Math 561 H Fall 2011 Homework 3 Solutions Drew
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چکیده
Proof. Suppose that φ; Z/nZ → Z/nZ is an automorphism with φ(1) = a. By the homomorphism property we have φ(x) = φ(1 + · · · + 1) = φ(1) + · · · + φ(1) = a + · · · + a = ax. Thus the image of φ is the (additive) cyclic subgroup 〈a〉 ≤ Z/nZ. Then since φ is surjective we must have 〈a〉 = Z/nZ. By the Lemma, this happens if and only if a and n are coprime, i.e. a ∈ (Z/nZ)×, in which case the map φ(x) = ax is also invertible with inverse φ−1(x) = a−1x. In summary, there is a bijection between (Z/nZ)× and Aut(Z/nZ) given by sending a ∈ (Z/nZ)× to the automorphism φa(x) = ax. Moreover, this bijection is a group isomorphism since for all a, b ∈ (Z/nZ)× and x ∈ Z/nZ we have
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Math 561 H Fall 2011
2. Let G be a set with binary operation (a, b) 7→ ab and consider the following possible axioms: (1) ∀ a, b ∈ G, a(bc) = (ab)c. (2) ∃ e ∈ G,∀ a ∈ G, ae = ea = a. (3) ∀ a ∈ G,∃ b ∈ G, ab = ba = e. (3’) ∀ a ∈ G,∃ b ∈ G, ab = e. Prove that the axioms (3) and (3’) are equivalent. That is, show that (1), (2), and (3) hold if and only if (1), (2), and (3’) hold. (One direction is easy. For the other ...
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