Question

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

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

Prove that G is a group of order 2n under matrix multiplication.

Short answer

It is proved that G is a group of order 2n under matrix multiplication.

Step-by-step solution

To show  is a group

Let $n\ge 3$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 $a\in {\mathrm{\mathbb{Z}}}_n$.

Here, G is a subset of the group $GL\left(2,{\mathrm{\mathbb{Z}}}_n\right)$containing the identity matrix

$\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$.

Then, $\left(\begin{array}{cc}{\left(-1\right)}^i& a\\ 0& 1\end{array}\right)\left(\begin{array}{cc}{\left(-1\right)}^i& b\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}{\left(-1\right)}^{i+j}& a+{\left(-1\right)}^{i}b\\ 0& 1\end{array}\right)$.

Hence, this subset is closed under multiplication.

Now, ${\left(\begin{array}{cc}{\left(-1\right)}^i& a\\ 0& 1\end{array}\right)}^{-1}=\left\{\begin{array}{l}\left(\begin{array}{cc}1& -1\\ 0& 1\end{array}\right)2\overline{)i}\\ \left(\begin{array}{cc}-1& a\\ 0& 1\end{array}\right)2\overline{)i}\end{array}\right.$

Hence. the subset is closed under inverse.

Hence, G is a subgroup and thus, in particular, a group.

To prove the order of G is 2n

Since the order of ${\mathrm{\mathbb{Z}}}_n$is n with two matrices $\left(\begin{array}{cc}1& a\\ 0& 1\end{array}\right)or\left(\begin{array}{cc}-1& a\\ 0& 1\end{array}\right)$with $a\in {\mathrm{\mathbb{Z}}}_n$,

it implies that $\left|G\right|=2n$.

Hence, G is a group of order 2n under matrix multiplication.