# 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 $f(x^{-1})=f(x)^{-1}.$

Github: ### References

#### Bibliography

1. Modler, Florian; Kreh, Martin: "Tutorium Algebra", Springer Spektrum, 2013