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Let K < H < G. Suppose (C : HJ = m ad [H : K] =n Prove that[C : K] = [G : HIIH : K] = mn.Do not assume G is finite....

Question

Let K < H < G. Suppose (C : HJ = m ad [H : K] =n Prove that[C : K] = [G : HIIH : K] = mn.Do not assume G is finite.

Let K < H < G. Suppose (C : HJ = m ad [H : K] =n Prove that [C : K] = [G : HIIH : K] = mn. Do not assume G is finite.



Answers

Let $\left\{b_{n}\right\}$ be a sequence and let $a_{n}=b_{n}-b_{n-1} .$ Show that $\sum_{n=1}^{\infty} a_{n}$ converges if and only if $\lim _{n \rightarrow \infty} b_{n}$ exists.

So the question is proved that every non identity element of a free group of is a finite order. Okay, so I have made the solution available for you. So this is the first step to show in a free group F any non identity element is a finite order. Or in finance in financial in finite order question is also in finance garden. So he had to prove begins let F be free on X. And let G is equal to expert. Excellent. And so we want to export X. M par accident. Yeah, it belongs to finite that F. Okay. But All the X. I belong two weeks now. X belongs to F. There's a function fire and X has a function which is that often. So we have a definition F. Such that X belongs to the F. In my F. Of X. I. Is required to G. F. I. The IAB standard basis vector of set of N. And ffx as you come to zero for all X belongs to the set X. That is excellent, comma. And so on to accent. Then there is a home on my shift. That is a fight. Such that F belongs to that often fission gentleman with all left. Okay. Is it going to fight? Oh hi mm. And here's your steps. So I just it is everything. You can have a look on it. And I'm following this solution. You will get proof here. The last step. Okay, fine.

So the question is proved that every non identity element of a free group of is a finite order. Okay, so I have made the solution available for you. So this is the first step to show in a free group F any non identity element is a finite order. Or in finance in financial in finite order question is also in finance garden. So he had to prove begins let F be free on X. And let G is equal to expert. Excellent. And so we want to export X. M par accident. Yeah, it belongs to finite that F. Okay. But All the X. I belong two weeks now. X belongs to F. There's a function fire and X has a function which is that often. So we have a definition F. Such that X belongs to the F. In my F. Of X. I. Is required to G. F. I. The IAB standard basis vector of set of N. And ffx as you come to zero for all X belongs to the set X. That is excellent, comma. And so on to accent. Then there is a home on my shift. That is a fight. Such that F belongs to that often fission gentleman with all left. Okay. Is it going to fight? Oh hi mm. And here's your steps. So I just it is everything. You can have a look on it. And I'm following this solution. You will get proof here. The last step. Okay, fine.

This problem covers limits. In order to solve part A and B of this problem, we will use the approach of contradiction. Let's look at part A. First. The claim is that if S. N. Is greater than equal to A. For all but finite lee many N than limit, S N is greater than equal to A. Let us assume for a contradiction that this limit Yeah, is less than let us assume. And let's say absalon is equal to a minus S. This is the assumption. By the definition of limits, we can say that. Bye. Definition of limit. We can see that there exists capital N. Substantive small and is greater than capital in then sn minus S is less than epsilon. This is what the definition says. I'll write it down there exists capital and such that Yeah. If and is greater than capital N. Then and send minus S. Yes, less than absolute. So if you open this inequality, you know, you will get minus epsilon Nice between s and minus S taking minus s on both sides. You get s minus absalon as less absalon. Now, you can see here that S plus epsilon is nothing but A. So you will use that factor and I can say that s minus epsilon. Uh it means that essen belongs to s minus absalon. Call my A. What does this mean? It means that S. N. Is less than a, S. N is less than A. But our problems said that our assumption was based that S and will be greater than equal to way, but we got the opposite. It means that for all but finite lee many end S. N. Is less than in this problem, as per the assumption, which is a contradiction. It means that our initial claim was correct that if S. N. Is greater than equal to A. For all but financially many in limit S and will be greater than equal to It similarly will solve part B of the problem. Part B says that if S. N is less than equal to be for all, but finitely many end then limit essen is less than equal to be. Let us assume again for contradiction. We will assume, Oh, that limit essen is greater than B. We're assuming as a contradiction. And let us take absalon as S minus B. Again, by the definition of limit there exists and such that if N is greater than in the sm minus S is less than absolute, just like this. We use their definition of LTD so we get yes and minus S is less than epsilon. Again, opening the inequality minus epsilon, S n minus S rather than absolute, this becomes s minus epsilon. Essen? S less. Absolutely. From here, we can see that S many cepsa loans nothing but B. So I can see here, B essen less than s plus absalon. So from here it is clearly seen that sns greater than B. Or I can say for all. But finitely many N. We have S. And greater than B, which is a contradiction according to our statement. Because we took us and less than equal to be. Hence the claim that if S. N. Is greater than less than equal to be for all. But finitely many end then limit us and less than equal to be was correct. And contradiction proved it false. Let's go to part C. Part C is a direct conclusion from part A and part B from part A. You know that? Yes. Which is the limit limit of essen is greater than equal to A. And from part B. This limit S it's less than equal to me, so S belongs to it to be. That's all. Yeah.

In this question here were given. Uh, I am equal to the B N minus. B a minus one. And we're interested in the submission that I am from one option infinity. So notice that they will explain this one when any could you want him to be one minus B zero, then plus B two minus, uh, be one plus the B three months, B two plus the big four minus P five plus up to the PM minus P m minus one. And so, um, and we see them, we can console the B one window. Be one here, be to win the B two B 300 b three. So we're missing the beat three years. So sorry. So let me in. Center doesn't before, uh, be 1234 minus b three here. And then we again this one will be canceled with this one and so on. And here we see that we will cancel this and so on on. We see we have left with only, uh if we consider only up to in the past with some s and for now, then we just stop up to here and then we see that this s and we ego Thio actually end here on me and then we see that this sn it will equal to the We have left with the minus p zero then plus with a p n. And we see that this is here as an we turned the limit here Angus t infinity And then he coaches limit on the industry Infinite. They on the minus B plus B n And we say this one we get echoed you This one will be p zero. So it will be the minus p zero and this leap plus the limit on the PM and goes to infinity. I doesn't implies that the I the submission on the I m converge even only if this limit here also. Uh, I already mitt because you some constant else model that infinity


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