Question

Let $\mathit{n}\mathbf{\ge }\mathbf{3}$be a positive integer and let G be the set of all matrices of the forms

$\left(\begin{array}{cc}1& a\\ 0& 1\end{array}\right)or\left(\begin{array}{cc}-1& a\\ 0& 1\end{array}\right)$ with $a\in {\mathrm{\mathbb{Z}}}_n$.

Prove that G is isomorphic to${\mathit{D}}_{\mathbf{n}}$.

Short answer

It is proved that G is isomorphic to$D_n$.

Step-by-step solution

To showφis surjective

Let us define a function $\phi :D_n\to G$such that $r^i{d}^j\to \left(\begin{array}{cc}{\left(-1\right)}^j& i\\ 0& 1\end{array}\right)$.

Now, find $\phi \left(dr\right)$as:

role="math" localid="1653053971207" $\begin{array}{rcl}\phi \left(dr\right)& =& \phi \left(r^{-1}d\right)\\ & =& \left(\begin{array}{cc}-1& -1\\ 0& 1\end{array}\right)\\ & =& \left(\begin{array}{cc}1& -1\\ 0& 1\end{array}\right)\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\\ & =& \phi \left(r^{-1}\right)\phi \left(d\right)\end{array}$

$\therefore \left(r^{-1}d\right)=\phi \left(r^{-1}\right)\phi \left(d\right)$.

Hence, this function is surjective since, for all $\left(\begin{array}{cc}{\left(-1\right)}^j& a\\ 0& 1\end{array}\right)\in G$, we have:

$\phi \left(r^a{d}^j\right)=\left(\begin{array}{cc}{\left(-1\right)}^j& a\\ 0& 1\end{array}\right)$.

To prove G is isomorphic toDn

Also, since G and $D_n$have the same cardinality, it implies that $\phi$is an isomorphism.

Hence, G is isomorphic to$D_n$.