# Proof

(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.

Github: ### References

#### Bibliography

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