*Q1* Let $\begin{pmatrix}1 & x_{1} & z_{1}\\0 &1 &y_{1} \\ 0 & 0 & 1\end{pmatrix},\begin{pmatrix}1 & x_{2} & z_{2}\\0 &1 &y_{2} \\ 0 & 0 & 1\end{pmatrix}\in G.$ Then $\begin{align} \phi\left( \begin{pmatrix}1 & x_{1} & z_{1}\\0 &1 &y_{1} \\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix}1 & x_{2} & z_{2}\\0 &1 &y_{2} \\ 0 & 0 & 1\end{pmatrix}\right)&=\phi\begin{pmatrix}1 & x_{1} + x_{2} & z_{1}+x_{1}y_{2}+z_{2}\\0 &1 &y_{1}+y_{2} \\ 0 & 0 & 1\end{pmatrix} \\ &= (x_{1}+x_{2},y_{1}+y_{2}) \\ &=(x_{1},x_{2})+(y_{1},y_{2}) \\ &=\phi\begin{pmatrix}1 & x_{1} & z_{1}\\0 &1 &y_{1} \\ 0 & 0 & 1\end{pmatrix} + \phi\begin{pmatrix}1 & x_{2} & z_{2}\\0 &1 &y_{2} \\ 0 & 0 & 1\end{pmatrix} \end{align}$That is for all $g_{1},g_{2}\in G,$ $\phi(g_{1}g_{2})=\phi(g_{1})+\phi(g_{2})$ and so $\phi$ is indeed a homomorphism. $\square$ Sufficient to show $\ker{\phi}=H$ since we know from lectures that $\text{ker}(\phi)\unlhd G.$ By definition, $\begin{align} \text{ker}(\phi) &=\{ g\in G : \phi(g)=(0,0) \} \\ &= \left\{ \begin{pmatrix}1 & 0 & z\\0 &1 &0 \\ 0 & 0 & 1\end{pmatrix} : z\in \mathbb{R} \right\} \\ &= H \tag*{$\square$} \end{align}$ Let $g=\begin{pmatrix}1 & x & z\\0 &1 &y \\ 0 & 0 & 1\end{pmatrix}\in G$ be an element of finite order $d>0.$ We get, $\begin{align} d\phi(g)&=\phi(g^d) &\text{since $\phi$ preserves powers}\\ (dx,dy)&= \phi(I_{3}) &\text{since $g$ has order $d$} \\ (dx,dy) &= (0,0) &\text{since $\phi$ preserves group identities} \\ (x,y) &= (0,0) & \text{since $d>0$} \end{align}$ Thus $g= \begin{pmatrix}1 & 0 & z\\0 &1 &0 \\ 0 & 0 & 1\end{pmatrix}.$ Note that $g^{0}=\begin{pmatrix}1 & 0 & 0\\0 &1 &0 \\ 0 & 0 & 1\end{pmatrix}.$ Suppose for some $k\in \mathbb{N},$ $g^{k}=\begin{pmatrix}1 & 0 & kz\\0 &1 &0 \\ 0 & 0 & 1\end{pmatrix}.$ Then $g^{k+1}=g g^{k}=\begin{pmatrix}1 & 0 & z\\0 &1 &0 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}1 & 0 & kz\\0 &1 &0 \\ 0 & 0 & 1\end{pmatrix}=\begin{pmatrix}1 & 0 & (k+1)z\\0 &1 &0 \\ 0 & 0 & 1\end{pmatrix}$So by mathematical induction, for all natural numbers $n,$ $g^n= \begin{pmatrix}1 & 0 & nz\\0 &1 &0 \\ 0 & 0 & 1\end{pmatrix}.$ Since $g$ has order $d,$ $I_{3}=g^{d}=\begin{pmatrix}1 & 0 & dz\\0 &1 &0 \\ 0 & 0 & 1\end{pmatrix}$which gives $dz=0$ and so $z=0$ since $d>0.$ Therefore $g=I_{3}.$ $\square$ *Q2* For all $\sigma\in A_{4},$ we must have $\sigma\in S_{4}$ since $A_{4} \subset S_{4}.$ Since $V_{4}$ is a normal subgroup of $S_{4},$ we have $\sigma V_{4}\sigma^{-1}=V_{4}$ and so $V_{4}$ is also a normal subgroup of $A_{4}.$ $\square$ By Lagrange's theorem, $\# (A_{4}/V_{4}) =\frac{\#A_{4}}{\#V_{4}}=\frac{12}{4}=3.$ Since $\sigma \not \in V_{4}$ ($V_{4}$ does not contain any $3$-cycles), we must have $\sigma V_{4}\neq V_{4}$ (we know from lectures that $\sigma V_{4}=V_{4} \iff\sigma\in V_{4}$). Suppose $\sigma=(a_{1},a_{2},a_{3}),$ then $\sigma^{2}=(a_{1},a_{3},a_{2})$ is also a $3$-cycle so the same argument gives $\sigma^{2}V_{4}\neq V_{4}.$ However, $\sigma^3 V_{4}=1V_{4}=V_{4}$ so $\sigma V_{4}$ has order $3$ (noting that $V_{4}$ is the identity of $S_{4}/V_{4}$), i.e. $\#\langle\sigma V_{4}\rangle=3$ . Also $\langle \sigma V_{4}\rangle = \{ V_{4}, \sigma V_{4}, \sigma^2 V_{4} \} \subseteq \ A_{4}/ V_{4}$ since $1,\sigma^{1},\sigma^{2}\in A_{4}.$ Since $\langle \sigma V_{4}\rangle \subseteq A_{4}/V_{4}$ and $\#\langle \sigma V_{4}\rangle=\#(A_{4}/V_{4})$, we must have $A_{4}/V_{4}=\langle \sigma V_{4}\rangle .$ $\square$ By Lagrange's theorem, $\#(S_{4}/V_{4})=\frac{\#S_{4}}{\#V_{4}}=\frac{4!}{4}=6$ for some $\sigma\in S_{4}.$ Suppose $\sigma$ is a permutation that has order $6,$ then it is either a $6$-cycle or a product of a $2$-cycle and a $3$-cycle that are disjoint. Such cycles don't exist in $S_{4}$ because it is the group of permutations of $4$ letters. It follows that $S_{4}/V_{4}$ does not contain any element of order $6$ since for all $\sigma\in S_{4},$ the order of $\sigma V_{4}$ is at most the order of $\sigma$ since $\sigma^{m}=e\implies(\sigma V_{4})^{m}=\sigma^mV_{4}=1V_{4}=V_{4}.$ Therefore, $S_{4}/V_{4}$ is indeed a non-cyclic group of order $6.$ $\square$