Let f : G → G0 be a group epimorphism, and let H be the normal subgroup that be the Kernal of the epimorphism. Then, prove that G0 is isomorphic to G/H.
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By the universal property of a quotient, there is a natural homorphism f : G/H G 0 .
Suppose that x is in the kernel of f. Then x has the form gH and by definition of f, f(x) = φ(g). Thus g is in the kernel of φ and so g ∈ H. In this case x = H, the identity of G/H. So the kernel of f is trivial and f is injective. Hence f is an isomorphism.