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Determine whether O not the group G = {L,i,-1,~i} (with i=V-1 under multiplication is isomorphic to the group U(8) = {1,3,5,7} with multiplication mod &. If the...

Question

Determine whether O not the group G = {L,i,-1,~i} (with i=V-1 under multiplication is isomorphic to the group U(8) = {1,3,5,7} with multiplication mod &. If the groups are not isomorphic, give a clear reason, justifying your statements If the groups are isomorphic, define a map from one to the other and prove it is an isomorphism: The Cayley table for U(8) is given to save time computing: You do not need to prove G is & group. (10 points) 3 5 7 3 5 3 5 5 5 7 3 7 7 |5 | 3 1

Determine whether O not the group G = {L,i,-1,~i} (with i=V-1 under multiplication is isomorphic to the group U(8) = {1,3,5,7} with multiplication mod &. If the groups are not isomorphic, give a clear reason, justifying your statements If the groups are isomorphic, define a map from one to the other and prove it is an isomorphism: The Cayley table for U(8) is given to save time computing: You do not need to prove G is & group. (10 points) 3 5 7 3 5 3 5 5 5 7 3 7 7 |5 | 3 1



Answers

Let f be an ordered field and x,y,z in F. Prove that if x<0 and y<z, then xy>xz.

In the problem we have to prove the properties of vector. So here it is given to be equal to we plus we. So it's a vector. Now here we will prove these vectors so late. Whether we equals a gap plus B J cap plus c kids gap. And then if we would apply the vector with two, therefore this twice of vector V equals two. Twice in two, A cap plus B in the cap plus c K cap. Now this is the left hand side, This is the left hand side. Now when we have the excess of right hand side so here I D. S victory plus victory. So it is a cab plus B, jakob plus C K Camp plus a cap plus B. The cap plus c K cap. Now this equals two, twice off A cap plus twice of B J cap plus twice off. See get gap. Now this is equivalent to twice. That is true. It's taken out and it is a cap plus me, Jacob plus C camp. Therefore it equals two price of feed and this is equal to the right hand side. Sorry, this equals two left hand side. That is allergies. So we have proved that right inside the left hand side are equal. Hence these victories satisfying the property. Now, further we have the other problem that is part B. So in part B we have to verify the end of V. This is written as we plus the blast dot dot dot last v. 10 times. Now this is written as similarly from the solution of that is the solution of jump. That is a problem. A. We have proved that twice of equals B plus B. So here envy equals two. We plus V plus v plus data dot V and times. So we have these are all vectors. This equals two N directory. So we have proved that left hand side because the right hand side, so this is the answer to the problem.

And this problem, we're talking about the properties of sets and specifically the relations between sets. We're talking about if a set or relation is reflexive, transitive and symmetric, and of course anti symmetric. The only thing in this problem is that we're not doing this proof based, we have to do this with diagrams. So what does that mean? We're going to have to analyze this problem using matrices. So let's first write our first matrix will call this AM sabar the matrix with respect to our relation. And our matrix would look like this. Who would have these entries? 010111011 So what does this mean? Well, are is going to be reflexive if each element in the diagonal of our matrix is one. And clearly we can see that is not the case. We have this first entry as zero. So our relation is not reflexive. And now let's square our matrix. So we basically take the first matrix and multiply it by that same matrix. When we do that we would get our square matrix equals this 11113 to 1 to two. So what does this mean? Well, the zero entries and our original matrix are non zero in our square matrix. So our isn't not to reflect part of me are is not transitive and I'm moving on to symmetry. How would we do this with matrices? We would have to transpose our original matrix. So are transposed matrix which we denote by this M. R. Raise to this. Tea is the exact same as our original matrix. So we can say that this is definitely symmetric because are transposed matrix is the same as our original now because our relation is symmetric, it's probably not anti symmetric. But we can check it proof based again, this is not anti symmetric because if we have some arbitrary matrix M. Sub I J equals zero, or an arbitrary matrix M sub J I equal to zero. Those don't exist for every I not equal to J. So we're not going to see Auntie Cemetery with with respect to this relation. So I hope this problem helped you understand how we can interpret the properties of the relations on sets, specifically using diagrams and matrices, as opposed to a proof based approach.

To check this subspace. What we need to do is we need to verify three exams. The first axiom is zero, vector should be belonging to a subspace. W. I mean the set W If you won't, you too belongs to W. Then we should verify that you want to see. You too also belongs to W. What is the closure property? Third, if you belongs to W then land up. You should also belongs to W Where lambda. Is it number from the field? Okay, if all these three exams satisfied, then we call W is a subspace of the factor space week. No, these set off all environ mattresses over the field kit. All right now W is the set of all symmetric matrices. So what do you mean by a symmetric matrix? The matrix was transposed the same as it's A So that's called a symmetric matrix. All right. Now if is symmetric then that's null. Matrix belongs to W. Yes, of course. Because the non matrix transpose his narrative so yes, it's true. Second axiom some of two symmetric matters is A and B. Supposed to be a symmetric B is also symmetric. Then a place be whole transpose is equal to the property of transport. It is a transport transport. But he transposes A. Because asymmetric transpose is B. Because B symmetric diet implies a place to be. He's also symmetric so that implies equals B belonged to W. So second exam is very fine. Now coming to Torrance if is symmetric lambda E. We'll transport the property of transport is lambda and a transport is equal to the lambda. A. Because the symmetric it is lambda lambda. He will transpose islam day. So that means lambda is also symmetric so it belongs to doug. So all the three exams are verified. So yes. Set of all symmetric matrices is a subspace. The second one upper triangular mattresses. So what do you mean by an upper triangular matics? All the elements below the main diagnosed should be zeroes, for example A B C 000 123 So this is an upper triangular matrix below this. All the elements are zeros of is a pet Wrangler. He's a pet Wrangler and he's also a strangler. Then it let's be is it a veranda? Of course, yes. Because these zeros get started with the other metrics in the same place. For example 123 4045006 If you add with 178 089 00 10. So what do you guys? 000000000 So some of the upper triangular mattresses. Also an upper triangular markets. So this is very frank. Null matrix is not trying biometrics because null metrics by default, all the elements below the main diagnosis anyway, zeros. In fact all the elements are real zeros. So it belongs to going mathematics, The Time magazine. If you want to play any number with the matrix scale are multiple, zero into any number of zero. So these zeros will still be retained. So lambda you Orlando A is also an upper triangular matrix. So yes, the set of a particular mattress is a subspace. The third diagonal matrix, basically, we can just individually say that the diagonal matrices from the subspace because lower triangular mattresses also from subspace. Because what do you mean by lower strangler? All the elements about domain that loves you with the same reasoning as a crime. And what is the diagonal matrix? It is both the lower triangle ring up strangler. Since both our Wrangler and Wrangler mattresses from subspace set of diagonal mattresses. Also from subsidies. Simple reason. Right. And skill harm metrics is a special case of diagonal matrix, scalar matrix means all the diagnosed elements should be zero, sorry, All the diagnostic elements should be same and other diagnosed elements other elements should be basically zeros. So this is a special case of diarrhea magics. And since diagnosed mattresses from subspace, the set of the subset of it also forms the suspects. Alright, so this is a prison. So

Hello. Real question. Envisages when that F B and ordered field and X. So I said enough. Okay. It has also given that if X less than zero and why less than that then we need to prove that X. Y greater than access it. So let us get to hear that if access less than zero, this can be written as minus Act should be greater than zero. Okay, now here, if y is less than that so Zach minus Y should be greater than zero. Okay, no, these two have become positive quantities. Some multiplication of two positive quantities should be always positive, should always be positive. So we stretch it as minus X. Which is a positive quantity. Now into that minus Y. Which is again a positive wants to know should be positive. Let us open the bracket minus X. Z bless X. Y should be positive. Let us add except to both the sides will be having X way this is minus exceed all. It is minus except plus X. Y. And we are adding acceptable the sides greater than exit. So these two will become zero. So from here we are getting X. Y greater than X zet. So this is the thing we need to prove. Thank you.


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