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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.

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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