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Definition: For integers x,y,n, we say x is equivalent to y modulo n if there exists an integer k such that x = y + nk. We denote this equivalence relation x ▷ y by x = y mod n. ☐ Equivalence relations are defined on a particular set and partition this set into several subsets. These subsets are mutually disjoint. If we examine one of these subsets, every element contained therein is equivalent to every other element in the subset. This is the idea of an equivalence class. Definition: Given a set S and element x ∈ S with equivalence relation ▷, the equivalence class of x is the subset of S that contains all elements of S that are equivalent to x.

We consider (x · y)2 = (x · y) · (x · y) = x · (y · x) · y = x · (x · y) · y = (x · x) · (y · y) = x2 · y 2 We are justified in making the statement x · y = y · x due to the hypothesis that G is abelian. ” It remains to show the other half of the statement. Now suppose that for all x, y ∈ G we have (xy)2 = x2 y 2 . Expanding this expression we see x · y · x · y = x · x · y · y . If we multiply on the left by x−1 and on the right by y −1 we have: x−1 · x · y · x · y · y −1 = x−1 · x · x · y · y · y −1 which simplifies to y · x = x · y .

Suppose by way of contradiction that g = g are both inverses of g . g = e · g = (g · g) · g = g · (g · g ) = g · e = g This shows that g = g , therefore inverses are unique. ☐ The previous theorem shows that for a group G and a, b, x ∈ G , the equation ax = b has a unique solution for x . Since a has a unique inverse, a−1 we may write x = a−1 ax = a−1 b . Theorem 6. If a, b are elements of a group G then (a · b)−1 = b−1 · a−1 . Proof. We prove this by using the definition of inverse. For a group G and a, b ∈ G , we consider (a · b) · (b−1 a−1 ) = a · ((b · b−1 ) · a−1 ) = a · (e · a−1 ) = a · a−1 = e A similar series of steps shows that (b−1 · a−1 ) · (a · b) = e .

### An Introduction to Abstract Algebra by Bookboon.com

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