Question

Write out the part of the proof of Theorem 9.21 showing that f is injective, including the reason for each step. Your answer should begin like this:

$f\left(Cy\right)=f\left(Cx\right)\Rightarrow y^{-1}ay=x^{-1}ax$ [definition of f]

$\Rightarrow a=y{x}^{-1}ax{y}^{-1}$[ Left multiply by y and right multiply by $y^{\mathbf{-}\mathbf{1}}$]

Short answer

It is proved that f in injective

Step-by-step solution

Injective function

Let A be function $f:A\to B$ is said to be injective (or one-to-one) if for all

$x_1,x_2\in A,f\left(x_1\right)=f\left(x_2\right)$implies$x_1=x_2$

Proving that that f is injective

Therefore we have to show,

$f\left(Cy\right)=f\left(Cx\right)\Rightarrow Cy=Cx$

Let, $f\left(Cy\right)=f\left(Cx\right)\Rightarrow y^{-1}ay=x^{-1}ax$ [definition of f]

$\Rightarrow a=y{x}^{-1}ax{y}^{-1}$[ Left multiply by y and right multiply by $y^{-1}$]

$\Rightarrow a={\left(x{y}^{-1}\right)}^{-1}a{\left(xy\right)}^{-1}$ [Corollary 7.6]

$\Rightarrow {\left(xy\right)}^{-1}a=a{\left(xy\right)}^{-1}$ [left multiply by ]

$\Rightarrow {\left(xy\right)}^{-1}\in C$ [ Definition of C]

$$\begin{gathered} \Rightarrow Cy=Cx \\ \Rightarrow f\left(Cy\right)=f\left(Cx\right)\Rightarrow Cy=Cx \end{gathered}$$

Hence f is injective.