Let *f* be a homomorphism from group G to group K.
Let *e*_{K} 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. $y=x$