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

By Mark Steinberger

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

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Additional resources for Algebra (Prindle, Weber and Schmidt Series in Advanced Mathematics)

Example text

CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 31 There are two subgroups associated with a homomorphism that can be useful in detecting its deviation from being an isomorphism. 8. Let f : G → G be a homomorphism. Then the kernel of f , written ker f , is given by ker f = {x ∈ G | f (x) = e}. 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.

Suppose that m and n are both exponents for an element g ∈ G. Then g = e. d. Since the order of an element is positive, g has order 1, and hence g 1 = e. Let fg : Z → G be the unique homomorphism with fg (1) = g. Then fg (n) = g n for all n ∈ Z. Thus, an integer n is in the kernel of fg if and only if g has exponent n. Also, im fg is precisely g . 2. 12. 14. Let G be a group and let g ∈ G. Let fg : Z → G be the unique homomorphism with fg (1) = g. 4 The image of fg is g , the cyclic subgroup of G generated by g.

2 The terminology that we’ve chosen here is not universal. 18. CHAPTER 2. 2. Let H ⊂ Z be a subgroup, and suppose that H = 0. Let n be the smallest positive element of H. Then H = n . Proof Since n ∈ H, n ⊂ H. Thus, it suffices to show that H ⊂ n . For m ∈ H, write m = qn + r with q and r integers and 0 ≤ r < n. Now m is in H, as is qn. Thus, m − qn = r must be in H. But r is non-negative and is less than n. Since n is the smallest positive integer in H, we must have r = 0. Thus, m = qn ∈ n . 3. Let m and n be integers.

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