the field? Okay, if all these three exams satisfied, then we call W is a subspace of the factor space week. No, these set off all environ mattresses over the field kit. All right now W is the set of all symmetric matrices. So what do you mean by a symmetric matrix? The matrix was transposed the same as it's A So that's called a symmetric matrix. All right. Now if is symmetric then that's null. Matrix belongs to W. Yes, of course. Because the non matrix transpose his narrative so yes, it's true. Second axiom some of two symmetric matters is A and B. Supposed to be a symmetric B is also symmetric. Then a place be whole transpose is equal to the property of transport. It is a transport transport. But he transposes A. Because asymmetric transpose is B. Because B symmetric diet implies a place to be. He's also symmetric so that implies equals B belonged to W. So second exam is very fine. Now coming to Torrance if is symmetric lambda E. We'll transport the property of transport is lambda and a transport is equal to the lambda. A. Because the symmetric it is lambda lambda. He will transpose islam day. So that means lambda is also symmetric so it belongs to doug. So all the three exams are verified. So yes. Set of all symmetric matrices is a subspace. The second one upper triangular mattresses. So what do you mean by an upper triangular matics? All the elements below the main diagnosed should be zeroes, for example A B C 000 123 So this is an upper triangular matrix below this. All the elements are zeros of is a pet Wrangler. He's a pet Wrangler and he's also a strangler. Then it let's be is it a veranda? Of course, yes. Because these zeros get started with the other metrics in the same place. For example 123 4045006 If you add with 178 089 00 10. So what do you guys? 000000000 So some of the upper triangular mattresses. Also an upper triangular markets. So this is very frank. Null matrix is not trying biometrics because null metrics by default, all the elements below the main diagnosis anyway, zeros. In fact all the elements are real zeros. So it belongs to going mathematics, The Time magazine. If you want to play any number with the matrix scale are multiple, zero into any number of zero. So these zeros will still be retained. So lambda you Orlando A is also an upper triangular matrix. So yes, the set of a particular mattress is a subspace. The third diagonal matrix, basically, we can just individually say that the diagonal matrices from the subspace because lower triangular mattresses also from subspace. Because what do you mean by lower strangler? All the elements about domain that loves you with the same reasoning as a crime. And what is the diagonal matrix? It is both the lower triangle ring up strangler. Since both our Wrangler and Wrangler mattresses from subspace set of diagonal mattresses. Also from subsidies. Simple reason. Right. And skill harm metrics is a special case of diagonal matrix, scalar matrix means all the diagnosed elements should be zero, sorry, All the diagnostic elements should be same and other diagnosed elements other elements should be basically zeros. So this is a special case of diarrhea magics. And since diagnosed mattresses from subspace, the set of the subset of it also forms the suspects. Alright, so this is a prison. So


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