The symmetric group S_{n} is not abelian for n ≥ 3. In other words, there are elements of S_{n} that do not commute each other. In this page, we will prove that the symmetric group S_{n} 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.

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## Symmetric Group is not Abelian Proof

Let us prove that the symmetric group S_{n} is not abelian whenever n≥3.

**Proof:**

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 S_{n} is not abelian provided that n≥3.

**Read Also:** Order of a Permutation in a Symmetric Group

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

Related Topics:

- Basics of Group Theory
- Group of order 4 is abelian: Proof
- Cyclic Group
- Group of prime order is cyclic
- Normal Subgroup
- Quotient Group
- Homomorphism of Groups
- First Isomorphism Theorem

## FAQs

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

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

**Q2: Is S**

_{3}abelian?**Answer:** No, S_{3} is not abelian. This is because (1 2) (2 3) ≠ (2 3) (1 2).

This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.