Elements of abstract algebra pdf

Get Cumulative Book Index Books now! Elements of Abstract and Linear Algebra Paley. Copyright Office. Elementary Matrix Algebra by Franz E. Abstract and Linear Algebra by David M. Catalog of Copyright Entries by Library of Congress. The American Mathematical Monthly by Anonim. I hope that students who use this book will obtain a solid understanding of the basic concepts of abstract algebra through doing problems, the best way to un- derstand this challenging subject.

So often I have encountered students who memorize a theorem without the ability to apply that theorem to a given problem. Therefore, my goal is to provide students with an array of the most typical problems in basic abstract algebra. At the beginning of each chapter, I state many of the major results in Group and Ring Theory, followed by problems and solutions.

I do not claim that the so- lutions in this book are the shortest or the easiest; instead each is based on certain well-known results in the field of abstract algebra. If you wish to comment on the contents of this book, please email your thoughts to abadawi aus. Publishers for their superb assistance in this book. It was a pleasure working with them. Ord a indicates the order of a in a group.

Let H be a subgroup of a group G. Let a be an element in a group G. Badawi C indicates the set of all complex numbers. Z indicates the set of all integers. Q indicates the set of all rational numbers. If A is a square matrix, then det A indicates the determinant of A. Aut G indicates the set of all isomorphisms automorphisms from G onto G. Sn indicates the group of all permutations on a finite set with n elements.

Then either p divides n or p divides m. If n divides c and m divides c, then nm divides c. Then H is a subgroup of G if and only if H is closed. If a has an infinite order, then all distinct powers of a are distinct elements. If n divides k and m divides k, then lcm n,m divides k. Then Ord H divides Ord G. Then Ord a divides Ord G. Then G contains an element of order p.

Then : 1. If pk divides Ord G , then G has a subgroup of order pk. Then any two Sylow-p- subgroups of G are conjugate, i. Moreover, the factorization is unique except for rearrangement of the factors. Then for each positive divisor k of n, there is a subgroup of G of order k. We say that a group is simple if its only normal subgroups are the identity subgroup and the group itself.

Then G is isomorphic to a subgroup of An. In particular, every normal subgroup of G of order p is contained in Z G the center of G. Chapter 2 Problems in Group Theory 2. We use math. Thus, m divides n by Theorem 1. Thus, n divides m. Badawi Solution: Deny. Thus, our denial is invalid. Thus, by Theorem 1.

Prove that G is Abelian. Now, let a and b be elements in G. Hence, n di- vides im again by Theorem 1. Find Ord a6 and Ord a Hence, c divides nm by Theorem 1. Hence, n divides mc. Solution: Let a and b be non identity elements of G. Then e, a, b,ab,and ba are elements of G. Thus, G is Abelian. A contradiction since nm is not prime.

Thus, every non identity element of G has the same order. Show that Ord g divides n. Solution: Deny. Hence, our denial is invalid. Then H and D are subgroups of Z. Then by the previous question, C is never a group since it is not closed. Solution: Let a and b be elements in C. Hence, C is a subgroup of G by Theorem 1. Prove that D is a subgroup of G. Thus, D is a subgroup of G by Theorem 1.

Prove that Z G is a subgroup of G. Thus, Z G is a subgroup of G by Theorem 1. Prove that C a is a subgroup of G. Hence, C a is a subgroup of G by Theorem 1. Using a similar argument as in Questions 2. Prove that N H is a subgroup of G. Solution: First, observe that H is a finite set with exactly elements. Thus, H is closed.

Prove that H is a subgroup of GL , Z Solution: First observe that H is a finite set. Hence, H is closed. How many distinct subgroups of order 7 does G have? Now, by Question 2. Hence, each subgroup of G of order 7 contains exactly 6 distinct elements of order 7. Prove that H is a subgroup of U Solution: Observe that H is a finite set. Thus, H is a subgroup of G by Theorem 1. Prove that H is a subgroup of G.

Since 13 is prime, 1 and 13 are the only divisors of Thus, 1,3,5,7,9,11,13,15,17,19,21 are all generators of Z List all generators for the subgroup of order 8. Solution: Let H be the subgroup of G of order 8.

How many subgroups does G have? Solution: Since for each positive divisor k of 48 there is a unique subgroup of order k by Theorem 1. Solution: Since a is cyclic and H is a subgroup of a , H is cyclic by Theorem 1.

By Theorem 1. Thus, i divides j and k divides j. Hence, lcm i,k divides j by Theorem 1. Thus, Ord a divides nm by Theorem 1. Hence, Ord a is finite, a contradiction. Hence, Our denial is invalid. Therefore, Ord am is infinite. Solution: Since G is cyclic, H is cyclic by Theorem 1.

Hence, k divides gcd n,m. Find the smallest subgroup of G containing a8 and a Prove that e is the only element in G of finite order. Hence, Ord a divides km by Theorem 1. Thus, e is the only element in G of finite order. Prove that G is a finite group. Solution: First, observe that H is cyclic by Theorem 1. Hence, Ord a divides nm by Theorem 1. Prove that G is never cyclic. Then G is cyclic.

Hence, a is a finite subgroup of G. Thus, G must be finite by the previous Question. Hence, by Theorem 1. A contradiction. Hence, G is never cyclic. Then H is a finite subgroup of G of order 4. It is easy to see that G is a noncyclic Abelian group. Prove that a , a2 , a3 , Prove that G has in- finitely many proper subgroups. Badawi Solution:Deny. Then G has finitely many proper subgroups. Also, by the previous Question, each element of G is of finite order. How many elements of order 16 does G have?

Solution: Since 16 divides Ord G , G is a finite group. Hence, by The- orem 1. Hence, By Theorem 1. Since each 3-cycle is an even cycle by the previous problem, a permutation that is a product of 3-cycles must be an even permutation. Prove that Sm contains a subgroup of order n. Prove that H is a subgroup of Sn. Solution: Let H be the subgroup of Sn described in the previous Question.

Thus by Question 2. Find all left cosets of H in G. Find all left cosets of a4 in a. What are the possible orders for the subgroups of G. Solution: Write 24 as product of distinct primes. Hence, We need only to find all divisors of Hence, possible orders for the subgroups of G are : 1,3,2,4,8,6,12, Prove that every proper subgroup of G is cyclic.

Solution: Let H be a proper subgroup of G. Then Ord H must divide pq by Theorem 1. Since H is proper, the possible orders for H are : 1, p,q. Then Ord h divide Ord H by Theorem 1.

Then by a similar argument as before, we conclude that H is cyclic. Hence, every proper subgroup of G is cyclic. Find in U Hence, 2 divides Ord G by Theorem 1. A contradiction since 2 is an even integer and Ord G is an odd integer. Prove that the product of all elements of G is the identity. Solution: By the previous Question, G does not have a non identity element that is the inverse of itself,i.

Hence, e, a1 aa 1a2 a2 a3 a Thus, Ord b divides 2Ord a. Solution:Since p divides the order of G, G contains an element a of order p by Theorem 1. Hence, H and bH are the only left cosets of H in G. Hence, G is Abelian. Thus, Ord b2 must be 1. Thus, each element of G that lies outside H is of order 2. Prove that G contains exactly 13 elements of order 2. Prove that G is cyclic. Badawi Solution: Since p divides Ord G and q divides Ord G , G contains an element, say, a, of order p and it contains an element, say,b, of order q.

But we know that S3 is not Abelian and hence it is not cyclic. Then G is a non-Abelian group of order 6. Let H be a proper subgroup of G. Hence, by the previous Question H is cyclic. This edition doesn't have a description yet. Can you add one? Add another edition? Copy and paste this code into your Wikipedia page.

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