Help with presentation of the symmetric group in the simplest cases

The symmetric group on n-letters, Sn, may be described as follows.

σi is the permutation that swaps the i:th element with the i+1 one.

With the generators: σ1, . . ., σn-1

And the following relations:

- (σi)(σj+1)(σi) =(σj+1)(σi)(σj+1) ∀ i
- σij = σjσi if |i - j| ≥ 2
- (σi)² =
**1**

How do I verify the above for the two simplest cases where n = 2 and where n = 3?

Can somebody help me out? I don't want anyone to solve this *for* me but rather if someone just could give me a hint and a push in the right direction.

Re: Help with presentation of the symmetric group in the simplest cases

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.