Algebra, 3rd Edition by Serge Lang (auth.)

By Serge Lang (auth.)

This ebook is meant as a simple textual content for a one-year path in Algebra on the graduate point, or as an invaluable reference for mathematicians and pros who use higher-level algebra. It effectively addresses the elemental innovations of algebra. For the revised 3rd variation, the writer has additional routines and made quite a few corrections to the text.

Comments on Serge Lang's Algebra:
Lang's Algebra replaced the way in which graduate algebra is taught, keeping classical themes yet introducing language and methods of pondering from class thought and homological algebra. It has affected all next graduate-level algebra books.
April 1999 Notices of the AMS, saying that the writer was provided the Leroy P. Steele Prize for Mathematical Exposition for his many arithmetic books.

The writer has a powerful knack for featuring the real and fascinating principles 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 edition

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Example text

For each x E G, l et cx : G � G be the map such that cx (Y) = xyx- 1 • Then it is immediately verified that the association x �----+ Cx is a homomorphism G � Aut( G), and so this map gives an operation of G on itself, called conjugation. e. all x E G which commute with every element of G. Thi s kernel is called the center of G. Automorphisms of G of the form Cx a re called inner. ' To avoid confusion about the operation on the left, we d o n t write xy for cx ( y) . Sometimes, one writes X Cx -t ( y ) == X - 1 yx == y , i .

There exist integers n, r with 0 < r < a such that y = na + r. Since H is a subgroup and r = y na, we have r E H, whence r = 0, and our assertion follows. Let G be a group. 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 (i n other words, if the map f : Z __.. G such that f(n) = an is surjective). Such an ele m ent 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 subgroup of G, which is cyclic.

Since f is surjective , we also have a surjective homomorphism m Z � H. Since mZ is cyclic (generated additively by m) , it follows that H is cyclic , thus proving the proposition . We observe that two cyclic groups of the same order m are isomorphic . Indeed , if G is cyclic of order m with generator a, then we have a surjective homomorphism f : Z � G such that f(n) = a n , and if kZ is the kernel , with k positive , then we have an isomorphism Z/kZ = G , so k = m . If u : G 1 � Z/mZ and v : G 2 � Z/mZ are isomorphisms of two cyclic groups with Z/mZ , then v - 1 o u : G 1 � G 2 is an isomorphism .

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