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    If G = <X> and $:G->G1 is an onto homomorphism, show that G1 = <$(X)>, where $(X) = the set of $(x) given that x belongs to X.

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    Let X={x1,...,xn}. G=<X>=<x1,...,xn>. So for any g in G, we can express g in form of x1^(k1)...xn^(kn).
    Since $ is onto homomorphism, ...

    Solution Summary

    This is a proof regarding an onto homomorphism.