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

= Theorem = 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.

= Proof =


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 * 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. $$ y = x$$
 * }
 * }