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, σ1, and the only relation that applies is:
(σ1)2 = 1
for the case n = 3, we have two generators: σ1 and σ2 (that is: (1 2) and (2 3)). the relation:
σiσj = σjσ1 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 (σ1)2 = (σ2)2 = 1 (and this is always true for any transposition), the only non-trivial relation we have is:
σ1σ2σ1 = σ2σ1σ2.
note the LHS is the conjugate of σ2 by σ1, and the RHS is the conjugate of σ1 by σ2.
that should be enough to get you started.