*Q1* Carry out Euclid's algorithm:$\begin{align}101 &= 9\times 11 + 2 \\11 &= 5\times 2 + 1 \end{align}$rearranging we get $\begin{align} 1&=11 - 5 \times 2 \\ &=11 - 5 (101 - 9\times 11) \\ &= 46\times 11 - 5 \times 101 \end{align}$that is $46 \times 11 \equiv 1 \pmod{101}$ and so $\overline{46}$ is the inverse of $\overline{11}$ in $\mathbb{Z}/101\mathbb{Z}.$ *Q2* (i) $\rho=(1 ,2, 3,5,4)$ and $\tau=(1,3,2)(4,5).$ (ii) $\rho=(1,4)(1,5)(1,3)(1,2)$ and is even while $\tau=(1,2)(1,3)(4,5)$ is odd. *Q3* Let $\rho=(1,2,3)(4,5)$ and $\tau=(1,2,3,4).$ Then $\begin{align} \rho^{-1} &= (4,5)(1,3,2); &\tau^{-1} &=(1,4,3,2) \\ \rho\tau &= (1,3,5,4,2); &\tau \rho^{2} &= (1,4) \end{align}$ *Q4* Define $H = \left\{ \begin{pmatrix}1 & r\\0 &s \end{pmatrix} : r\in \mathbb{R}, s\in \mathbb{R}^{*}\right\}$ Clearly $I_{2}\in H.$ Let $\begin{pmatrix}1 & r_{1} \\0 & s_{1}\end{pmatrix},\begin{pmatrix}1 & r_{2} \\0 & s_{2}\end{pmatrix}\in H$ then $\begin{pmatrix}1 & r_{1} \\0 & s_{1}\end{pmatrix}\cdot\begin{pmatrix}1 & r_{2} \\0 & s_{2}\end{pmatrix}= \begin{pmatrix}1 & r_{2}+r_{1}s_{2} \\0 &s_{1}s_{2}\end{pmatrix}\in H$since for all $s_{1},s_{2}\in \mathbb{R}^{*},$ $s_{1}s_{2}\in \mathbb{R}^{*}.$ Take $\begin{pmatrix}1 & r \\0 & s\end{pmatrix}\in H$ then $\begin{pmatrix}1 & r \\0 & s\end{pmatrix}^{-1}= \begin{pmatrix}1 & -rs^{-1} \\0 & s^{-1}\end{pmatrix}\in H$since for all $s\in \mathbb{R}^*,$ by definition $s^{-1}\in \mathbb{R}^*$ $H$ is a group since it satisfies [[Two-Step Subgroup Test]]. $\square$ For example, $\begin{align} \begin{pmatrix}1 & 1 \\0 & 1\end{pmatrix}\begin{pmatrix}1 & 0 \\0 & -1\end{pmatrix}&=\begin{pmatrix}1 & -1 \\0 & -1\end{pmatrix} \\ &\neq \begin{pmatrix}1 & 1 \\0 & -1\end{pmatrix}=\begin{pmatrix}1 & 0 \\0 & -1\end{pmatrix}\begin{pmatrix}1 & 1 \\0 & 1\end{pmatrix} \end{align}$so $H$ is non-abelian. $\square$ We can check that for all $r\in \mathbb{R},$ $\begin{pmatrix}1 & r \\ 0 &-1 \end{pmatrix}\in H$ has order $2$ so there are infinitely many order $2$ elements of $H.$ $\square$ *Q5*