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The Timpson Lyceum · Discrete Math (2025)

Permutations: Example of a Group - Lecture 13 (Discrete Math)

Overview

A permutation of a set is simply a bijection from the set to itself — a reordering of its elements. This lecture introduces permutations as the central example of an algebraic group. After a quick review of functions, one-to-one and onto properties, and function composition, we collect all permutations of {1,2,…,n}\{1,2,\dots,n\} into the set SnS_n and verify the four group axioms (closure, associativity, identity, inverses). We then develop efficient ways to write permutations — two-row notation and the favored cyclic notation — and study transpositions (single swaps). The deepest result is that the number of transpositions needed to build a given permutation always has a fixed parity (always even or always odd), which lets us define the sign of a permutation and read it straight off the cycle structure.

Key Concepts

Function review. A function f:A→Bf:A\to B assigns each input in AA exactly one output in BB (equal inputs force equal outputs). The domain is all of AA; the image is a subset of BB.

Permutation. For a set AA, a permutation on AA is a function π:A→A\pi:A\to A that is a bijection. Equivalently, it is a reordering of the elements of AA. Permutation variables use lowercase Greek letters: π,σ,τ,ι\pi,\sigma,\tau,\iota.

Counting permutations. A set of nn elements has exactly n!n! permutations (choose the first image in nn ways, the next in n−1n-1, …, down to 11).

The set SnS_n. SnS_n denotes the set of all permutations of {1,2,…,n}\{1,2,\dots,n\}. So ∣Sn∣=n!|S_n| = n!. The identity permutation ι\iota maps every element to itself; ι\iota is always a bijection, hence always lies in SnS_n.

SnS_n is a group. SnS_n together with composition ∘\circ satisfies the group axioms:

Notation for a permutation.

Transposition. A permutation τ∈Sn\tau\in S_n is a transposition (a single swap) if there exist distinct i,ji,j with τ(i)=j\tau(i)=j, τ(j)=i\tau(j)=i, and τ(k)=k\tau(k)=k for every other kk. In cyclic notation a transposition is written as its one nontrivial cycle, e.g. (2  4)(2\;4).

Decomposition into transpositions. Every permutation can be written as a composition of transpositions (the "swaps" that turn the identity ordering into the target ordering). This is like factoring a number into primes — but the factorization is not unique, and you can always pad it (e.g. (1  3)(1  3)=ι(1\;3)(1\;3)=\iota does nothing), so a permutation can be expressed using infinitely many different counts of transpositions.

Even / odd / sign. A permutation is even if it can be written as an even number of transpositions, odd if as an odd number. This parity is well-defined: if π\pi can be written with aa transpositions one way and bb another, then a≡b(mod2)a\equiv b \pmod 2. The sign is

sgn⁡(π)=(−1)a={+1π even−1π odd.\operatorname{sgn}(\pi)=(-1)^{a}=\begin{cases}+1 & \pi \text{ even}\\ -1 & \pi \text{ odd.}\end{cases}

Cycle-counting lemma. Composing a permutation with one transposition changes its cycle count by exactly ±1\pm 1: it glues two cycles into one (if the swapped pair lies in different cycles) or splits one cycle into two (if they lie in the same cycle). This is the engine behind the parity proof.

Worked Example

Decompose a permutation into transpositions and find its sign. Take π=(1  2  3  4  5)∈S5\pi=(1\;2\;3\;4\;5)\in S_5 (the 5-cycle 1→2→3→4→5→11\to2\to3\to4\to5\to1). Build it as a chain of swaps, then apply the parity formula.

π=(1  2)∘(1  3)∘(1  4)∘(1  5)4 transpositionssgn⁡(π)=(−1)a=(−1)4=+1⇒π is evenn−c(π)=5−1=4≡0(mod2)matches a=4 (one cycle)π(3)=((1  2)∘(1  3)∘(1  4)∘(1  5))(3)=4trace 3→3→3→3→4\begin{aligned} \pi &= (1\;2)\circ(1\;3)\circ(1\;4)\circ(1\;5) &&\text{4 transpositions} \\ \operatorname{sgn}(\pi) &= (-1)^{a} = (-1)^{4} = +1 &&\Rightarrow \pi \text{ is even} \\ n - c(\pi) &= 5 - 1 = 4 \equiv 0 \pmod 2 &&\text{matches } a=4 \text{ (one cycle)} \\ \pi(3) &= \big((1\;2)\circ(1\;3)\circ(1\;4)\circ(1\;5)\big)(3) = 4 &&\text{trace } 3\to3\to3\to3\to4 \end{aligned}

The chain reorders 1 2 3 4 51\,2\,3\,4\,5 into 2 3 4 5 12\,3\,4\,5\,1 exactly as π\pi does; reading 33 right-to-left through the swaps lands on 44, confirming π(3)=4\pi(3)=4. Both the count of transpositions (44) and the formula n−c(π)=4n-c(\pi)=4 give the same even parity, so sgn⁡(π)=+1\operatorname{sgn}(\pi)=+1.

Summary