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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

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