The finite symmetric group $S_n$, one-line notation, and cycle notation

Let $n\in \mathbb{N}$, so that $n=\{0,1,\dots ,n-1\}$ (The natural numbers $\mathbb{N}$ (von Neumann)). The symmetric group on $n$ letters is

$S_n:=\mathrm{Sym}(n)=\mathrm{Sym}(\{0,1,\dots ,n-1\}),$

the group of all bijections of $n$ under composition (The symmetric group $\mathrm{Sym}(X)$: the bijections of a set $X$ under composition), with the composition convention

$(\sigma \tau )(i):=(\sigma \circ \tau )(i)=\sigma (\tau (i))(i\in n),$

so that in a product the right-hand factor acts first. An element of $S_n$ is named by either of the two notations below.

One-line notation. For $\sigma \in S_n$, its one-line form is the list of its values in order of their arguments,

$\sigma =[\sigma (0),\sigma (1),\dots ,\sigma (n-1)]\mathrm{.}$

This list has length $n$ and its entries are $0,1,\dots ,n-1$, each occurring once, because $\sigma$ is a bijection of $n$. Conversely, a list $[b_0,b_1,\dots ,b_{n-1}]$ whose entries are $0,1,\dots ,n-1$ each occurring once is the one-line form of exactly one element of $S_n$, namely the map sending each $i\in n$ to $b_i$: that map is injective because the entries are distinct, and surjective because every element of $n$ occurs among them. So one-line notation is a bijection from $S_n$ to the arrangements of $0,1,\dots ,n-1$ in a list. For $n=0$ the one-line form of the unique element of $S_0$ is the empty list.

Cycle notation. For distinct $a_0,a_1,\dots ,a_{k-1}\in n$ with $k\ge 2$, the symbol $(a_0{a}_1\cdots a_{k-1})$ denotes the element of $S_n$ that sends $a_i$ to $a_{i+1}$ for each $i<k-1$, sends $a_{k-1}$ to $a_0$, and fixes every element of $n$ outside $\{a_0,\dots ,a_{k-1}\}$ (The symmetric group $\mathrm{Sym}(X)$: the bijections of a set $X$ under composition); it is called a $k$-cycle, and a $2$-cycle is a transposition. Writing cycle symbols side by side means composing them, so $(ab)(cd)$ is $(ab)\circ (cd)$, and the empty juxtaposition of cycle symbols is the identity $\mathrm{id}$.

Unlike one-line notation, cycle notation does not name each permutation once: the symbol may be started at any of its entries, so

$(a_0{a}_1\cdots a_{k-1})=(a_1\cdots a_{k-1}{a}_0)$

and each $k$-cycle is written by exactly $k$ symbols of this shape. A cycle symbol also does not record $n$, which must be supplied by the context.

• The brackets carry the meaning, so the same list of numbers reads two different ways. Square brackets are one-line notation and round brackets are cycle notation. In $S_3$ the one-line form $[1,2,0]$ and the cycle symbol $(012)$ happen to name the same permutation, the one sending $0\mapsto 1$, $1\mapsto 2$, $2\mapsto 0$; but $[0,1,2]$ is the identity while $(012)$ is not, and $[2,1,0]$ is the transposition exchanging $0$ and $2$ while $(210)$ is a $3$-cycle. Inside a cycle symbol this library separates the entries by thin spaces rather than by commas, which keeps the two notations apart on the page.

• Relation to the two-row form. Many texts write a permutation as the array $\sigma =\left(\begin{array}{cccc}0& 1& \cdots & n-1\\ \sigma (0)& \sigma (1)& \cdots & \sigma (n-1)\end{array}\right)$, whose first row lists the arguments and whose second row lists their images. One-line notation is that array with its first row deleted, which loses nothing because the first row is the same for every $\sigma \in S_n$.

• Why the identity is a product of no cycles rather than a cycle. The cycle symbols are restricted to $k\ge 2$, so a fixed point is never written. The identity is therefore the empty product, and a permutation is written by listing only the cycles that move something. Which permutations admit such a factorisation, and in how many ways, is Every permutation of a finite set is a product of pairwise disjoint cycles, uniquely up to reordering and cyclic rotation; the fixed points that cycle notation suppresses are restored as one-cycles when a cycle type is recorded (Support, fixed points, disjoint cycles, cycle length, disjoint-cycle decompositions, and cycle type).