Step 1: Concept
A group is cyclic if there exists an element $a$ such that every element can be written as $a^n$.
Step 2: Meaning
We test the powers of elements in each set under the given operation.
Step 3: Analysis
In $(\{1, 3, 5, 7\}, \times_8)$, the square of every element is 1: $3^2=9 \equiv 1$, $5^2=25 \equiv 1$, $7^2=49 \equiv 1$. No single element can generate the whole set $\{1, 3, 5, 7\}$.
Step 4: Conclusion
Since no generator exists, this group (known as the Klein 4-group $V_4$) is non-cyclic.
Final Answer: (C)