Step 1: Analyze cyclic subgroups.
A 6-cycle would have order 6, but \(S_5\) has only 5 symbols.
However, a permutation composed of a 3-cycle and a disjoint 2-cycle, such as \((1\,2\,3)(4\,5)\),
has order \(\mathrm{lcm}(3,2) = 6\). Hence, it generates a cyclic subgroup of order 6.
Step 2: Check other options.
(A) Not always true.
(B) False, since abelian subgroups like \(\langle (1\,2) \rangle\) exist.
(D) Index 7 would imply order \(|S_5|/7 = 120/7\), not an integer, impossible.
Step 3: Conclusion.
Hence, only (C) is correct.