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