Prove Symmetric Group S_n is not Abelian for n ≥ 3

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.

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:

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

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