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

By Mark Steinberger

The rationale of this booklet is to introduce readers to algebra from some extent of view that stresses examples and category. every time attainable, the most theorems are taken care of as instruments that could be used to build and learn particular varieties of teams, jewelry, fields, modules, and so on. pattern structures and classifications are given in either textual content and workouts.

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1. For any two groups, G and G , there is a homomorphism f : G → G obtained by setting f (g) = e for all g ∈ G. We call this the trivial homomorphism from G to G. 2. Let H be a subgroup of the group G. Then the inclusion i : H ⊂ G is a homomorphism, because the multiplication in H is inherited from that of G. 3. Recall that the canonical map π : Z → Zn is defined by π(k) = k for all k ∈ Z. Thus, π(k + l) = k + l = k + l = π(k) + π(l), and hence π is a homomorphism. For some pairs G, G of groups, the trivial homomorphism is the only homomorphism from G to G .

Show that a group G is cyclic if and only if there is a surjective homomorphism f : Z → G. † 8. Let f : G → G be a homomorphism. (a) Let H ⊂ G be a subgroup. Define f −1 (H ) ⊂ G by f −1 (H ) = {x ∈ G | f (x) ∈ H }. Show that f −1 (H ) is a subgroup of G which contains ker f . (b) Let H ⊂ G be a subgroup. Define f (H) ⊂ G by f (H) = {f (x) | x ∈ H}. Show that f (H) is a subgroup of G . If ker f ⊂ H, show that f −1 (f (H)) = H. (c) Suppose that H ⊂ im f . Show that f (f −1 (H )) = H . Deduce that there is a one-to-one correspondence between the subgroups of im f and those subgroups of G that contain ker f .

17. Show that every submonoid of a finite group is a group. 3 The Subgroups of the Integers One of the simplest yet most powerful results in mathematics is the Euclidean Algorithm. We shall use it here to identify all subgroups of Z and to derive the properties of prime decomposition in Z. 1. (The Euclidean Algorithm2 ) Let m and n be integers, with n > 0. Then there are integers q and r, with 0 ≤ r < n, such that m = qn + r. Proof First, we assume that m ≥ 0, and argue by induction on m. If m < n, we may take q = 0 and r = m.

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