Show that a group G is simple if and only if every nontrivial group homomorphism G -> G1 is one-to-one.
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Proof:
If G is simple, then G has no nontrivial normal subgroup. Then for any nontrivial group homomorphism f: G->G1, by the isomorphism theory, we have ...
... set of all elements of each cyclic group whose order ... be isomorphic to H, since direct products preserve isomorphism. ... Let G and H denote finite abelian groups. ...
... then H must be a normal subgroup of G. 7. Let G be an abelian group of order ... Write down all direct products of cyclic groups that could be isomorphic to G ...
Give an example of groups H_i, K_j such that H_1xH_2 is isomorphic... Let G be the additive group Q of rational numbers ... then we have , but is not isomorphic to any ...
... For example, consider Ab, the category of (additive) abelian groups. ... If Φ is a group isomorsphism, then : must be a bijective map ... Then : is an isomorphism in ...
... (c) Let G be the group of 2x2 ... d)For the two homomorphisms in (c), prove that the two quotient groups G/N, where N is the corresponding kernel, are isomorphic. ...
... in the group, not the order of the group itself. ... 2. Homomorphism: Let G and H be two groups. ... 4. Isomorphism: An isomorphism is a homomorphism that is bijective ...
... The only groups of order 6 are and . ... Then So taking . Thus and the isomorphism theorem gives us. ... 1) Let G be a finite group with n elements. ...
