5

(#) Lct G b group' . Let G.G denote the swallect subgroup of G containing5 = {ryr 'y :I"€G}(The subgroup [G,G] called the commutator subgroup o...

Question

(#) Lct G b group' . Let G.G denote the swallect subgroup of G containing5 = {ryr 'y :I"€G}(The subgroup [G,G] called the commutator subgroup of G )Show: thatryr-'w-1, (u-I s")(u-'yu)(u-'ru)-!(u-'yu)-1 for all r,y; eG Deduce that [G,G]<G Show that the quotient group G {{G,G) abelian. I6 NaG and G}N is alxelian, show that (G,G] € N. (In other words, G/IG.G] tluee largest alwlian quotient of G.) Slow that il k G a [G.G c k thcn K <6.

(#) Lct G b group' . Let G.G denote the swallect subgroup of G containing 5 = {ryr 'y :I"€G} (The subgroup [G,G] called the commutator subgroup of G ) Show: that ryr-'w-1, (u-I s")(u-'yu)(u-'ru)-!(u-'yu)-1 for all r,y; eG Deduce that [G,G]<G Show that the quotient group G {{G,G) abelian. I6 NaG and G}N is alxelian, show that (G,G] € N. (In other words, G/IG.G] tluee largest alwlian quotient of G.) Slow that il k G a [G.G c k thcn K <6.



Answers

I need help solving number 5. How do show that the equation does not preserve structure? or it is not homomorphism?

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.

Okay. So the question is as follows, where we look at the status, which we say is the smallest sigma algebra, What is a signal of zero generated by the sigma algebra in in our right, generated by the set are Come on R Plus one. Where are is a rational number. So this is the problem statement should be on our Okay, so moreover, we need to show that S is the collection of moral subsets of our So we need to say that S is the borough subsets. Okay, well the barrel uh, the barrels sat on our is you can think of it as the smallest. Take my algebra generated by uh let's say a basis for the topology of our and a basis for the topology of our is just as I said, A B where am br elements of the reels. Okay, so the topology, the standard topology and our is generated by such sets. And if we can show that S contains uh a set of the form, A comma, B where A and B are real numbers, then that's sufficient to demonstrate. That. Is is uh wow the moral sigma algebra on our and the way that we do this is what we grab one of these. Uh the goal here is to make that said a B. Right. Where A and B are real numbers. You're seeing or from Sets of the form are comma are plus one where are is irrational. Uh taking these sets to these such sets to be an element of sigma algebra. Okay then the way that we do this is let me make some space for these. Oops. So let's look at this set A Excuse me? A comma eight plus one. Right. And uh let me let me do something. A sub N. Coma A sub N plus one. Where S A Ben is a series of rational numbers, right? Such that he saw Ben converges to a rational number. A uh sorry, a real number A From the right. Okay. Then if we take the union over end of all. A seven come on in some men Plus one, guess what? We're going to get this set A comma a Plus one. And this set will be an element of S. And it's important to note this because A Now is a real number. Right? And we want to do something similar. Right? Now we look at the set B. Seven come up beasts up and plus one where of course be summoned once again is rational, right? And we say that Visa ban converges to be from the left and B is a real number. Then it turns out that the complement of this set complement of this set, which will be let's look at let me copy this. I don't have to rewrite it. Once again, the complement of this set will be the same from negative infinity coma be suburban Union be. So then plus one comma infinity. Right? And this set will be in S because see my algebra. Czar close with respect to compliments. And now if we grab this set and we intersected with the set that we found down here, so let me grab it and copy they grabbed the set right, Win or sickbed. We'd are set here. It's what we're going to get. We're making the assumption. Obviously that is less than strictly less than the So keep that in mind. Well, if we do these, we're going to get this set a comma B. So then closed. And if we take the union over end of these sets, we will get a comma B, which is the goal that we were trying to achieve. So the sigma algebra, in conclusion, the sigma algebra generated by this set contains sets of this, the following form and all sets of the following form. And we know that all such sets for my basis for the topology of our therefore the city. S. Is the world sigma algebra owner

So to show that this is not a hormone dwarfism. Um We essentially just have to show that one of these statements here is not true because these are the two criteria for something to be a home abort is um for some function and for this, I think the bottom one will be the easiest to actually go about doing. So How I'll do this is let's first plug in. Just half of zero or a half of the zero factor, I should say. So 00. And if we were to plug that in, so this is going to be zero, this is going to be zero and then that's going to be negative too. So we get that this outputs negative too. Well, if this is a home of dwarfism, if I were to multiply the zero vector by anything, then it should give me the same output as if I were to just multiplied on the inside and then go about doing So Let's say I were to just multiply the outside of this by two, two, two. That's going to give -4. So let's see if we do the same thing. If we could out um negative 44 and you'll end up saying that we won't because F of two times 00. Well, that's still just F of zero. And we just showed that is negative two. So essentially we just showed that two times F of 0, 0 is not equal to F of two times 00. Which would then tell us that is not a home amorphous. Um So implies not homo morph ism Um There is something a little bit more streets or we can do depending on if you have talked about it or not. Um And the thing is is that your identities should get taken to another identity. So for addition, remember this here is the additive identity. Because if we add it to anything, nothing changes. But in just the real numbers, -2 is not the additive identity, it would just be zero. So if you have where an identity does not get taken to another identity, then you can say that it's not a home a morph ISM by that, but uh that's more of a just like theorem that people often have to be taught um or go about proving I should say. But as long as you just show something like this, that would also be a valid way of showing that is not a homo dwarfism.


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