# Proof

• 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