Proof
(related to Proposition: Properties of a Group Homomorphism)
Proof of (1)
 Since \(f\) is a group homomorphism, it follows for the identity \(e_G\) of the group \((G,\ast)\) that $f(e_G)=f(e_G\ast e_G)=f(e_G)\cdot f(e_G).$
 The multiplication (inside the group \(H\)) of the equation with the element and \(f(e_G)^{1}\) gives us $f(e_G)\cdot f(e_G)^{1}=f(e_G)\cdot f(e_G)\cdot f(e_G)^{1}.$
 This can be simplified to $e_H=f(e_G)\cdot e_H=f(e_G).$
Proof of (2)
 In order to show \(f(x^{1})=f(x)^{1}\), we multiply the equation (inside the group \(H\)) with \(f(x)\), use the homomorphism property of \(f\) and the result in (1), which gives us $f(x)\cdot f(x^{1})=f(x\ast x^{1})=f(e_G)=e_H.$
 After multiplying both sides of the equation with \(f(x)^{1}\), we get $f(x)^{1}\cdot f(x)\cdot f(x^{1})=f(x)^{1}\cdot e_H.$
 This can be simplified to $e_H\cdot f(x^{1})=f(x)^{1},$ and finally to $f(x^{1})=f(x)^{1}.$
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References
Bibliography
 Modler, Florian; Kreh, Martin: "Tutorium Algebra", Springer Spektrum, 2013