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 onesided versions of (ii) and (iii) are suﬃcient to guarantee that G is a group under · , the onesided 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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 Spring '08
 ctw
 Math, Algebra, Identity element, Inverse element, Monoid, Binary operation

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