Show that Sn is non-abelian for n>=3

google "Sn non abelian" and the first hit is a nice explanation of the solution.

Thank you very much!!!!!!!

To show that the symmetric group Sn is non-abelian for n≥3, we can demonstrate a counterexample where the group's operation does not commute.

The symmetric group Sn consists of all possible permutations of n distinct elements, and the group operation is the composition of permutations. That means, if σ and τ are two permutations in the symmetric group, their composition σ∘τ is obtained by applying τ first and then σ.

Now, let's choose three distinct elements a, b, and c from a set of n elements, where n≥3. We can define two permutations σ and τ as follows:

σ: Swap the positions of a and b, keeping all other elements fixed.
τ: Swap the positions of b and c, keeping all other elements fixed.

For example, if n=4 and a=1, b=2, and c=3, then σ would swap 1 and 2, while τ would swap 2 and 3, giving us the permutations σ=(12) and τ=(23).

Let's consider the composition σ∘τ:

σ∘τ: Apply τ first and then σ. This means we swap 2 and 3 (τ), and then swap 1 and 2 (σ).

If we evaluate this composition, we get:
σ∘τ: (23)(12) = (123)

Now, let's consider the composition τ∘σ:

τ∘σ: Apply σ first and then τ. This means we swap 1 and 2 (σ), and then swap 2 and 3 (τ).

If we evaluate this composition, we get:
τ∘σ: (12)(23) = (132)

Notice that (123) and (132) are different permutations. This implies that σ∘τ is not equal to τ∘σ, and the group operation does not commute. Thus, we have shown that Sn is non-abelian for n≥3.