Read: Continue with 4.1.
Turn in: 4.4(a,b,d), 4.5, 4.10, 4.12(d,f,g,l)
- On 4.5, you are assuming that \(G\) is cyclic, so \(G\) can be generated by some \(a\in G\). Thus, you can write \(G = \langle a \rangle\). To show \(G\) is abelian, take arbitrary \(x,y \in G\); the goal is to show \(xy =yx\). To get started, use that \(G = \langle a \rangle\) to first write \(x\) and \(y\) in terms of \(a\).
- 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 \(\langle a \rangle\)?