A bijective mapping on a finite set S is called a permutation on S. In this post, we will discuss the order of a permutation, how to find the order of a permutation with examples, and related theorems.

## Definition of Order of a Permutation

Order of permutation:- The order of a permutation σ on a finite set S is the least positive integer n such that **σ ^{n}=i** where i denotes the identity permutation on S.

By definition, the order of the identity permutation is 1.

Example of Order of permutation:- Let σ= $\left( {\begin{array}{*{20}{c}}1&2&3 \\ 2&3&1 \end{array}} \right)$ be a permutation on {1, 2, 3}.

Now, σ^{2} = $\left( {\begin{array}{*{20}{c}}1&2&3 \\ 2&3&1 \end{array}} \right)$ $\left( {\begin{array}{*{20}{c}}1&2&3 \\ 2&3&1 \end{array}} \right)$

= $\left( {\begin{array}{*{20}{c}}1&2&3 \\ 3&1&2 \end{array}} \right)$

⇒ σ^{3} = σ^{2} ⋅ σ

= $\left( {\begin{array}{*{20}{c}}1&2&3 \\ 3&1&2 \end{array}} \right)$ $\left( {\begin{array}{*{20}{c}}1&2&3 \\ 2&3&1 \end{array}} \right)$

= $\left( {\begin{array}{*{20}{c}}1&2&3 \\ 1&2&3 \end{array}} \right)$

= i, the identity permutation on {1, 2, 3}

Thus 3 is the least positive integer such that σ^{3}=i. So the order of σ is 3.

## How to Find Order of a Permutation

The order of a given permutation is determined by the least common multiple of the lengths of the cycles in the decomposition of the given permutation into disjoint cycles. This will be proved in the theorem below.

**Theorem:** The order of a permutation on a finite set is the least common multiple (lcm) of the lengths of its disjoint cycles.

*Proof:*

Let σ be a permutation on a set S={1, 2, 3, …, n}. We assume that

σ = σ_{1} σ_{2} … σ_{m}

is the decomposition of σ as the product of disjoint cycles σ_{1}, σ_{2}, …, σ_{m} of lengths r_{1}, r_{2}, …, r_{m}. We have

σ_{1}^{r1} = i, σ_{2}^{r2} = i, …, σ_{m}^{rm} = i.

As the multiplication of disjoint cycles is commutative, we obtain that

σ^{n} = σ_{1}^{n} σ_{2}^{n} … σ_{m}^{n}

Let t : =lcm( r_{1}, r_{2}, …, r_{m}). Then σ^{t} = σ_{1}^{t} σ_{2}^{t} … σ_{m}^{t} = i, the identity permutation. Since t is the lcm, we can easily check that t is the smallest positive integer such that σ^{t} = i. Therefore,

t= order of σ.

In other words, the order of σ is the lcm of the lengths of its disjoint cycles.

## Solved Examples on Order of Permutation

**Example 1:** Find the order of (1 4 5 7) (2 6 3).

*Solution:*

See that σ = (1 4 5 7) (2 6 3) is the product of two disjoint cycles. Here (1 4 5 7) is a cycle of length 4 and (2 6 3) is a cycle of length 3.

By the above theorem on orders of permutations, we deduce that:

The order of σ is

= lcm(4, 3)

= 12.

∴ The order of (1 4 5 7) (2 6 3) is 12.

## FAQs on Order of Permutation

**Q1: What is the order of (2 3 4)?**

Answer: As (2 3 4) is a 3-cycle, the order of (2 3 4) is 3.

**Q2: What is the order of (1 2 3) (1 3 2)?**

Answer: By multiplication of permutations, we have (1 2 3) (1 3 2) = i, where i is the identity permutation. So the order of (1 2 3) (1 3 2) is 1.