# Prove Symmetric Group S_n is not Abelian for n ≥ 3

The symmetric group Sn is not abelian for n ≥ 3. In other words, there are elements of Sn that do not commute each other. In this page, we will prove that the symmetric group Sn is not commutative. Before that let us recall abelian group.

A group (G, ∗) is called an abelian group if a∗b = b∗a ∀ a,b ∈ G, that is, the elements of G commute each other.

## Symmetric Group is not Abelian Proof

Let us prove that the symmetric group Sn is not abelian whenever n≥3.

Since n≥3, let us consider two permutations σ= $\left( {\begin{array}{*{20}{c}}1&2&3 \\ 3&2&1 \end{array}} \right)$ and τ = $\left( {\begin{array}{*{20}{c}}1&2&3 \\ 1&3&2 \end{array}} \right)$ on {1, 2, 3, …, n}.

Now στ = $\left( {\begin{array}{*{20}{c}}1&2&3 \\ 3&1&2 \end{array}} \right)$

τσ = $\left( {\begin{array}{*{20}{c}}1&2&3 \\ 2&3&1 \end{array}} \right)$

Hence, στ ≠ τσ.

That is, σ and τ do not commute each other. This proves that the symmetric group Sn is not abelian provided that n≥3.

Remark: If G is an abelian group, then its center Z(G) = G. But, the center of the symmetric group Sn is trivial for n ≥ 3. Thus, Sn is not abelian for n ≥ 3.

## FAQs

Q1: Why symmetric group Sn is not abelian/commutative?

Answer: We know that the center of Sn is trivial. If the symmetric group is abelian then its center would have been non-trivial. This makes Sn non-abelian.

Q2: Is S3 abelian?

Answer: No, S3 is not abelian. This is because (1 2) (2 3) ≠ (2 3) (1 2).

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