Proof
(related to Lemma: Kernel and Image of a Group Homomorphism are Subgroups)
 Given a group homomorphism \(f:(G,\ast)\mapsto (H,\cdot)\), we first show that \(\ker(f)\) is a subgroup of \(G\).
 Note that from the properties of a group homomorphism, it follows that \(f(e_G)=e_H.\)
 Thus, the kernel \(\ker(f)\) is not empty.
 Therefore assume that \(x,y\in \ker(f)\), i.e. \(f(x)=f(y)=e_H\).
 It follows $f(x\ast y^{1})=f(x)\cdot f(y^{1})=e_H\cdot (f(y))^{1}=e_H\cdot (e_H)^{1}=e_H.$
 We have found that \(x\ast y^{1}\in ker(f)\), which is a subgroup criterion.
 Thus, \(\ker(f)\) is a subgroup of \(G\).
 We now show analogously that \(\operatorname{im}(f)\) is a subgroup of \(H\).
 Again from \(e_H=f(e_G)\) it follows that the image \(\operatorname{im}(f)\) is not empty.
 Therefore assume that \(a,b\in im(f)\), i.e. that there exist some \(x,y\in G\) with \(f(x)=a\) and \(f(y)=b\).
 It follows that $a\cdot b^{1}=f(x)\cdot (f(y))^{1}=f(x)\cdot f(y^{1})=f(x\ast y^{1}).$
 We have found that \(a\cdot b^{1}\in H\) is the image of \(x\ast y^{1}\in G\), which is a subgroup criterion.
 Thus, \(\operatorname{im}(f)\) is a subgroup of \(H\).
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References
Bibliography
 Kramer Jürg, von Pippich, AnnaMaria: "Von den natürlichen Zahlen zu den Quaternionen", SpringerSpektrum, 2013