well the case n = 2 is fairly trivial, there is only one non-identity element, the permutation (1 2), of order 2. this is the only cyclic symmetric group, there is but one generator, σ_{1}, and the only relation that applies is:

(σ_{1})^{2}= 1

for the case n = 3, we have two generators: σ_{1}and σ_{2}(that is: (1 2) and (2 3)). the relation:

σ_{i}σ_{j}= σ_{j}σ_{1}if |i - j| ≥ 2 also does not apply here (as |i - j| is either 0 or 1). i note in passing that what this captures is "disjoint cycles commute".

so besides the fact that (σ_{1})^{2}= (σ_{2})^{2}= 1 (and this is always true for any transposition), the only non-trivial relation we have is:

σ_{1}σ_{2}σ_{1}= σ_{2}σ_{1}σ_{2}.

note the LHS is the conjugate of σ_{2}by σ_{1}, and the RHS is the conjugate of σ_{1}by σ_{2}.

that should be enough to get you started.