Randall R. Holmes's Abstract Algebra I PDF

By Randall R. Holmes

Show description

Read or Download Abstract Algebra I PDF

Similar abstract books

Download e-book for iPad: Cohomology of finite groups by Alejandro Adem

Adem A. , Milgram R. J. Cohomology of finite teams (Springer, 1994)(ISBN 354057025X)

Download e-book for kindle: Syzygies and Homotopy Theory by F.E.A. Johnson

An important invariant of a topological area is its basic workforce. whilst this is often trivial, the ensuing homotopy thought is easily researched and customary. within the basic case, even if, homotopy idea over nontrivial primary teams is far extra difficult and much much less good understood. Syzygies and Homotopy idea explores the matter of nonsimply attached homotopy within the first nontrivial circumstances and provides, for the 1st time, a scientific rehabilitation of Hilbert's approach to syzygies within the context of non-simply attached homotopy idea.

Extra info for Abstract Algebra I

Sample text

32 G1 × G2 is the direct product of the groups G1 and G2 . , the binary operations are both +), then the direct product is called the direct sum and it is denoted G1 ⊕ G2 . In this case, the operation is denoted + and it is called componentwise addition: (x1 , x2 ) + (y1 , y2 ) = (x1 + y1 , x2 + y2 ). 10 Isomorphism Consider the group Z = {. . , −2, −1, 0, 1, 2, . . } under addition. If we create a new group Z = {. . , −2, −1, 0, 1, 2, . . } by putting bars over each element of Z and by using a binary operation that acts just like + acts on the unadorned integers (so, for instance, 2 3 = 5), then the group (Z, ) is essentially the same as the original group (Z, +).

The center of G, denoted Z(G), is the set of those elements of G that commute with every element of G: Z(G) = {z ∈ G | zx = xz for all x ∈ G}. Prove that Z(G) is a subgroup of G. 5–2 Fix n ∈ N. Let SLn (R) be the set of all n × n matrices over R having determinant 1: SLn (R) = {A ∈ Matn (R) | det(A) = 1}. Prove that SLn (R) is a subgroup of GLn (R) (= invertible n × n matrices over R). ) Hint: From linear algebra, we know that a square matrix is invertible if and only if its determinant is nonzero.

Then m = km and n = kn for some integers m and n . We have km n (a, b) = (km n a, km n b) = (n ma, m nb) = (0, 0). Since (a, b) has order mn and km n > 0 it follows that mn ≤ km n . So mn ≤ km n ≤ km kn = mn. Since the ends of this string are the same, the intermediate ≤ signs must actually be equalities. Thus km n = km kn which implies k = 1 as desired. (⇐) Assume that gcd(m, n) = 1. Let k be the order of the element (1, 1) of Zm ⊕ Zn . Then (k1, k1) = k(1, 1) = (0, 0). 3, m divides k and n divides k.

Download PDF sample

Abstract Algebra I by Randall R. Holmes


by Charles
4.5

Rated 4.95 of 5 – based on 9 votes