The center of the symmetric group S_{n} is trivial if n≥3, that is, Z(S_{n}) = {e} where e denotes the identity element of S_{n}. The center of a group G, denoted by Z(G), is defined as follows:

Z(G) = {g ∈ G : ag = ga ∀ a ∈ G}.

In this article, we will show that the centre of S_{n} is trivial.

Table of Contents

## Center of S_{n} is Trivial Proof

Theorem: For n ≥ 3 prove that Z(S_{n}) = {e}, the center of S_{n} is trivial. |

**Proof:**

For a contradiction, we assume that the center of S_{n} is non-trivial. So the center Z(S_{n}) contains a non-trivial element, say σ.

Therefore, σ commutes all elements of S_{n}, that is,

στ = τσ ∀ τ ∈ S_{n} …(∗) |

As σ is non-trivial, so there exists i, j ∈ {1, 2, …, n} (both are not equal) such that σ(i) = j.

Also the assumption n ≥ 3 implies that there is a number k ∈ {1, 2, …, n} different from i and j. Now we consider the permutation τ = (i k) ∈ S_{n}. That is, τ sends i to k, k to i, and fixes all other numbers.

We have the following:

τσ (i) = τ(j) = j

στ (i) = σ(k) ≠ j.

Here σ(k) ≠ j follows from the fact that σ already sends i to j, so it cannot send any other element to j.

Thus, we deduce that στ ≠ τσ, contradicting the above fact (∗).

So our assumption was wrong. This proves that the center of the symmetric group S_{n} is trivial for n≥3.

**You can also Read:** Order of a Permutation

Prove that Symmetric Group S_{n} is non-abelian for n≥3

Related Topics:

- Basics of Group Theory
- Abelian Group
- Cyclic Group
- Group of prime order is cyclic
- Normal Subgroup
- Quotient Group
- Homomorphism of Groups
- First Isomorphism Theorem

## FAQs

**Q1: What is the center of the symmetric group S**

_{n}for n≥3?**Answer:** The center of the symmetric group S_{n} is trivial for n≥3, that is, Z(S_{n}) = {e}.

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.