Algebra (Prindle, Weber and Schmidt Series in Advanced - download pdf or read online

By Mark Steinberger

ISBN-10: 0534936784

ISBN-13: 9780534936785

The purpose of this e-book is to introduce readers to algebra from some degree of view that stresses examples and class. each time attainable, the most theorems are taken care of as instruments which may be used to build and research particular sorts of teams, jewelry, fields, modules, and so forth. pattern buildings and classifications are given in either textual content and routines.

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Extra info for Algebra (Prindle, Weber and Schmidt Series in Advanced Mathematics)

Example text

O  o o  ?  oo o  o o b O a ab ab2 OOO ?  OOO  OOO ???  OOO ??  OOO?  e Here, b has order 3, while a , ab , and ab2 have order 2. The upward-slanted lines represent the inclusions of subgroups. Because the elements of monoids don’t necessarily have inverses, the definition of a submonoid will have to be different from that of a subgroup. 17. A submonoid of a monoid M is a subset which is closed under multiplication and contains the identity element. 18. 1. Show that in the real numbers R, the cyclic subgroup generated by 1 is the integers.

Recall that the image of a function f : X → Y is the subset im f ⊂ Y given by im f = {y ∈ Y | y = f (x) for some x ∈ X}. 4, shows that the kernel and image of a homomorphism f : G → G are subgroups of G and G , respectively. 9. Let f : G → G be a homomorphism. Then ker f detects the deviation of f from being injective: we have f (x) = f (x ) if and only if x−1 x ∈ ker f . In particular, f is injective if and only if the kernel of f is the trivial subgroup, e, of G. The image detects whether f is onto: it is onto if and only if im f = G .

But then q and r are positive. , n > n, which is impossible. Thus, q = 1, and hence m = n. 7. Let H be a subgroup of Z. Then there is a unique non-negative integer which generates H. 8. Let m and n be integers. We write m + n = {rm + sn | r, s ∈ Z}. 9. m + n is a subgroup of Z. Proof The inverse of rn + sm is (−r)m + (−s)n. The result follows since (rm + sn) + (r m + s n) = (r + r )m + (s + s )n ∈ m + n . In fact, it is easy to see that m + n is the subgroup of Z generated by the set {m, n}. 10.

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Algebra (Prindle, Weber and Schmidt Series in Advanced Mathematics) by Mark Steinberger

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