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    Epimorphism proof

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    Prove: A group G is the union of 3 proper subgroups iff there exists an epimorphism between G and the Klein 4-group.

    © BrainMass Inc. brainmass.com October 9, 2019, 10:32 pm ad1c9bdddf
    https://brainmass.com/math/discrete-math/existence-epimorphism-g-klein-4-group-223666

    Solution Preview

    Equivalently, a group is a union of three proper subgroups iff it has a quotient isomorphic to C_2 x C_2.

    Observe that C_2 x C_2 is already a union of three of its proper subgroups, namely:

    H_1 = <(1, 0)>

    H_2 = <(0, 1)>

    H_3 = <(1, 1)>;

    so if there is a surjective homomorphism f : G --> C_2 x C_2, we have

    G = f^{-1} (C_2 x C_2) = f^{-1} (H_1) U f^{-1}(H_2) U f^{-1} (H_3)

    We prove the converse under the additional assumption that G is finite. So, assume that G can be written as
    a union of 3 proper subgroups A, B, C. Notice that any one of these subgroups cannot be contained in the union
    of the other two, for otherwise G is a union of two proper subgroups, impossible.

    Now, ...

    Solution Summary

    This provides an example of proving a union between proper subgroups.

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