Question

Prove that the function $\mathit{\phi }\mathbf{:}\mathit{G}\mathbf{\to }\mathit{A}\mathbf{\left(}\mathbf{T}\mathbf{\right)}$ given by$\mathit{\phi }\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{=}{\mathit{f}}_{{\mathbf{a}}^{\mathbf{-}\mathbf{1}}}$ is a homomorphism of groups whose kernel is contained in K.

Short answer

It is proved that the function$\phi :G\to A\left(T\right)$ given by$\phi \left(a\right)=f_{a^{-1}}$ is a homomorphism of groups whose kernel is contained in K.

Step-by-step solution

Step by Step Solution Step 1: Referring to part (a) of the question 

${\mathit{f}}_{\mathbf{a}}\mathbf{\left(}\mathbf{K}\mathbf{b}\mathbf{\right)}\mathbf{=}\mathit{K}\mathit{b}\mathit{a}$

Proving that the function φ: G→ A(T) given by φ(a) = fa-1 is a homomorphism of group whose kernel is contained in K

Consider $a,b\in K$as arbitrary elements.

$\begin{array}{c}\phi (a,b)=f_{{\left(ab\right)}^{-1}}\\ =f_{b^{-1}{a}^{-1}}\\ =f_{b^{-1}}{f}_{a^{-1}}\\ =\phi \left(a\right)\phi \left(b\right)\end{array}$

Because we know that for any arbitrary $Kg\in T,$

$\begin{array}{c}{f}_{a^{-1}{b}^{-1}}\left(Kg\right)=Kg{b}^{-1}{a}^{-1}\\ =f_{a^{-1}}\left(f_{b^{-1}}Kg\right)\\ =f_{a^{-1}}{f}_{b^{-1}}\left(kg\right)\end{array}$

Therefore, it is proved that $\phi$is a homomorphism.

Let a be any arbitrary element $a\in \ker \phi$. Then,

$\begin{array}{c}\phi \left(a\right)=Kba{a}^{-1}\\ =f_a\left(Kb\right)\\ =Kb\end{array}$

Therefore, $Kba=Kb\Rightarrow Ka=K$.

For , $a\in K,\ker \phi \in K$that means $\ker \phi \subset K$.

Hence, it is proved that the function $\phi :G\to A\left(T\right)$given by $\phi \left(a\right)=f_{a^{-1}}$ is a homomorphism of group whose kernel is contained inK.