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 ...
... 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 ...
... (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. ...
... the first isomorphism theorem: Z30/ker(phi) is isomorphic to G ... Since Z30 is a cyclic group, and ker(phi) is ... by the fundamental theorem of cyclic groups: ker(phi ...
... Some concepts of groups and rings are given. ... The group of automorphisms of a finite cyclic group is studied ... Since it is also homomorphism, it is an isomorphism. ...
... 4=id φ(b)^2=id φ(b)^(-1) φ(a)φ(b)=φ(a)^(-1). Since is isomorphic to permutation group of the four vertices of the square, the group homomorphism is just ...
