5

Let G = {( :) ez} Notice that G is a group under matrix multiplication (you do not have to prove this) Prove that the map $ : G + Z defined via a isan isomorphism ...

Question

Let G = {( :) ez} Notice that G is a group under matrix multiplication (you do not have to prove this) Prove that the map $ : G + Z defined via a isan isomorphism (Z is under addition):

Let G = {( :) ez} Notice that G is a group under matrix multiplication (you do not have to prove this) Prove that the map $ : G + Z defined via a isan isomorphism (Z is under addition):



Answers

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

And it is given to us that and uh order of a. Yes I mean to him and or no BNC. It's and in two. Okay we know that but tradition. Mhm. We'll see what oh same order only. So after leading the N. C. We get the orders that is and into No we have to multiply with the the so you can see that this product is possible. Yes. No number Oh all of in a is it close to No not at all go in the floor here. Number of problem in A and and I'm gonna going to be placido and the product is possible. And after making the product can see that in order that was a family too well and then into a after multiplying them we get order of them into okay so we can say that kind of no order at all angel, let's see. Mhm. And growth. No we should move to the right side. In the right side. There is Eddie lot it's we can see that order eight I'm into an order of these and into the similarly order of age. I mean well order of these and into after making the product we can see that we remain with I'm going to be and we have to add yeah the product O. N. C. And it will be also I mean to and all that and due to this event on order after adding them the guilt again. I'm sorry I'm going to be. I mean so we can see that the orders in both the case here also it is I'm going to be and he had also did them into so we can say came to be. Let's see it, baby. Plus it's okay. Thank you.

And it is given to us that and uh order of a. Yes I mean to him and or no BNC. It's and in two. Okay we know that but tradition. Mhm. We'll see what oh same order only. So after leading the N. C. We get the orders that is and into No we have to multiply with the the so you can see that this product is possible. Yes. No number Oh all of in a is it close to No not at all go in the floor here. Number of problem in A and and I'm gonna going to be placido and the product is possible. And after making the product can see that in order that was a family too well and then into a after multiplying them we get order of them into okay so we can say that kind of no order at all angel, let's see. Mhm. And growth. No we should move to the right side. In the right side. There is Eddie lot it's we can see that order eight I'm into an order of these and into the similarly order of age. I mean well order of these and into after making the product we can see that we remain with I'm going to be and we have to add yeah the product O. N. C. And it will be also I mean to and all that and due to this event on order after adding them the guilt again. I'm sorry I'm going to be. I mean so we can see that the orders in both the case here also it is I'm going to be and he had also did them into so we can say came to be. Let's see it, baby. Plus it's okay. Thank you.

In this exercise, we're asked to find all subgroups of Paris to find all subgroups of the group Z nine. And of the secret groups 18 to do that. It's important to remember one theory. All subgroups of a secret group are sickly. That's one. And the order of the subgroups our divers are defenders of the earth as a group and there is only one supper for each division. So let's start with Z nine. Z nine is a secular group and The only dessert of nine is 3. There was the only difference of 9, 1 and three. Therefore there are subgroups A further nine and 3 Of Under 1. 3. Yeah. Uh huh. Moreover, they're both cyclic and there is only one sub group for each user. Therefore There's a lot 49 look like reunited. It's the next supper Uh says the nine. It seems that the nine is generated by X. The next separable T is a Murphy to 33. And it will be generated by excuse. And last one is a trivial group generated by the identity. Oh, now let's move on to Z to finance all subgroups of CNT. The users are one to divide it. Three divides 18, 16 points 18, 1918. And that's all. Oh, therefore there will be some groups of others 12369 and one of the one subgroup for charter and they're all going to be sick. So first we start with Z 18 And assumed the 18th generated by X. So from C 18 we will have to suburbs. One generated by that's cute. That's going to be the suburb of order six and the other generated by X squared is going to be subgroup of all your life. Then they both have the six. We'll have. Is he too is a suburb generated by X. Does that mean? And Z three generated by extra six and jeanine will also contain the three is sub because three D. White snake, but it won't contain Z. two is a suburb because two does not divide And both the two and the three will contain the identity group. This up.

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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