Question

Show that $\mathbf{\{}\mathbf{\left(}\begin{array}{cc}\mathbf{a}& \mathbf{b}\\ \mathbf{-}\mathbf{b}& \mathbf{a}\end{array}\mathbf{\right)}\mathbf{|}\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{\in }\mathbf{ℝ}\mathbf{,}\mathbf{n}\mathbf{o}\mathbf{t}\mathbf{b}\mathbf{o}\mathbf{t}\mathbf{h}\mathbf{0}\mathbf{\}}$is an abelian group under

matrix multiplication.

Short answer

It is shown that $G$is an abelian group.

Step-by-step solution

Determine the conditions 

Consider,$\left[\begin{array}{cc}a& b\\ -b& a\end{array}\right],\left[\begin{array}{cc}c& d\\ -d& c\end{array}\right]\in G$.

Then define composition,

$\left[\begin{array}{cc}a& b\\ -b& a\end{array}\right]\left[\begin{array}{cc}c& d\\ -d& c\end{array}\right]=\left[\begin{array}{cc}ac-bd& ad+bc\\ -ad-bc& ac-bd\end{array}\right]$

If $ac-bd=0$and$ad+bc=0$ then,$ac=bd$. Thus second equation is equivalent to both$a(c^2+d^2)=0$ and$b(c^2+d^2)=0$ .

Suppose, $c^2+d^2>0$thus as$ℝ$ is an integral domain.

Now,$a=0$ and$b=0$ , contradiction. Thus$G$ is closed under multiplication.

$G$is associative because the product in $G$is matrix multiplication.

The identity is in$G$ ,

$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

Thus$G$ contains identity element.

Determine the is an abelian group 

Now,

${\left[\begin{array}{cc}a& b\\ -b& a\end{array}\right]}^{-1}=\frac{1}{a^2+b^2}\left[\begin{array}{cc}a& b\\ b& a\end{array}\right]$

Thus $G$is closed under inverse. Therefore,$G$ satisfied the group axioms.

The multiplication is also given as,

$\begin{array}{c}\left[\begin{array}{cc}a& b\\ -b& a\end{array}\right]\left[\begin{array}{cc}c& d\\ -d& c\end{array}\right]=\left[\begin{array}{cc}ac-bd& ad+bc\\ -ad-bc& ac-bd\end{array}\right]\\ =\left[\begin{array}{cc}ca-db& cb+da\\ -cb-da& ca-db\end{array}\right]\\ =\left[\begin{array}{cc}c& d\\ -d& c\end{array}\right]\left[\begin{array}{cc}a& b\\ -b& a\end{array}\right]\end{array}$

Therefore, $G$is an abelian group.

Hence it is proved.