Question

(b) Use Theorem $\mathbf{8}\mathbf{.}\mathbf{11}$to show that $\mathit{N}$is normal in$\mathit{G}$.

Short answer

It is proved that$\mathit{N}$is normal in$\mathit{G}$.

Step-by-step solution

Assume

To prove that $\mathit{N}$is a normal subgroup, we need to show that role="math" localid="1657373957339" $\mathit{A}\mathbf{\in }\mathit{G}\mathbf{,}\mathit{A}\mathit{N}{\mathit{A}}^{\mathbf{-}\mathbf{1}}\mathbf{\subset }\mathit{N}$.

Assumerole="math" localid="1657373705255" $\mathit{A}\mathbf{=}\mathbf{\left(}\begin{array}{ll}\mathbf{a}& \mathbf{b}\\ \mathbf{0}& \mathbf{d}\end{array}\mathbf{\right)}$withrole="math" localid="1657373708742" $\mathit{a}\mathit{d}\mathbf{\ne }\mathbf{0}$, androle="math" localid="1657373713587" $\mathit{B}\mathbf{\in }\mathit{N}\mathbf{}$is in the form$\mathit{A}\mathbf{=}\mathbf{\left(}\begin{array}{ll}\mathbf{1}& \mathbf{c}\\ \mathbf{0}& \mathbf{1}\end{array}\mathbf{\right)}$.

Prove that N is normal subgroup

Determine the value of $\mathit{A}\mathit{B}{\mathit{A}}^{\mathbf{-}\mathbf{1}}$as :

localid="1657373593074" $\mathit{A}\mathit{B}{\mathit{A}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\mathbf{\left(}\begin{array}{ll}\mathbf{a}& \mathbf{b}\\ \mathbf{0}& \mathbf{d}\end{array}\mathbf{\right)}\mathbf{\left(}\begin{array}{ll}\mathbf{1}& \mathbf{c}\\ \mathbf{0}& \mathbf{1}\end{array}\mathbf{\right)}\mathbf{\left(}\begin{array}{cc}{\mathbf{a}}^{\mathbf{-}\mathbf{1}}& \mathbf{-}\mathbf{b}\\ \mathbf{0}& {\mathbf{d}}^{\mathbf{-}\mathbf{1}}\end{array}\mathbf{\right)}$

$\mathbf{=}\left(\begin{array}{cc}a& ac+b\\ 0& d\end{array}\right)\left(\begin{array}{cc}{a}^{-1}& -b\\ 0& d^{-1}\end{array}\right)$

$\mathbf{=}\left(\begin{array}{cc}1& -ab+ac{d}^{-1}+b{d}^{-1}\\ 0& 1\end{array}\right)$

This implies that localid="1657374134155" $\mathit{A}\mathit{B}{\mathit{A}}^{\mathbf{-}\mathbf{1}}\mathbf{\in }\mathit{N}$

Hence, it is proved that$\mathit{N}$is a normal subgroup.