Proof: (related to "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 $f(x^{-1})=f(x)^{-1}.$

| | | | created: 2014-06-09 22:17:38 | modified: 2019-07-27 15:48:42 | by: bookofproofs | references: [677]