Let $\displaystyle G=C_8\times C_8$ and $\displaystyle H\leq G$. $\displaystyle H$ is cyclic of order 8.
Show that $\displaystyle G/H$ is cyclic.
I'm not sure my solution is correct, but here it goes.
I assume $\displaystyle C_8$ is a cyclic group of order 8. So H is isomorphic $\displaystyle Z_8$, or to $\displaystyle \{0\} \times Z_8$. So $\displaystyle (C_8 \times C_8)/H$ is isomorphic to $\displaystyle (Z_8 \times Z_8)/(\{0\} \times Z_8)$, which is isomorphic to $\displaystyle (Z_8/ \{0\}) \times (Z_8/Z_8)$, which is isomorphic to $\displaystyle Z_8$, which is a cyclic group.
Edited to, hopefuly, correct the mistakes.