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(32) Let G, and Gz be isomorphic groups. Prove that if G has G2 has a subgroup of order n; then subgroup of order nN'^^ !~ tn dinli : | "1 '~/' ...

Question

(32) Let G, and Gz be isomorphic groups. Prove that if G has G2 has a subgroup of order n; then subgroup of order nN'^^ !~ tn dinli : | "1 '~/' | "I tahb %0 !v | {To 2 MD; :1<" Dm WUrJa Tfun X0 Ke Sa 02

(32) Let G, and Gz be isomorphic groups. Prove that if G has G2 has a subgroup of order n; then subgroup of order n N '^^ !~ tn dinli : | "1 '~/' | "I tahb %0 !v | {To 2 MD; : 1<" Dm WUrJa Tfun X0 Ke Sa 0 2



Answers

Determine whether the f and g are the inverses

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.

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.

The problem we're being asked to determine if the two functions are in verses of each other. We'll remember in order for them to be in verses of each other. When we find their compositions, both F of G of X and G of F of X, it has to equal to X. So let's first start by finding F of G of X. Well, remember you always start with the inside function which in this case is G of X. So that means we're now going to have F of G of X function which is six over X. So now to find F of six over racks, we need to substitute six over X in place of X in R F function. Well, that would give us six divided by six over rocks. Well, in this case we're dividing fractions. Remember to divide fractions, we multiply by the reciprocal. So in other words we have six getting multiplied by X over six. Well, in this case the 60s are gonna cancel each other out and were simply left with X. So now we show the F of G of X is equal to X. So if they are in verses of each other. Now we have to find G of F of X. See if that also equals X. Well, F of X, that's our inside function which if you recall is six over X. So that's going to give us G of six over X. So now just to simplify G of six over X, what we need to do is substitute six over X. In place of X and G of X function. So that will give us six divided by six over X. Well just like with the previous example to divide fractions, we multiply by the reciprocal. So we have six times X over six which we already established to be X. So now we know that F of G of X is X and G of F of X is also X. So because both of these compositions equal to X, they are in fact in verses of each other. Okay, so now let's move on to part B. So in part B we're given F of X is equal to negative X plus four And that G of X Let's put the four here and that G of X is equal to X plus four. So again we're going to first start by finding F of G of X. So remember we start with the inside function which is G of X, which is X plus four. So that means we're now going to have F of X plus four. So now to find F of X plus four we need to substitute X plus four in place of X in R. F of X function. So that means y minus X plus four plus four. And now we just need to simplify, so first we'll distribute the negative sign so they'll give us negative X minus four plus four and negative four plus 40. So they cancel each other up. So we're just left with negative X. So that's F of G of X. Now we have to find gff a bex. So remember we start with our inside function which is F. Of X. Which is so that means we're gonna have G of negative X plus four. And then to find geo negative X plus four we substitute negative X plus four in place of X. And our G function. So that means we're gonna have negative X plus four plus four. Well now we'll combine the like terms four plus four is equal to eight. So off with negative X plus C. And we can't simplify any further. So now we have F of G of X and G of F. Of X. Because they are not both equal to X. That means these two functions are not in verses of each other.


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