algebraHWs

algebraHWs - e is a left identity too and hence unique...

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In the following problems, G is a nonempty set with an associative binary operation · (a semigroup), that is, · : G × G G satisfies ( a, b, c G )( a · b ) · c = a · ( b · c ). As customary, we will write ab for a · b . All quantifiers below refer to the universe G , that is, we simply write ( x ) and ( x ) for ( x G ) and ( x G ). Recall that ( G, · ) is a group if the following two additional properties hold: (ii) G contains an identity [for · ] : ( e )( g ) ge = g = eg [two-sided identity], (iii) every element of G has an inverse: ( g )( h ) gh = e = hg [two-sided inverse]. Problem 1. Prove that right identity and right inverses are sufficient. More precisely, if there exists an element e such that ( g ) ge = g and ( g )( h ) gh = e , then ( G, · ) is a group, that is, ( g ) eg = g and ( g )( h ) hg = e also hold. [Hint: Firstly, left multiply gh = e with h to show that a right inverse is a left inverse too. Then, left-multiply gh = e with e to show that
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Unformatted text preview: e is a left identity too, and hence unique.] Clearly, all linear equations are solvable in a group: ( ∀ a, b )( ∃ x ) ax = b and ( ∀ a, b )( ∃ y ) ya = b . The following problem states the converse. Problem 2. Show that if all linear equations are solvable in G then ( G, · ) is a group: If (iv) ( ∀ a, b )( ∃ x ) ax = b , and (v) ( ∀ a, b )( ∃ y ) ya = b , then ( G, · ) is a group. While the one-sided versions of (ii) and (iii) are sufficient to guarantee that G is a group under · , the one-sided condition (iv) alone - without the matching (v) - is not sufficient: Problem 3. Find a set G with an associative binary operation · : G × G → G such that the operation · satisfies (iv) yet ( G, · ) is not a group....
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