(related to Lemma: Kernel and Image of Group Homomorphism)

- By hypothesis, \((G,\ast)\) and \((H,\cdot)\) are groups with the respective identities \(e_G\) and \(e_H\) and \(f:G\rightarrow H\) is a group homomorphism.
- We first show that \(ker(f)=\{e_G\}\) \(\Longleftrightarrow f\) is injective.
- "$\Rightarrow$" * Assume \(\ker(f)=\{e_G\}\). * This means that the fiber of \(e_H\) under \(f\) equals \(\{e_G\}\), i.e. \(f^{-1}(e_H)=e_G\). * This means that \(f(e_G)=e_H\). * Because \(f\) is a group homomorphism, it follows that \[\begin{array}{cl} f(x)=f(x\ast e_G)=e_H\cdot x=x\\ f(y)=f(y\ast e_G)=e_H\cdot y=y\\ \end{array}\] for any two elements \(x,y\in H\). * Therefore, from \(f(x)=f(y)\) it follows that \(x=y\), so \(f\) is injective.
- "$\Leftarrow$" * Assume \(f\) is injective. * This means that from \(f(x)=f(y)\) it follows that \(x=y\). * Because \(f\) is a group homomorphism, we have \(f(x)\cdot(f(y))^{-1}=e_H\) and it follows further from its properties that \(f(x)\cdot f(y^{-1})=e_H\). * Applying the homomorphism property again, we get \(f(x\ast y^{-1})=e_H\). * Now, because \(x=y\), we have \(f(e_G)=e_H\), i.e. \(f^{-1}(e_H)=e_G\), thus \(ker(f)=\{e_G\}\).

- We now show that \(\operatorname{im}(f)=H\) \(\Longleftrightarrow f\) is surjective.
- We just have to apply the definition of surjectivity.∎

- We just have to apply the definition of surjectivity.

**Kramer Jürg, von Pippich, Anna-Maria**: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013