By Serge Lang
This ebook is meant as a uncomplicated textual content for a one-year direction in Algebra on the graduate point, or as an invaluable reference for mathematicians and pros who use higher-level algebra. It effectively addresses the fundamental suggestions of algebra. For the revised 3rd version, the writer has further workouts and made quite a few corrections to the text.
Comments on Serge Lang's Algebra:
"Lang's Algebra replaced the best way graduate algebra is taught, conserving classical subject matters yet introducing language and methods of pondering from class concept and homological algebra. It has affected all next graduate-level algebra books."
-April 1999 Notices of the AMS, asserting that the writer used to be presented the Leroy P. Steele Prize for Mathematical Exposition for his many arithmetic books.
"The writer has a magnificent knack for proposing the $64000 and engaging rules of algebra in precisely the "right" method, and he by no means will get slowed down within the dry formalism which pervades a few elements of algebra."
-MathSciNet's evaluation of the 1st variation
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Additional resources for Algebra
We shall say that G is cyclic if there exists an element a of G such that every element x of G can be written in the form an for some n E Z (in other words, if the map f : Z -+ G such that f(n) = an is surjective). Such an element a of G is then called a generator of G. Let G be a group and a E G. The subset of all elements an (n E Z) is obviously a cyclic subgroup of G. If m is an integer such that am = e and m > 0 then we shall call m an exponent of a. m = e for all x E G. Let G be a group and a E G.
E . Since 0 ~ s - r < d we must have s - r = O. The cyclic subgroup generated by a has order d. 1. Let G be a finite group of order n > 1. Let a be an element of G , a e. Then the period of a divides n. If the order ofG is a prime number p, then G is cyclic and the period of any generator is equal to p. 2. Let G be a cyclic group . Then every subgroup ofG is cyclic. Iff is a homomorphism of G, then the image off is cyclic . Proof. If G is infinite cyclic, it is isomorphic to Z, and we determined above all subgroups of Z, finding that they are all cyclic .
B; such that B1 + ... + B; = A and Bi +. n (B. + ... + BJ = O. 38 I, §7 GROUPS In that case we write Let A be an abelian group . ] (i E l) be a family of elements of A . We say that this family is a basis for A if the family is not empty , and if every element of A has a unique expression as a linear combination with X i E Z and almost all X i = O. Thus the sum is actually a finite sum. An abelian group is said to be free ifit has a basis. If that is the case, it is immediate that if we let Z, = Z for all i, then A is isomorphic to the direct sum A"'" EBZi ' iE!