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    Group Theory: Formation of a Group under an Associative Product

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    Question: Let G be a nonempty set closed under an associative product, which in addition satisfies:
    (a) There exists an e in G such that e.a = a for all a in G.
    (b) Given a in G, there exists an element y(a) in G such that y(a).a = e.
    Then G is a group under this product.

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    Let be a nonempty set closed under an associative product, which in addition satisfies:
    (a) There exists an such that for all
    (b) Given , there exists an element such that .
    Prove that must be a group under this product.

    Solution: Let be a nonempty set closed under an associative product, which in addition satisfies:
    (a) There exists an such that for all
    (b) Given , there exists an element such that .

    Since is closed under an associative product,
    Closure Property holds in
    and associative law also holds in .

    Let be any ...

    Solution Summary

    This solution provides a response which shows the formation of a group under an associative product. The solution is detailed, well presented and has been completed in a Word document file which is attached.

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