Read: Start Section 4.1.
Turn in: 4.1, 4.2, 4.4(a,b,e), 4.5, 4.8, 4.10
For 4.5, start with “Since $G$ is cyclic, $G = \langle g \rangle$ for some $g\in G$.” This means that every $x\in G$ can be written as $x=g^m$ for some $m\in \mathbb{Z}$. Use this idea to write out expressions for arbitrary $x,y\in G$ and prove $xy=yx$.
On 4.10, the hypothesis is that $G$ has no proper nontrivial subgroups—this means that every subgroup of $G$ must be equal to either ${e}$ or $G$. First treat the case when $G = {e}$; here you have $G = \langle e \rangle$. Now you can just focus on the case when $G \neq \langle e \rangle$, which means there exists some $a\in G$ with $a\neq e$. What can you say about $\langle a \rangle$?
Extra practice: 3.66, 3.67, 4.3