hw1sol_Math113_Fall2008 - 4.29 . Let S := { x G | x-1 6 = x...

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Homework 1, selected solutions Math 113: Introduction to Abstract Algebra (Sections 2, 4) 0.18 . We define an injective (one-to-one) map ϕ of B A onto P ( A ). Let f B A and let ϕ ( f ) := { x A | f ( x ) = 1 } . Suppose that ϕ ( f ) = ϕ ( g ). Then f ( x ) = 1 if and only if g ( x ) = 1. Because the only possible values for f ( x ) and g ( x ) are 0 and 1, we see that f ( x ) = 0 if and only if g ( x ) = 0. Consequently f ( x ) = g ( x ) for all x A so f = g and ϕ is injective. To show that ϕ is surjective (onto P ( A )), let S A , and let h : A → { 0 , 1 } be defined by h ( x ) = 1 if x S and h ( x ) = 0 otherwise. Clearly ϕ ( h ) = S , showing that ψ is indeed surjective.
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Unformatted text preview: 4.29 . Let S := { x G | x-1 6 = x } . Then S has an even number of elements because its elements can be grouped in pairs x, x-1 . Because G has an even number of elements, the number of elements in G but not in S (the set G \ S ) must be even. The set G \ S is nonempty because it contains e . Thus there is at least one element of G \ S other than e , that is, at least one element other than e that is its own inverse....
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