# 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.

## Center of Sn is Trivial Proof

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,

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

## FAQs

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}.

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