Read: Read Section 7.1.
Turn in: 6.35, 6.41, 7.11, 7.12, 7.15, 7.16(a,b), 7.20
For 6.41, you are assuming that $G$ is cyclic, and want to prove that $G/H$ is cyclic. You should first try to guess a generator for $G/H$ (using what you know about $G$), and then show an arbitrary $aH \in G/H$ can be written as a power of the generator.
For 7.11, there are two things to prove: (1) $\ker(\phi)$ is a subgroup of $G_1$, and (2) $a\ker(\phi) = \ker(\phi)a$ for all $a\in G_1$. You already proved that $\ker(\phi)$ is a subgroup of $G_1$ on Exam 2.
Please prove 7.15 WITHOUT using 7.13 Showing that “$\phi$ is one-to-one $\implies$ $\operatorname{ker}(\phi) = \{e_1\}$” should follow quickly from the definition of the kernel. To prove “$\operatorname{ker}(\phi) = \{e_1\}$ $\implies$ $\phi$ is one-to-one”, you should begin by assuming that $\phi(x) = \phi(y)$ (and you want to show $x=y$). To make use of your hypothesis that $\operatorname{ker}(\phi) = \{e_1\}$, try manipulating the equation $\phi(x) = \phi(y)$ to get the identity on one side.
On 7.16, the idea is that if you know that $\phi(i)= h$ and $\phi(j) = v$, then, for example, you can figure out $\phi(k) = \phi(ij) = \phi(i)\phi(j)=hv$.
Extra practice: 6.40, 7.5, 7.13
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