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But (12) does not commute with elements of H: (12)(123) = (23) = (132)(12) = (123)2 (12). 34. Let N be a subgroup of G. The following are equivalent: (1) (2) (3) (4) (5) N G. g −1 N g ⊆ N gN g −1 ⊆ N g −1 N g = N gN g −1 = N for for for for each each each each g g g g ∈ G. ∈ G. ∈ G. ∈ G. Proof. (2) ⇐⇒ (3) and (4) ⇐⇒ (5) because g −1 runs through all elements of G as g does. (1) =⇒ (2) Let n ∈ N . We must show that g −1 ng ∈ N . We know that N g = gN , so there exists some n1 ∈ N with ng = gn1 .

Congruence modulo I is an equivalence relation. Proof. reflexive: a − a = 0 ∈ I since I is a subring. symmetric: Assume a ≡ b (mod I). Then a − b ∈ I. Since I is a subring, its additive inverse, b − a is also in I, and so b ≡ a (mod I). transitive: Assume a ≡ b (mod I) and b ≡ c (mod I). Then a − b ∈ I and b − c ∈ I, hence the sum a − c = (a − b) + (b − c) ∈ I, so a ≡ c (mod I). We use this to show that arithmetic works “modulo I”. 5. Let I be an ideal of a ring R. If a ≡ b (mod I) and c ≡ d (mod I), then (1) a + c ≡ b + d (mod I); (2) ac ≡ bd (mod I).

As for rings, isomorphism is an equivalence relation. Examples: 1. Let G be a group and g ∈ G. Define a function φ : G → G by φ(h) = ghg −1 . φ is an automorphism of G (check) of a special type called an inner automorphism. If g ∈ / Z(G), then φ is not the identity. 9 2. log : (R+ , ·) → (R, +) is an isomorphism. 3. Every ring homomorphism f : R → S induces an group homomorphism of their additive groups. Just ignore the multiplication. 4. Let G = g be a cyclic group. Define f : Z → G by f (k) = g k .

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Abstract algebra [Lecture notes] by Thomas C. Craven

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