*Q1* By Lagrange, every element of $G$ has order $1$ or $p.$ The identity is the only order $1$ element so every non-identity element of $G$ has order $p$/ generates $G$. *Q2* (I) Let $G$ be a group. Let $g,h\in G$ with orders $m,n$ respectively such that $\gcd(m,n)=1$. Suppose $g^r=h^s$ for some $r,s\in \mathbb{Z}.$ Taking the $n$th powers of both sides, we get $g^{nr}=h^{ns}=1_{G}.$Since $g$ has order $m,$ we must have $m\mid nr$ which gives $m\mid r$ since $m$ and $n$ are coprime. Similarly, $n\mid s$ and so $g^{r}=h^{s}=1_{G}.$ $\square$ (II) Suppose $g^ah^b=g^ch^d$ for some $a,b,c,d\in \mathbb{Z}.$ Then $g^{a-c}=h^{d-b}$ and so $m|(a-c)$ and $n|(d-b)$ i.e. $a \equiv c \pmod{m}$ and $b\equiv d \pmod{n}.$ $\square$ (III) Let $d$ be the order of $gh.$ First $(gh)^{mn}=1_{G}$ so $d \mid mn.$ Since $d$ is the order of $gh,$ $1_{G}=(gh)^d=g^dh^d$ since $G$ is abelian. Using (I), $m\mid d$ and $n \mid d$ which gives $mn \mid d$ since $m$ and $n$ are coprime. $\square$ (IV) Let $g=(1,2)$ and $h=(1,2,3)$ in $S_{3}.$ Then these have coprime orders $2$ and $3$ respectively. However, $gh=(1,2)(1,2,3)=(2,3)$ has order $2\neq6.$ $\square$ *Q3* (I) Let $p_{1}p_{2}\dots p_{m}$ be the order of $G.$ By Cauchy's theorem, there exists $g_{1},g_{2},\dots,g_{m}\in G,$ with orders $p_{1},p_{2},\dots,p_{m}$ respectively. Then using Q1, it follows inductively that $g_{1}g_{2}\dots g_{m}$ has order $p_{1}p_{2}\dots p_{m}$ and so is a generator for $G.$ (II) $S_{3}$ has order $6$ which is squarefree but is non-cyclic. Indeed every element of $S_{3}$ has order $1,2$ or $3.$ *Q4* (i)