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

${\displaystyle {\text{ker}}~f={\lbrace e_{G}\rbrace }}$ means f is injective.
 0. Choose ${\displaystyle x,y\in G}$ such that ${\displaystyle f(x)=f(y)}$ 1. ${\displaystyle f(y\ast x^{-1})=f(y)\circledast f(x^{-1})}$ f is a homomorphism 2. ${\displaystyle =f(x)\circledast f(x^{-1})}$ 0. 3. ${\displaystyle =f(x\ast x^{-1})}$ f is a homomorphism 4. ${\displaystyle =f(e_{G})=e_{K}}$ homomorphism maps identity to identity 5. ${\displaystyle y\ast x^{-1}\in {\text{ker}}~f}$ 1,2,3,4. 6. ${\displaystyle y\ast x^{-1}=e_{G}}$ given ${\displaystyle {\text{ker}}~f={\lbrace e_{G}\rbrace }}$ 7. ${\displaystyle y=x}$