Homework 15

Read: Continue with Section 4.1.

Turn in: 4.5, 4.8, 4.10, 4.12(f,g,k,l), 4.14, 4.17

• 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 $m\in \mathbb{Z}$, which you can use to prove that $G$ is abelian.

• 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 $G$. Now, if $G = {e}$, then $G = \langle e \rangle$. Thus, we can now focus on the case when $G \neq \langle e \rangle$, so we know there exists some $g\in G$ with $g\neq e$. What can you say about $g$ given the hypothesis that $G$ has no proper nontrivial subgroups?

• When thinking about 4.17, make sure to read the hint in the footnote. Remember that $\langle g\rangle = \{\ldots,g^{-2},g^{-1},e,g,g^2,\ldots\}$, so if $\langle g\rangle$ is finite, there must be repetitions in the list. For the reverse direction, assume that $g^n = e$, and try to use this to prove that for any $m\in \mathbb{Z}$, $g^m$ is equal to one of $e,g,g^2,\ldots,g^{n-1}$. This would show that $\langle g\rangle \subseteq \{e,g,g^2,\ldots,g^{n-1}\}$, implying that $\langle g\rangle$ is finite. The Division Algorithm (Theorem 4.16) is very helpful: notice that $g^{nq +r} = (g^{n})^qg^r$.

Extra practice: 4.6, 4.7, 4.9, 4.11, 4.12(all parts you didn’t turn in), 4.13