Prove that Center of Symmetric Group S_n is Trivial

The center of the symmetric group Sn is trivial if n≥3, that is, Z(Sn) = {e} where e denotes the identity element of Sn. 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 Sn is trivial.

Table of Contents

Center of Sn is Trivial Proof

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

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

Therefore, σ commutes all elements of Sn, that is,

στ = τσ ∀ τ ∈ Sn …(∗)

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) ∈ Sn. 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 Sn is trivial for n≥3.

You can also Read: Order of a Permutation

Prove that Symmetric Group Sn is non-abelian for n≥3

  1. Basics of Group Theory
  2. Abelian Group
  3. Cyclic Group
  4. Group of prime order is cyclic
  5. Normal Subgroup
  6. Quotient Group
  7. Homomorphism of Groups
  8. First Isomorphism Theorem


Q1: What is the center of the symmetric group Sn for n≥3?

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

Share via:
WhatsApp Group Join Now
Telegram Group Join Now