A square has eight geometric symmetries, but the moment we label its vertices those symmetries become rearrangements of the set $\{1,2,3,4\}$. This translation is powerful and dangerous at the same time: powerful because it turns motion into algebra, dangerous because the same abstract group can look different when it acts on a different set. A rotation group may act faithfully on vertices, less faithfully on diagonals, and in a completely different way on cosets of a subgroup.

Permutation groups answer the central question: what can be learned about a group by watching how it moves points of a set? Instead of treating a group only as a multiplication table, we study its orbits, stabilisers, cycles, signs, and invariant decompositions. The payoff is that many abstract questions become concrete questions about rearrangements.

Here is the notation used in the first example. A permutation of a set is a bijective rearrangement of that set. The [symmetric group](/page/Symmetric%20Group) on $\{1,2,3,4\}$ is denoted $S_4$ and consists of all permutations of the four labels. Cycle notation records repeated images: $(1\ 2\ 3\ 4)$ means $1\mapsto 2$, $2\mapsto 3$, $3\mapsto 4$, and $4\mapsto 1$. If $q$ is a group element, then $\langle q\rangle$ denotes the subgroup generated by $q$, meaning all powers of $q$; the notation $H\le G$ means that $H$ is a subgroup of $G$.

[example: Rotations of a Square as Permutations] Let $X=\{1,2,3,4\}$ label the vertices of a square in cyclic order, and let $q$ be the quarter-turn rotation sending each vertex to the next labelled vertex. Thus \begin{align*} q(1)=2,\quad q(2)=3,\quad q(3)=4,\quad q(4)=1. \end{align*} By the definition of the cycle $(1\ 2\ 3\ 4)$, this is exactly the permutation \begin{align*} q=(1\ 2\ 3\ 4)\in S_4. \end{align*} We compute the powers of $q$ by applying $q$ repeatedly to each element of $X$. For the square $q^2$, we get \begin{align*} q^2(1)=q(q(1))=q(2)=3,\quad q^2(3)=q(q(3))=q(4)=1. \end{align*} Also, \begin{align*} q^2(2)=q(q(2))=q(3)=4,\quad q^2(4)=q(q(4))=q(1)=2. \end{align*} Thus $q^2$ sends $1\mapsto 3\mapsto 1$ and $2\mapsto 4\mapsto 2$, so \begin{align*} q^2=(1\ 3)(2\ 4). \end{align*} For the third power, use the values of $q^2$ just computed: \begin{align*} q^3(1)=q(q^2(1))=q(3)=4,\quad q^3(4)=q(q^2(4))=q(2)=3. \end{align*} Similarly, \begin{align*} q^3(3)=q(q^2(3))=q(1)=2,\quad q^3(2)=q(q^2(2))=q(4)=1. \end{align*} Therefore $q^3$ sends $1\mapsto 4$, $4\mapsto 3$, $3\mapsto 2$, and $2\mapsto 1$, so \begin{align*} q^3=(1\ 4\ 3\ 2). \end{align*} For the fourth power, apply $q$ once more: \begin{align*} q^4(1)=q(q^3(1))=q(4)=1,\quad q^4(2)=q(q^3(2))=q(1)=2. \end{align*} Also, \begin{align*} q^4(3)=q(q^3(3))=q(2)=3,\quad q^4(4)=q(q^3(4))=q(3)=4. \end{align*} Since $q^4$ fixes every element of $X$, we have \begin{align*} q^4=\operatorname{id}_X. \end{align*} The subgroup generated by $q$ consists of the distinct powers before the identity repeats: \begin{align*} \langle q\rangle=\{\operatorname{id}_X,q,q^2,q^3\}. \end{align*} Substituting the computed cycle forms gives \begin{align*} \langle q\rangle=\{\operatorname{id}_X,(1\ 2\ 3\ 4),(1\ 3)(2\ 4),(1\ 4\ 3\ 2)\}\le S_4. \end{align*} Labelling the vertices has turned the geometric rotation group of the square into a concrete subgroup of the symmetric group on the four vertex labels. [/example]

The same square also shows the first failure mode. If the rotations act on the two diagonals instead of the four vertices, the half-turn becomes invisible, so the action no longer remembers the full cyclic rotation group. We need definitions that keep track of both the group and the set being moved.

## Definition

Before a group of rearrangements can be studied, we need the individual rearrangements. A permutation of a set $X$ is a bijective function

\begin{align*} \sigma: X \to X. \end{align*}

A single permutation is only one motion, so it is not enough for studying a whole symmetry system. The symmetric group on $X$, denoted $\operatorname{Sym}(X)$, collects all permutations of $X$ into one group under composition. For $X=\{1,\dots,n\}$, this group is denoted $S_n$. The full symmetric group is usually too large: a square does not admit every rearrangement of its vertices as a geometric symmetry. The next definition names the smaller collection of rearrangements that are actually allowed in a given problem.

[definition: Permutation Group] A permutation group on a set $X$ is a subgroup \begin{align*} G \le \operatorname{Sym}(X). \end{align*} The set $X$ is called the permutation domain of $G$. [/definition]

This definition is deliberately concrete. It remembers the ambient set, so it can distinguish two actions of the same abstract group when they move different domains or forget different elements. To explain where such subgroups come from, we now pass from listed permutations to actions of abstract groups.

[example: Two Actions of $C_4$] Let $C_4=\langle r:r^4=e\rangle$. We compare two ways for the same abstract group to act by permutations. First let $C_4$ act on $X=\{1,2,3,4\}$ by sending the generator $r$ to \begin{align*} \sigma=(1\ 2\ 3\ 4). \end{align*} By the definition of this cycle, \begin{align*} \sigma(1)=2,\quad \sigma(2)=3,\quad \sigma(3)=4,\quad \sigma(4)=1. \end{align*} Compute the powers by applying $\sigma$ repeatedly. For $\sigma^2$, \begin{align*} \sigma^2(1)=\sigma(\sigma(1))=\sigma(2)=3,\quad \sigma^2(3)=\sigma(\sigma(3))=\sigma(4)=1. \end{align*} Also, \begin{align*} \sigma^2(2)=\sigma(\sigma(2))=\sigma(3)=4,\quad \sigma^2(4)=\sigma(\sigma(4))=\sigma(1)=2. \end{align*} Thus $\sigma^2$ swaps $1$ with $3$ and swaps $2$ with $4$, so \begin{align*} \sigma^2=(1\ 3)(2\ 4). \end{align*} For the third power, \begin{align*} \sigma^3(1)=\sigma(\sigma^2(1))=\sigma(3)=4,\quad \sigma^3(4)=\sigma(\sigma^2(4))=\sigma(2)=3. \end{align*} Similarly, \begin{align*} \sigma^3(3)=\sigma(\sigma^2(3))=\sigma(1)=2,\quad \sigma^3(2)=\sigma(\sigma^2(2))=\sigma(4)=1. \end{align*} So $\sigma^3$ sends $1\mapsto 4\mapsto 3\mapsto 2\mapsto 1$, and therefore \begin{align*} \sigma^3=(1\ 4\ 3\ 2). \end{align*} For the fourth power, \begin{align*} \sigma^4(1)=\sigma(\sigma^3(1))=\sigma(4)=1,\quad \sigma^4(2)=\sigma(\sigma^3(2))=\sigma(1)=2. \end{align*} Also, \begin{align*} \sigma^4(3)=\sigma(\sigma^3(3))=\sigma(2)=3,\quad \sigma^4(4)=\sigma(\sigma^3(4))=\sigma(3)=4. \end{align*} Hence $\sigma^4$ fixes every element of $X$, so \begin{align*} \sigma^4=\operatorname{id}_X. \end{align*} The defining relation $r^4=e$ is therefore respected by the assignment $r\mapsto\sigma$. The induced permutation group is \begin{align*} \langle \sigma\rangle=\{\operatorname{id}_X,(1\ 2\ 3\ 4),(1\ 3)(2\ 4),(1\ 4\ 3\ 2)\}\le S_4. \end{align*} These four permutations are distinct because their values at $1$ are respectively $1,2,3,4$. Thus this action distinguishes the four elements $e,r,r^2,r^3$ of $C_4$. Now let $C_4$ act on $Y=\{a,b\}$ by sending $r$ to \begin{align*} \tau=(a\ b). \end{align*} Then \begin{align*} \tau(a)=b,\quad \tau(b)=a. \end{align*} Applying $\tau$ twice gives \begin{align*} \tau^2(a)=\tau(\tau(a))=\tau(b)=a,\quad \tau^2(b)=\tau(\tau(b))=\tau(a)=b. \end{align*} Thus $\tau^2$ fixes every element of $Y$, so \begin{align*} \tau^2=\operatorname{id}_Y. \end{align*} Consequently, \begin{align*} \tau^4=\tau^2\tau^2=\operatorname{id}_Y\operatorname{id}_Y=\operatorname{id}_Y. \end{align*} The relation $r^4=e$ is again respected. The four powers of the generator act as \begin{align*} e\mapsto \operatorname{id}_Y,\quad r\mapsto (a\ b),\quad r^2\mapsto \operatorname{id}_Y,\quad r^3\mapsto (a\ b), \end{align*} because $\tau^0=\operatorname{id}_Y$, $\tau^1=\tau$, $\tau^2=\operatorname{id}_Y$, and $\tau^3=\tau^2\tau=\operatorname{id}_Y\tau=\tau$. Therefore the induced permutation group is \begin{align*} \{\operatorname{id}_Y,(a\ b)\}\le \operatorname{Sym}(Y). \end{align*} The first action remembers all four powers of the generator, while the second identifies $e$ with $r^2$ and identifies $r$ with $r^3$; it records only whether the exponent of $r$ is even or odd. [/example]

The example raises a structural question: how does an abstract group produce permutations of a set without first being given as a subgroup of a symmetric group? The next definition is needed because it records the compatibility between group multiplication and the motion of points.

[definition: Group Action] Let $G$ be a group and let $X$ be a set. A left action of $G$ on $X$ is a map \begin{align*} G \times X \to X \end{align*} written $(g,x) \mapsto g \cdot x$, such that for all $g,h \in G$ and all $x \in X$, \begin{align*} e \cdot x = x \end{align*} and \begin{align*} g \cdot (h \cdot x) = (gh) \cdot x. \end{align*} [/definition]

The action axioms say exactly that each group element acts as a permutation and that multiplication in $G$ matches composition of the resulting permutations. This motivates the formal bridge from actions to symmetric groups.

[quotetheorem:5004]

The homomorphism $\rho$ is called the permutation representation associated with the action. Because a representation may collapse non-identity elements, the next question is when the action loses no group information.

[definition: Faithful Action] Let $G$ be a group acting on a set $X$. The action is faithful if the associated homomorphism \begin{align*} \rho: G \to \operatorname{Sym}(X) \end{align*} is injective. [/definition]

Faithfulness is often checked by asking which elements fix every point of the domain. If a non-identity element fixes every point, then the action cannot distinguish it from the identity permutation. The obstruction to faithfulness is therefore the subgroup of elements that act invisibly on all of $X$, and this subgroup should coincide with the kernel of the permutation representation.

[quotetheorem:10134]