Repaso de estructuras algebraicas. Herstein Algebra Abstracta. Algebra moderna: grupos, anillos, campos, teoria de Galois. Trillas, — Algebra — pages.

Author:Kazijin Kigazshura
Language:English (Spanish)
Published (Last):5 August 2009
PDF File Size:1.22 Mb
ePub File Size:7.54 Mb
Price:Free* [*Free Regsitration Required]

Learn more about Scribd Membership Home. Much more than documents. Discover everything Scribd has to offer, including books and audiobooks from major publishers.

Start Free Trial Cancel anytime. Solucionario Herstein. Uploaded by Juan Camilo Cala. Document Information click to expand document information Description: Soluciones a los ejercicios del libro Algebra Abstracta de Herstein Manual solutions for Abstract Algebra by Herstein. Date uploaded Jan 04, Did you find this document useful? Is this content inappropriate? Report this Document. Flag for Inappropriate Content.

Download Now. Related titles. Carousel Previous Carousel Next. Jump to Page. Search inside document. The chiet thing is that you take a stab at a good eross-saction of thom, In tis addendum to the book you wil ind solutions io many of the problems. Roughyy one-all of the so-called Easier Problems are solved hora; a slighty highor tracton ot tho Middlo-Lovel Problems also have their solution gwen.

The choice was made on the basis of how instructive of usolul a given solution would be for the student who tempted a given problom and di not succood in coving it Even if you do soe a particular ono, it might bo of interst to you to compare your approach and mine to that exercise. This is deliberate on my part The ability to. Solutions fo some problems depend on those of other ones. There is often some cross-reference Indicated inthe statement ofthe problem.

Some problems come a litle before the required material needed to handle them has been ciscussed. This serves two purposes. First it gives you a chance to develop some ofthe neecied notions on your own. Easier Problems. We thus get the required equality of these two sets, They are: S, 1 , 2 , 3 , 4 , 1,,, 41, 2,3 , 2,4 , 3,4 , 1,2,5 ,1,2,4 , 1,3,4 , 2,3,4 , and the empty set Middle-Level Problems. This finishes the proof. The formal proof can be carried out after we have studied induction in Section 6.

Let Abe the set of Americans that have gone to high-school and let B be the set of all Americans that read a daily newspaper.

Then AuB ' consists of at,most all, Americans. On the other hand, given a 5S-digit binary number It 1s the number assigned to some subset of S.

For instance, would come from the subset a,,a5,a5. Since the number of S-digit binary numbers is 32 there are 52 subsets in S. Of course we could solve s the problem by enumerating all 32 subsets of S D. If isa mapping of S onto T then 7! Furthermore, for this x andy, fa. To show that f defines a function we must show that to every s Ins, f assigns a unique element of T. Suppose that the elements of T are ty We are considering the elements f'n!

Also, fh! Finally,since we are talking about the product of mappings, the product 1s associative. Thus G is a group. Therefore a to some positive power falls inH. As is immediate, if x and y are in H then xy and x"! Thus G is cyclic with a as generator Middle-Level Problems. By the result of Problem 14 any element of G other than e generates G. Thus 6 consists of the Pdistinct elements e, a, a2,.. We claim that p is a prime. Since S is finite A S is a finite group.

Thus AB is a subgroup of 6. We saw in Problem 19 that AB is a subgroup of G. How can AB -[ab a. But this implies that ap7! This solves Problem Note that the argument used for [ABI did not depend on G being abelian. HCxy and as x runs over 6 with y fixed then xy runs over all the elements of 6. Harder Problems. Let Se the set of all the integers and let x be the set of positive Integers. Thus, given s in S there is one and only one S,.

Suppose every left coset of H in G is aright coset of H in G. So, even for infinite Groups there is a correspondence of M onto N; for finite groups this translates into: Ml and N have the same number of elements.

Thus there are the same number of left cosets of H in G as there are right cosets of H inG The elements of Ug are 1 , 51, 71, , , Thus the only terms remaining uncancelled in ay Recall the basic multiplication rule: T. So there 1s no such a. Middle-Level Problems. Thus a! So w ts a monomorphism of G into A G.

It is because 2? Thus G is a subgroup of G. Computing, fg! Hence H is 2 normal subgroup of G. We already know that is a homomorphism of G into itself by e of Problem 1. This happens if and only if m and n are relatively prime 16, This problem occurred as Problem 28 in Section 3; see the solution there.

Lets, t, and v be 3 distinct elements of S. Also fey,e9 , where e, is the unit element of Gand e, that of Go ts the unit element of G since gy. The associative law easily checks out as a consequence of the fact that the associative law holdsin G, and Go, Thus G fs a group b.

Trivially the similar argument works for Gy. If is an automorphism of 6 then, since M CM. Thus MN is 2 characteristic subgroup of G. See Problem 16 to see that HH is a subgroup of G; to see why it nas order p? Thus H is a characteristic subgroup of 6. Ma CM for all a in. Thus M is normal in G. Then Let G be a non-abelian group of order 6. If every element were of order 2 then, by Problem 9 of Section I, G would be abelian.

Also, if there were an element of order 6 in G then G would be cyclic. Since a? Suppose that H is a subgroup of G, IG! Thus the mapping w of Problem 38 cannot be an isomorphism. Hence H is normal in G. If G ts cyclic then we are done, for then G Is abelian. See Problem 29 of Section 4 for the kind of argument needed for this last step Since a"!

The outcome of all this 15 that 1 mod p.


Abstract Algebra I. N. Herstein (solution).



Solucionario Herstein




Related Articles