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Problem 8: Find two rings of cardinality 125 of the form Zplz]/(q(z)) that are not isomorphic to each other; and prove that they are not isomorphic...

Question

Problem 8: Find two rings of cardinality 125 of the form Zplz]/(q(z)) that are not isomorphic to each other; and prove that they are not isomorphic

Problem 8: Find two rings of cardinality 125 of the form Zplz]/(q(z)) that are not isomorphic to each other; and prove that they are not isomorphic



Answers

Find all the subgroups of (a) Z9, (b) Z18 and write down the
subgroup lattices

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.

In this problem, we have been asked to prove that the subsets of finite non empty set X are partially ordered by set inclusion. So what we need to do is prove that this relation of set inclusion is a fractional order. So, first of all we need to prove that it is reflexive. This relation will be reflexive because let us consider any subsidy of X. And we can say that he is a subset of E. It is a subset of itself and thus is related to enhance. This relation is reflexive. Next we need to show that it is anti symmetric. Now, for that let us consider any two subsets of X, A and B. And let us assume that he is related to B and B is related to A. So that means that is a subset of B, and B is a subset of A. Now if is included in B and bs included an A. Then the only possible conclusion is that the two sets A and B are equal. So, since this condition implies that is equal to be. Hence we can say that the relation is anti symmetric. And the third condition that we need to show is that it is transitive. So for that let us consider three subsets of X, A, B and C. And let us suppose that is a subset of B, and B is a subset of. See. Now, if A is included in B and bs included in C, then we can conclude that is included in C. For example, if we consider a Venn diagram, this is the set E. This is included in the set B, and that's it is included in the set C. So from this diagram we can see that the set E is included in this set. See So this is what we get. So if it is related to B and B is related to see, we get the day is related to. See. Hence it is transitive. Thus, set inclusion is reflexive, it is anti symmetric, and it is also transitive, thus it is a partial order, and thus we approve that the subsets of a given finite non implicit X are partially ordered by set inclusion.

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.


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