Abstract Algebra/Group Theory/Homomorphism/A Homomorphism with Trivial Kernel is Injective

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Let f be a homomorphism from group G to group K. Let eK be identity of K.

 {\text{ker}}~ f = {\lbrace e_{G} \rbrace} means f is injective.


0. Choose x, y \in G such that  f(x) = f(y)

1. f(y\ast x^{-1}) = f(y) \circledast  f(x^{-1}) f is a homomorphism
2.  = f(x) \circledast f(x^{-1}) 0.
3.  = f(x \ast x^{-1}) f is a homomorphism
4.  = f(e_{G}) = e_K homomorphism maps identity to identity

5.  y\ast x^{-1} \in \text{ker}~f 1,2,3,4.
6.  y\ast x^{-1} = e_{G} given {\text{ker}}~ f = {\lbrace e_{G} \rbrace}
7.  y = x