Theorem

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.

Proof

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. $y=x$