Question

(a) Prove that $\mathit{G}\mathbf{=}\mathbf{\{}\left(\begin{array}{ll}a& b\\ 0& d\end{array}\right)\mathbf{\mid }\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{,}\mathbf{d}\mathbf{\in }\mathbf{R}\mathbf{\text{and}}\mathbf{ad}\mathbf{\ne }\mathbf{0}\mathbf{\}}$is a group under matrix multiplication and that $\mathit{N}\mathbf{=}\left\{\left(\begin{array}{cc}1& b\\ 0& 1\end{array}\right)\mid b\in R\right\}$is a subgroup ofrole="math" localid="1657372781318" $\mathit{G}$.

Short answer

It is proved that $\mathit{G}$is a group under the matrix multiplication and$\mathit{N}$ is a subgroup of $\mathit{G}$.

Step-by-step solution

Prove that G is group under matrix multiplication

Let$\mathit{A}\mathbf{,}\mathit{B}\mathbf{\in }\mathit{G}$be two matrices, consider $\mathit{A}\mathbf{=}\mathbf{\left(}\begin{array}{ll}\mathbf{a}& \mathbf{b}\\ \mathbf{0}& \mathbf{d}\end{array}\mathbf{\right)}$,and $\mathit{B}\mathbf{=}\mathbf{\left(}\begin{array}{ll}\mathbf{e}& \mathbf{f}\\ \mathbf{0}& \mathbf{h}\end{array}\mathbf{\right)}$, where role="math" localid="1657372858846" $\mathit{a}\mathit{d}\mathbf{\ne }\mathbf{0}$and$\mathit{e}\mathit{h}\mathbf{\ne }\mathbf{0}$.

This implies that:

$\mathit{A}\mathit{B}\mathbf{=}\left(\begin{array}{ll}a& b\\ 0& d\end{array}\right)\left(\begin{array}{ll}e& f\\ 0& h\end{array}\right)$

$\mathbf{=}\left(\begin{array}{cc}ae& af+bh\\ 0& dh\end{array}\right)$

It is observed that $\mathit{a}\mathit{e}\mathit{d}\mathit{h}\mathbf{=}\mathbf{\left(}\mathit{a}\mathit{d}\mathbf{\right)}\mathbf{\left(}\mathit{e}\mathit{h}\mathbf{\right)}\mathbf{\ne }\mathbf{0}$.This implies that $\mathit{A}\mathit{B}\mathbf{\in }\mathit{G}$.

Assume that$\mathit{I}\mathbf{=}\mathbf{\left(}\begin{array}{ll}\mathbf{1}& \mathbf{0}\\ \mathbf{0}& \mathbf{1}\end{array}\mathbf{\right)}$,it is the identity element of $\mathit{G}$.Consider$\mathit{B}\mathbf{=}\left(\begin{array}{cc}{a}^{-1}& -b\\ 0& d^{-1}\end{array}\right)$.

This implies that $\mathit{A}\mathit{B}\mathbf{=}\mathit{I}\mathbf{=}\mathit{B}\mathit{A}$.

Thus, every element of$\mathit{G}$ has unique inverse. Therefore, $\mathit{G}$is a group under the matrix multiplication.

Prove that N is normal subgroup

To prove that $\mathit{N}$is a subgroup, we need to show thatlocalid="1657372452255" $\mathit{A}\mathbf{,}\mathit{B}\mathbf{\in }\mathit{N}\mathbf{,}\mathit{A}{\mathit{B}}^{\mathbf{-}\mathbf{1}}\mathbf{\in }\mathit{N}$

Assume that $\mathit{A}\mathbf{,}\mathit{B}\mathbf{\in }\mathit{N}$as :

$\mathit{A}\mathbf{=}\left(\begin{array}{ll}1& a\\ 0& 1\end{array}\right)\mathbf{,}\mathbf{\text{and}}\mathit{B}\mathbf{=}\left(\begin{array}{ll}1& b\\ 0& 1\end{array}\right)$

Therefore,

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

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

localid="1657372918067" $\mathbf{\in }\mathit{N}$

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