A group action begins with a promise: a group $G$ moves a set $X$ around while respecting multiplication in $G$. The first question is not only where a point goes, but how much of the group fails to move it at all. Those unmoving elements carry information. They measure symmetry left over after choosing a point, and they explain why orbits often have sizes smaller than the group itself. The stabiliser is the answer to that question. It turns an action into a family of subgroups, one subgroup for each point of the set being acted on. When the action comes from geometry, the stabiliser is the symmetry group of a chosen object. When the action comes from algebra, it records conjugation symmetries, fixed vectors, and automorphisms preserving extra structure. [example: Rotations of a Square and a Chosen Vertex] Let $G$ be the rotational symmetry group of a square, so \begin{align*} G=C_4=\{e,r,r^2,r^3\}, \end{align*} where $r$ is rotation by $90$ degrees counterclockwise and $r^4=e$. Label the vertices cyclically by $x_0,x_1,x_2,x_3$, and choose $x=x_0$. Then \begin{align*} e\cdot x_0 &= x_0, \\ r\cdot x_0 &= x_1, \\ r^2\cdot x_0 &= x_2, \\ r^3\cdot x_0 &= x_3. \end{align*} Hence the orbit is \begin{align*} G\cdot x_0 &= \{g\cdot x_0 : g\in G\} \\ &= \{e\cdot x_0,r\cdot x_0,r^2\cdot x_0,r^3\cdot x_0\} \\ &= \{x_0,x_1,x_2,x_3\} = X. \end{align*} For the stabiliser, we test the same four elements: \begin{align*} e\cdot x_0 &= x_0, \\ r\cdot x_0 &= x_1 \ne x_0, \\ r^2\cdot x_0 &= x_2 \ne x_0, \\ r^3\cdot x_0 &= x_3 \ne x_0. \end{align*} Therefore \begin{align*} G_{x_0} &= \{g\in G : g\cdot x_0=x_0\} \\ &= \{e\}. \end{align*} Now let $G$ act on the set $Y$ of the two diagonals. Write \begin{align*} d_1 &= \{x_0,x_2\}, & d_2 &= \{x_1,x_3\}. \end{align*} The action on $d_1$ is \begin{align*} e\cdot d_1 &= \{e\cdot x_0,e\cdot x_2\}=\{x_0,x_2\}=d_1, \\ r\cdot d_1 &= \{r\cdot x_0,r\cdot x_2\}=\{x_1,x_3\}=d_2, \\ r^2\cdot d_1 &= \{r^2\cdot x_0,r^2\cdot x_2\}=\{x_2,x_0\}=d_1, \\ r^3\cdot d_1 &= \{r^3\cdot x_0,r^3\cdot x_2\}=\{x_3,x_1\}=d_2. \end{align*} Thus \begin{align*} G_{d_1} &= \{g\in G : g\cdot d_1=d_1\} \\ &= \{e,r^2\}. \end{align*} The chosen vertex has no residual rotational symmetry, while the chosen diagonal is still preserved by the half-turn $r^2$. [/example] The example shows the basic tension. The orbit tells us how many positions a chosen object can occupy; the stabiliser tells us which group elements are invisible from that object's point of view. The main theorem of the subject says that these two measurements multiply to the size of the acting group when $G$ is finite. ## Definition Before naming the stabiliser, we need the action itself to be part of the data. A group does not act merely because its elements look like transformations; we need a specified rule telling each group element how it moves each point, and that rule must respect the group law. [definition: Group Action] Let $G$ be a group and let $X$ be a set. A group action of $G$ on $X$ is a map \begin{align*} G \times X &\to X \\ (g,x) &\mapsto g \cdot x \end{align*} such that for all $g,h \in G$ and all $x \in X$, \begin{align*} e \cdot x &= x, \\ g \cdot (h \cdot x) &= (gh) \cdot x, \end{align*} where $e$ is the identity element of $G$. [/definition] Once the action is fixed, it becomes meaningful to ask which group elements become undetectable at a chosen point. We do not ask which elements are the identity in $G$; we ask which elements behave like the identity on that particular point. [definition: Stabiliser] Let $G$ be a group acting on a set $X$, and let $x \in X$. The stabiliser of $x$ in $G$ is \begin{align*} G_x = \{g \in G : g \cdot x = x\}. \end{align*} [/definition] For stabilisers to be useful in group theory, they must inherit the algebraic structure of $G$ rather than remain arbitrary subsets. The possible obstruction is that two elements may each fix $x$ for unrelated reasons, so it is not automatic from the definition alone that their product still fixes $x$, or that undoing one of them still fixes $x$. The action axioms are exactly what remove this obstruction. [quotetheorem:5000] This theorem converts geometric or combinatorial questions about points into algebraic questions about subgroups. To use it for counting, we also need the complementary object: the set of all positions that the point can reach under the group action. [definition: Orbit] Let $G$ be a group acting on a set $X$, and let $x \in X$. The orbit of $x$ is \begin{align*} G \cdot x = \{g \cdot x : g \in G\}. \end{align*} [/definition] The orbit is a subset of $X$, while the stabiliser is a subgroup of $G$. The two live in different places, but they are linked by a simple idea: two group elements send $x$ to the same point exactly when they differ by an element that fixes $x$. ## Orbits and Cosets ### The Orbit-Stabiliser Count The stabiliser first pays off when we count. If $G$ is finite, a point cannot usually have $|G|$ distinct images, because several elements of $G$ may produce the same image. The stabiliser measures exactly how much repetition occurs. To make that repetition visible, compare the map $g \mapsto g \cdot x$ with cosets of $G_x$. Elements in the same left coset of $G_x$ send $x$ to the same place, and elements in different left cosets send $x$ to different places. [quotetheorem:845] This is the main structural statement about stabilisers in finite group actions. It says that a point with a large stabiliser has a small orbit, and a point with a small stabiliser has a large orbit. The theorem also explains why orbit sizes must divide $|G|$. [example: Reflections of a Triangle] Let $G=D_6$ be the symmetry group of an equilateral triangle, using the convention that $D_{2n}$ has $2n$ elements. Label the vertices cyclically by $x_0,x_1,x_2$, let $r$ be rotation by $120$ degrees counterclockwise, and let $s$ be the reflection through the altitude passing through $x_0$. Then \begin{align*} G=\{e,r,r^2,s,rs,r^2s\}. \end{align*} For the chosen vertex $x=x_0$, the action of these six elements is \begin{align*} e\cdot x_0 &= x_0, \\ r\cdot x_0 &= x_1, \\ r^2\cdot x_0 &= x_2, \\ s\cdot x_0 &= x_0, \\ rs\cdot x_0 &= r\cdot(s\cdot x_0)=r\cdot x_0=x_1, \\ r^2s\cdot x_0 &= r^2\cdot(s\cdot x_0)=r^2\cdot x_0=x_2. \end{align*} Hence the orbit is \begin{align*} G\cdot x_0 &=\{g\cdot x_0:g\in G\} \\ &=\{x_0,x_1,x_2,x_0,x_1,x_2\} \\ &=\{x_0,x_1,x_2\}, \end{align*} so $|G\cdot x_0|=3$. The stabiliser consists exactly of those elements in the same list that send $x_0$ back to $x_0$: \begin{align*} G_{x_0} &=\{g\in G:g\cdot x_0=x_0\} \\ &=\{e,s\}. \end{align*} Thus $|G_{x_0}|=2$, and the computed sizes satisfy \begin{align*} |G\cdot x_0|\,|G_{x_0}| &=3\cdot 2 \\ &=6 \\ &=|D_6|. \end{align*} The non-identity element in the stabiliser is the reflection through the altitude passing through $x_0$, so it is precisely the symmetry left after that vertex has been singled out. [/example] ### Decomposition into Orbits Counting a single orbit is often not enough, because an action may split $X$ into several disjoint regions that never communicate under the action. We need a structural statement saying that these regions account for every point exactly once, so stabiliser counts can be applied orbit by orbit. [quotetheorem:5001] Together with orbit-stabiliser, orbit decomposition turns a group action into a counting machine. A finite $G$-set is assembled from pieces whose sizes are indices of stabilisers. [example: A Non-Transitive Action] Let $G=C_2=\{e,s\}$ act on $X=\{1,2,3\}$ by \begin{align*} e\cdot 1&=1, & e\cdot 2&=2, & e\cdot 3&=3,\\ s\cdot 1&=2, & s\cdot 2&=1, & s\cdot 3&=3. \end{align*} For the point $1$, its orbit is \begin{align*} G\cdot 1 &=\{g\cdot 1:g\in G\}\\ &=\{e\cdot 1,s\cdot 1\}\\ &=\{1,2\}. \end{align*} Its stabiliser is \begin{align*} G_1 &=\{g\in G:g\cdot 1=1\}\\ &=\{e\}, \end{align*} because $e\cdot 1=1$ while $s\cdot 1=2\ne 1$. For the point $3$, its orbit is \begin{align*} G\cdot 3 &=\{g\cdot 3:g\in G\}\\ &=\{e\cdot 3,s\cdot 3\}\\ &=\{3,3\}\\ &=\{3\}. \end{align*} Its stabiliser is \begin{align*} G_3 &=\{g\in G:g\cdot 3=3\}\\ &=\{e,s\}\\ &=G, \end{align*} because both $e\cdot 3=3$ and $s\cdot 3=3$. Thus the action has two orbits, $\{1,2\}$ and $\{3\}$: the moving points have stabiliser of order $1$, while the fixed point has stabiliser of order $2$. This separates belonging to the same underlying set from being reachable by the group action. [/example] ## Conjugate Stabiliser Subgroups The stabiliser depends on the point, so a natural question is how it changes as the point moves around its orbit. If $y = g \cdot x$, then the symmetries fixing $y$ should be the transported symmetries fixing $x$. Algebraically, transported symmetry means conjugation. This is the point where stabilisers connect group actions to the internal subgroup structure of $G$. Moving a basepoint should not produce an unrelated subgroup; the next result says that it produces a conjugate subgroup. [quotetheorem:5002] When a stabiliser is as large as possible, conjugation no longer changes the question: every group element fixes the point already. This extreme case deserves a name because it separates points moved by the action from points on which the whole action vanishes. [definition: Fixed Point] Let $G$ be a group acting on a set $X$. A point $x \in X$ is a fixed point of the action if \begin{align*} g \cdot x = x \end{align*} for every $g \in G$. [/definition] The fixed point definition identifies one point whose stabiliser is all of $G$, but many arguments need to separate the part of $X$ where the action is genuinely visible from the part where every group element fixes the point. Treating those points one at a time would obscure this unmoved region of the action, so we collect them into a single subset of $X$. [definition: Fixed-Point Set] Let $G$ be a group acting on a set $X$. The fixed-point set of the action is \begin{align*} X^G = \{x \in X : g \cdot x = x \text{ for all } g \in G\}. \end{align*} [/definition] The fixed-point set is not the same as a stabiliser. It lives inside $X$, whereas a stabiliser lives inside $G$. The relation is \begin{align*} x \in X^G \quad \iff \quad G_x = G. \end{align*} [example: Conjugation and the Centre] Let a group $G$ act on itself by conjugation: \begin{align*} g \cdot x = gxg^{-1}. \end{align*} Fix $x \in G$. By the definition of stabiliser for this action, \begin{align*} G_x &= \{g \in G : g \cdot x = x\} \\ &= \{g \in G : gxg^{-1} = x\}. \end{align*} For any $g \in G$, the condition $gxg^{-1}=x$ is equivalent to commutation with $x$: \begin{align*} gxg^{-1} &= x \\ gxg^{-1}g &= xg \\ gx &= xg, \end{align*} where the second line multiplies both sides on the right by $g$ and uses $g^{-1}g=e$. Conversely, if $gx=xg$, then \begin{align*} gx &= xg \\ gxg^{-1} &= xgg^{-1} \\ gxg^{-1} &= x. \end{align*} Therefore \begin{align*} G_x &= \{g \in G : gxg^{-1}=x\} \\ &= \{g \in G : gx=xg\} \\ &= C_G(x), \end{align*} the centraliser of $x$ in $G$. The fixed-point set consists of the elements of $G$ fixed by every conjugating element: \begin{align*} G^G &= \{x \in G : g \cdot x=x \text{ for all } g \in G\} \\ &= \{x \in G : gxg^{-1}=x \text{ for all } g \in G\}. \end{align*} Using the equivalence just proved, for each fixed pair $g,x \in G$, \begin{align*} gxg^{-1}=x \quad \Longleftrightarrow \quad gx=xg. \end{align*} Hence \begin{align*} G^G &= \{x \in G : gx=xg \text{ for all } g \in G\} \\ &= Z(G), \end{align*} the centre of $G$. Thus conjugation stabilisers recover centralisers one element at a time, while the global fixed-point set recovers the centre. [/example] The conjugation example is a useful warning. Stabiliser notation may look external, but in many important actions it reconstructs familiar internal subgroups. ## Free and Transitive Actions ### Freeness Stabilisers measure how far an action is from having no redundancy. The cleanest possible action is one where no non-identity element fixes any point. In such an action, every group element is visible from every point. [definition: Free Action] Let $G$ be a group acting on a set $X$. The action is free if for every $x \in X$, \begin{align*} G_x = \{e\}. \end{align*} [/definition] The definition of freeness says that every stabiliser has the smallest possible size. For a finite action, the natural counting question is then whether each orbit must have the largest possible size. [quotetheorem:796] Freeness controls stabilisers but not the number of orbits. An action may be free on many separate copies of the group, so we need a different condition to say that all points lie in one orbit. [definition: Transitive Action] Let $G$ be a group acting on a set $X$. The action is transitive if for all $x,y \in X$, there exists $g \in G$ such that \begin{align*} g \cdot x = y. \end{align*} [/definition] A natural next problem is to identify actions where motion from any $x$ to any $y$ exists and is unique. Freeness supplies uniqueness, transitivity supplies existence, and the combined condition receives its own name. [definition: Regular Action] Let $G$ be a group acting on a set $X$. The action is regular if it is both free and transitive. [/definition] The definition of a regular action has two parts, but computations usually encounter a different question: given $x$ and $y$, can we solve $g \cdot x=y$, and if so is the solution forced? Transitivity gives at least one solution, while freeness prevents two different group elements from giving the same solution. Thus regularity can be recognized by uniqueness of the transporter from each point to each other point. [quotetheorem:3240] This formulation is often the most useful one in computations: existence comes from transitivity, uniqueness comes from identity-only stabilisers. [example: Left Translation on a Group] Let $G$ be a group, and let $G$ act on itself by left translation: \begin{align*} g \cdot x = gx. \end{align*} For a fixed $x \in G$, the stabiliser is \begin{align*} G_x &= \{g \in G : g \cdot x = x\} \\ &= \{g \in G : gx = x\}. \end{align*} If $g \in G_x$, then $gx=x$, and multiplying both sides on the right by $x^{-1}$ gives \begin{align*} gx &= x, \\ (gx)x^{-1} &= x x^{-1}, \\ g(xx^{-1}) &= e, \\ ge &= e, \\ g &= e. \end{align*} Conversely, the identity element fixes $x$ because \begin{align*} e \cdot x &= ex \\ &= x. \end{align*} Therefore \begin{align*} G_x=\{e\} \end{align*} for every $x \in G$, so the action is free. The action is transitive because, for any $x,y \in G$, the element $g=yx^{-1}$ sends $x$ to $y$: \begin{align*} (yx^{-1}) \cdot x &= (yx^{-1})x \\ &= y(x^{-1}x) \\ &= ye \\ &= y. \end{align*} Thus left translation is both free and transitive, hence regular. Choosing one point of the acted-on copy of $G$ identifies every other point uniquely by the group element that translates the chosen point to it. [/example] ### Non-Free Transitive Actions Not every natural transitive action is free. In geometry, points with larger stabilisers are often the most interesting points, because they have extra symmetry. [example: Rotation Action on the Sphere] Let $G=SO(3)$ act on $S^2 \subsetneq \mathbb{R}^3$ by matrix multiplication. Fix $x \in S^2$, and choose $Q \in SO(3)$ with $Qe_3=x$. For $R \in SO(3)$, the condition $R \in G_x$ is \begin{align*} Rx&=x,\\ R(Qe_3)&=Qe_3,\\ Q^{-1}RQe_3&=e_3. \end{align*} Write $A=Q^{-1}RQ$. Since $A \in SO(3)$ and $Ae_3=e_3$, the third column of $A$ is $e_3$. Orthogonality of the columns then forces the first two columns to have third coordinate $0$, so \begin{align*} A= \begin{pmatrix} B & 0\\ 0 & 1 \end{pmatrix} \end{align*} for some real $2\times 2$ matrix $B$. From $A^\top A=I_3$ and $\det A=1$, we get \begin{align*} \begin{pmatrix} B^\top & 0\\ 0 & 1 \end{pmatrix} \begin{pmatrix} B & 0\\ 0 & 1 \end{pmatrix} &= \begin{pmatrix} I_2 & 0\\ 0 & 1 \end{pmatrix},\\ B^\top B&=I_2, \end{align*} and \begin{align*} 1=\det A=\det(B)\cdot 1=\det(B). \end{align*} Thus $B\in SO(2)$. Conversely, every $B\in SO(2)$ gives an element \begin{align*} Q \begin{pmatrix} B&0\\ 0&1 \end{pmatrix} Q^{-1}\in SO(3) \end{align*} which fixes $x=Qe_3$. Hence \begin{align*} G_x = Q \left\{ \begin{pmatrix} B&0\\ 0&1 \end{pmatrix} :B\in SO(2) \right\} Q^{-1}, \end{align*} so $G_x \cong SO(2)$. These are exactly the rotations around the axis through $x$ and $-x$. The action is transitive. Given $x,y\in S^2$, choose $Q_x,Q_y\in SO(3)$ with $Q_xe_3=x$ and $Q_ye_3=y$. Then \begin{align*} (Q_yQ_x^{-1})x &=(Q_yQ_x^{-1})(Q_xe_3)\\ &=Q_ye_3\\ &=y. \end{align*} The action is not free, because $G_x\cong SO(2)$ contains non-identity rotations, for example rotation by angle $\pi$ around the axis through $x$. Thus choosing a point on the sphere leaves a full circle of rotational symmetry. [/example] ## Coset Spaces and Homogeneous Spaces ### Coset Actions A transitive action can be reconstructed from one stabiliser. This is one of the main reasons stabilisers appear in algebra, topology, and differential geometry: they turn spaces with symmetry into quotient spaces of groups. The construction starts from a subgroup. Given $H \le G$, the group $G$ acts on the set of left cosets by left multiplication. The stabiliser of the distinguished coset $H$ is exactly $H$. [definition: Coset Action] Let $G$ be a group and let $H \le G$. The coset action of $G$ on the left coset space $G/H$ is the action \begin{align*} G \times G/H &\to G/H \\ (g,aH) &\mapsto gaH. \end{align*} [/definition] This definition packages a subgroup as the stabiliser of a point in a transitive action. The converse question is now unavoidable: if we start with an arbitrary transitive action, can it always be recovered from the stabiliser of one point? [quotetheorem:3240] This theorem is a classification statement. A transitive $G$-set is not mysterious once a basepoint has been chosen: it is a coset space, and the stabiliser of the basepoint is the subgroup used in the quotient. [example: Projective Lines as Coset Spaces] Let $G=GL(2,\mathbb{R})$ act on $\mathbb{RP}^1$ by sending a line $W\subset \mathbb{R}^2$ to its image $A(W)$ under an invertible matrix $A$. Let \begin{align*} L=\operatorname{span}(e_1). \end{align*} We compute the stabiliser of $L$ and then identify the orbit with a coset space. Write \begin{align*} A= \begin{pmatrix} a & b\\ c & d \end{pmatrix} \in GL(2,\mathbb{R}). \end{align*} Since $L=\{\lambda e_1:\lambda\in\mathbb{R}\}$, \begin{align*} A(L) &=\{A(\lambda e_1):\lambda\in\mathbb{R}\}\\ &=\{\lambda Ae_1:\lambda\in\mathbb{R}\}\\ &=\operatorname{span}(Ae_1). \end{align*} Now \begin{align*} Ae_1 &= \begin{pmatrix} a & b\\ c & d \end{pmatrix} \begin{pmatrix} 1\\ 0 \end{pmatrix}\\ &= \begin{pmatrix} a\\ c \end{pmatrix}. \end{align*} Therefore $A(L)=L$ exactly when $\operatorname{span}\begin{pmatrix}a\\c\end{pmatrix}=\operatorname{span}\begin{pmatrix}1\\0\end{pmatrix}$, which is equivalent to $c=0$ and $a\ne 0$. With $c=0$, \begin{align*} \det A &= \det \begin{pmatrix} a & b\\ 0 & d \end{pmatrix}\\ &=ad, \end{align*} so invertibility is equivalent to $ad\ne 0$, that is, $a\ne 0$ and $d\ne 0$. Hence \begin{align*} G_L = \left\{ \begin{pmatrix} a & b\\ 0 & d \end{pmatrix} : a,d\in\mathbb{R}\setminus\{0\},\ b\in\mathbb{R} \right\}. \end{align*} The action is transitive. Indeed, if $M\in\mathbb{RP}^1$, choose a nonzero vector $w\in M$ and choose $z\in\mathbb{R}^2$ not lying in $M$. Then $w,z$ are linearly independent, so the matrix $A$ with first column $w$ and second column $z$ is invertible. Since $Ae_1=w$, \begin{align*} A(L) &=\operatorname{span}(Ae_1)\\ &=\operatorname{span}(w)\\ &=M. \end{align*} Thus every projective line lies in the orbit of $L$. Now define \begin{align*} \Phi:GL(2,\mathbb{R})/G_L&\to \mathbb{RP}^1\\ AG_L&\mapsto A(L). \end{align*} By *Transitive Actions as Coset Actions*, this map is a well-defined bijection of $GL(2,\mathbb{R})$-sets. Therefore \begin{align*} \mathbb{RP}^1 \cong GL(2,\mathbb{R})/G_L \end{align*} as a $GL(2,\mathbb{R})$-set. The quotient records the fact that choosing the base line $L$ leaves exactly the upper triangular subgroup $G_L$ as residual symmetry. [/example] ### Isotropy in Geometry In Lie theory and differential geometry, the same coset description becomes the language of homogeneous spaces. To emphasize that the stabiliser is now a Lie subgroup controlling local geometry, it is given the geometric name isotropy subgroup. [definition: Isotropy Subgroup] Let $G$ be a Lie group, let $M$ be a smooth manifold, and let \begin{align*} G \times M &\to M \\ (g,p) &\mapsto g \cdot p \end{align*} be a smooth action. For $p \in M$, the isotropy subgroup at $p$ is the stabiliser \begin{align*} G_p = \{g \in G : g \cdot p = p\}. \end{align*} [/definition] The word isotropy emphasizes geometry rather than set theory. Once the orbit is regarded as a manifold, the natural question is whether the algebraic quotient $G/G_p$ is not only a set model for the orbit but also a smooth model for it. [quotetheorem:5003] This is the bridge from algebraic stabilisers to geometry. Choosing a point in a symmetric space replaces the space by a group modulo the symmetries that remain after the point is fixed. ## Counting with Stabiliser Data Many counting arguments ask for the number of orbits, not just the size of a single orbit. Stabilisers appear again through fixed-point counts: instead of fixing a point and asking which group elements stabilise it, we fix a group element and ask which points it stabilises. This reversal leads to Burnside's lemma, but it also creates a new local object. A stabiliser $G_x$ keeps the point $x$ fixed and varies the group element; now we keep the group element $g$ fixed and collect all points on which it acts as the identity. That collection is the fixed set of $g$, and it is the unit of data that Burnside's average will sum over. [definition: Fixed Set of a Group Element] Let $G$ be a group acting on a set $X$, and let $g \in G$. The fixed set of $g$ is \begin{align*} X^g = \{x \in X : g \cdot x = x\}. \end{align*} [/definition] Averaging the sizes of these single-element fixed sets should count orbits because points with larger stabilisers are seen more often in the fixed-set sum. The obstruction is that this sum counts fixed incidences rather than orbits directly, so we need a reason that each orbit contributes exactly the same normalized amount. Burnside's lemma supplies that double-counting bridge between fixed incidences and orbit count. [quotetheorem:3241] Burnside's lemma counts orbits by averaging stabilised points. The hidden double count is the set of pairs $(g,x)$ with $g \cdot x = x$: counted by group elements it gives fixed sets, and counted by points it gives stabilisers. [example: Necklaces with Two Colours and Three Beads] Let the two colours be $0$ and $1$, so a labelled colouring is a triple $(c_0,c_1,c_2)\in\{0,1\}^3$. Hence \begin{align*} |X|=|\{0,1\}^3|=2^3=8. \end{align*} Let $G=C_3=\{e,r,r^2\}$ act by cyclic rotation, with \begin{align*} r\cdot(c_0,c_1,c_2)&=(c_2,c_0,c_1),\\ r^2\cdot(c_0,c_1,c_2)&=(c_1,c_2,c_0). \end{align*} The identity fixes every colouring, so \begin{align*} X^e=X \end{align*} and therefore \begin{align*} |X^e|=8. \end{align*} Now compute the fixed colourings of $r$. A colouring $(c_0,c_1,c_2)$ lies in $X^r$ exactly when \begin{align*} r\cdot(c_0,c_1,c_2)&=(c_0,c_1,c_2),\\ (c_2,c_0,c_1)&=(c_0,c_1,c_2). \end{align*} Equality of triples gives \begin{align*} c_2&=c_0,\\ c_0&=c_1,\\ c_1&=c_2. \end{align*} Thus $c_0=c_1=c_2$, so the fixed colourings are precisely \begin{align*} (0,0,0)\quad\text{and}\quad(1,1,1). \end{align*} Hence \begin{align*} |X^r|=2. \end{align*} Similarly, a colouring lies in $X^{r^2}$ exactly when \begin{align*} r^2\cdot(c_0,c_1,c_2)&=(c_0,c_1,c_2),\\ (c_1,c_2,c_0)&=(c_0,c_1,c_2), \end{align*} which gives \begin{align*} c_1&=c_0,\\ c_2&=c_1,\\ c_0&=c_2. \end{align*} Again $c_0=c_1=c_2$, so \begin{align*} |X^{r^2}|=2. \end{align*} By *Burnside's Lemma*, the number of orbits, which is the number of necklaces up to rotation, is \begin{align*} \frac{1}{|G|}\sum_{g\in G}|X^g| &=\frac{1}{3}\bigl(|X^e|+|X^r|+|X^{r^2}|\bigr)\\ &=\frac{1}{3}(8+2+2)\\ &=\frac{12}{3}\\ &=4. \end{align*} The two constant colourings are exceptional because their stabiliser is all of $C_3$, while every nonconstant colouring is fixed by no non-identity rotation and therefore has stabiliser $\{e\}$. [/example] The necklace example uses Burnside's lemma globally, but the mechanism can be isolated on a single orbit. We need the following orbit-level identity to explain why each orbit contributes exactly one unit to Burnside's average. [quotetheorem:845] This theorem is a compact way to see why each orbit contributes exactly one to Burnside's average. It also reinforces the point that stabilisers are not auxiliary objects; they are the local weights that make orbit counting work. ## Normality, Kernels, and Faithfulness A stabiliser fixes one point. Sometimes we need the group elements that fix every point. This gives a subgroup of $G$ obtained by intersecting all point stabilisers, and it measures whether the action represents $G$ without collapsing any non-identity element. [definition: Kernel of an Action] Let $G$ be a group acting on a set $X$. The kernel of the action is \begin{align*} \ker(G \curvearrowright X) = \{g \in G : g \cdot x = x \text{ for all } x \in X\}. \end{align*} [/definition] Unlike a point stabiliser, the kernel is unchanged by conjugating inside $G$, because fixing every point is a global condition. The clean way to see the normality is to turn the action into a homomorphism. Define \begin{align*} \rho: G &\to \operatorname{Sym}(X),\\ \rho(g)(x)&=g\cdot x. \end{align*} The action law gives $\rho(gh)=\rho(g)\circ \rho(h)$, and the ordinary kernel of $\rho$ is exactly $\ker(G\curvearrowright X)$. Thus the action kernel is the kernel of a [group homomorphism](/page/Group%20Homomorphism), so the general kernel theorem below applies directly. [quotetheorem:788] Applied to the homomorphism $\rho$, the theorem proves that $\ker(G\curvearrowright X)$ is a [normal subgroup](/page/Normal%20Subgroup) of $G$. The same subset can also be read pointwise as the intersection of all point stabilisers: \begin{align*} \ker(G \curvearrowright X) = \bigcap_{x \in X} G_x. \end{align*} Point stabilisers need not be normal subgroups, even though the kernel always is. This distinction is a common source of mistakes: fixing one point is not invariant under changing basepoint unless the subgroup survives conjugation. [example: A Stabiliser Need Not Be Normal] Let $G=S_3$ act on $X=\{1,2,3\}$ by evaluation of permutations. For the point $1$, the stabiliser is \begin{align*} G_1 &=\{\sigma\in S_3:\sigma\cdot 1=1\}. \end{align*} The six elements of $S_3$ act on $1$ as follows: \begin{align*} e\cdot 1&=1,\\ (1\ 2)\cdot 1&=2,\\ (1\ 3)\cdot 1&=3,\\ (2\ 3)\cdot 1&=1,\\ (1\ 2\ 3)\cdot 1&=2,\\ (1\ 3\ 2)\cdot 1&=3. \end{align*} Hence exactly $e$ and $(2\ 3)$ fix $1$, so \begin{align*} G_1=\{e,(2\ 3)\}. \end{align*} This subgroup has order $2$. To show that $G_1$ is not normal in $S_3$, conjugate the non-identity element of $G_1$ by $\tau=(1\ 2)$. Since $\tau^{-1}=\tau$, we have \begin{align*} \tau(2\ 3)\tau^{-1} &=(1\ 2)(2\ 3)(1\ 2). \end{align*} Applying the rightmost factor first gives \begin{align*} 1&\mapsto 2\mapsto 3\mapsto 3,\\ 3&\mapsto 3\mapsto 2\mapsto 1,\\ 2&\mapsto 1\mapsto 1\mapsto 2. \end{align*} Thus this conjugate sends $1$ to $3$, sends $3$ to $1$, and fixes $2$, so \begin{align*} (1\ 2)(2\ 3)(1\ 2)=(1\ 3). \end{align*} But \begin{align*} (1\ 3)\notin \{e,(2\ 3)\}=G_1, \end{align*} and therefore \begin{align*} (1\ 2)G_1(1\ 2)^{-1} &=\{(1\ 2)e(1\ 2)^{-1},(1\ 2)(2\ 3)(1\ 2)^{-1}\}\\ &=\{e,(1\ 3)\}\\ &\ne G_1. \end{align*} So $G_1$ is not a normal subgroup of $S_3$. The other point stabilisers are obtained by the same definition: \begin{align*} G_2 &=\{\sigma\in S_3:\sigma\cdot 2=2\}\\ &=\{e,(1\ 3)\}, \end{align*} because $e$ and $(1\ 3)$ fix $2$, while $(1\ 2)$, $(2\ 3)$, $(1\ 2\ 3)$, and $(1\ 3\ 2)$ move $2$. Similarly, \begin{align*} G_3 &=\{\sigma\in S_3:\sigma\cdot 3=3\}\\ &=\{e,(1\ 2)\}. \end{align*} The kernel of the action consists of the permutations fixing every point, hence it is the intersection of the three point stabilisers: \begin{align*} \ker(S_3\curvearrowright X) &=G_1\cap G_2\cap G_3\\ &=\{e,(2\ 3)\}\cap\{e,(1\ 3)\}\cap\{e,(1\ 2)\}\\ &=\{e\}. \end{align*} Thus fixing one point can give a non-normal stabiliser, while fixing every point gives the global kernel of the action. [/example] When the kernel is identity-only, the action lets us view $G$ as a group of permutations of $X$ without losing information. This property is global: it asks whether each non-identity element moves at least one point somewhere in $X$. [definition: Faithful Action] Let $G$ be a group acting on a set $X$. The action is faithful if \begin{align*} \ker(G \curvearrowright X) = \{e\}. \end{align*} [/definition] Faithfulness is global; freeness is pointwise. To make the distinction precise, we package an action as a homomorphism into a symmetric group and read faithfulness as injectivity of that homomorphism. [quotetheorem:5004] This theorem is often the best conceptual definition of the kernel. A group action is a representation of $G$ by permutations, and the kernel consists of the elements represented by the identity permutation. ## Stabiliser in Linear and Representation-Theoretic Settings When $X$ has extra structure, stabilisers usually preserve that structure. In linear algebra and representation theory, the acted-on set may be a [vector space](/page/Vector%20Space), a set of vectors, a set of subspaces, or a set of tensors. The stabiliser then becomes the subgroup preserving the chosen object. The first version is the stabiliser of a vector under a linear representation. It records group elements that leave a vector unchanged, not merely elements that preserve its span. [definition: Stabiliser of a Vector] Let $G$ be a group, let $V$ be a vector space over a field $k$, and let \begin{align*} \rho: G &\to GL(V) \end{align*} be a representation. For $v \in V$, the stabiliser of $v$ is \begin{align*} G_v = \{g \in G : \rho(g)v = v\}. \end{align*} [/definition] This definition is the ordinary stabiliser for the induced action of $G$ on the underlying set of $V$. It becomes especially meaningful when $v$ encodes a structure, such as a [bilinear form](/page/Bilinear%20Form), a tensor, or a polynomial. [example: Stabiliser of a Vector in the Standard Representation] Let $G=GL(n,\mathbb{R})$ act on $V=\mathbb{R}^n$ by matrix multiplication, and let $e_1=(1,0,\dots,0)^\top$. We compute the stabiliser \begin{align*} G_{e_1} &=\{A\in GL(n,\mathbb{R}):Ae_1=e_1\}. \end{align*} Write the columns of $A$ as \begin{align*} A=\begin{pmatrix} | & | & & |\\ v_1&v_2&\cdots&v_n\\ | & | & & | \end{pmatrix}. \end{align*} Since $e_1=(1,0,\dots,0)^\top$, multiplication by $e_1$ selects the first column: \begin{align*} Ae_1 &= \begin{pmatrix} | & | & & |\\ v_1&v_2&\cdots&v_n\\ | & | & & | \end{pmatrix} \begin{pmatrix} 1\\ 0\\ \vdots\\ 0 \end{pmatrix}\\ &=1v_1+0v_2+\cdots+0v_n\\ &=v_1. \end{align*} Therefore $Ae_1=e_1$ exactly when the first column of $A$ is $e_1$. Thus any matrix in $G_{e_1}$ has the form \begin{align*} A= \begin{pmatrix} 1 & a\\ 0 & B \end{pmatrix}, \end{align*} where $a\in \mathbb{R}^{1\times(n-1)}$, the lower-left block is the zero column in $\mathbb{R}^{n-1}$, and $B$ is an $(n-1)\times(n-1)$ real matrix. Since $A\in GL(n,\mathbb{R})$, it must be invertible. Expanding the determinant along the first column gives \begin{align*} \det A &= \det \begin{pmatrix} 1 & a\\ 0 & B \end{pmatrix}\\ &=1\cdot \det(B)\\ &=\det(B). \end{align*} Hence $\det A\ne 0$ exactly when $\det(B)\ne 0$, so the invertibility condition is precisely $B\in GL(n-1,\mathbb{R})$. Conversely, if \begin{align*} A= \begin{pmatrix} 1 & a\\ 0 & B \end{pmatrix} \quad\text{with}\quad B\in GL(n-1,\mathbb{R}), \end{align*} then $\det A=\det(B)\ne 0$, so $A\in GL(n,\mathbb{R})$, and \begin{align*} Ae_1 &= \begin{pmatrix} 1 & a\\ 0 & B \end{pmatrix} \begin{pmatrix} 1\\ 0\\ \vdots\\ 0 \end{pmatrix}\\ &= \begin{pmatrix} 1\\ 0\\ \vdots\\ 0 \end{pmatrix}\\ &=e_1. \end{align*} Therefore \begin{align*} G_{e_1} = \left\{ \begin{pmatrix} 1 & a\\ 0 & B \end{pmatrix} : a\in \mathbb{R}^{1\times(n-1)},\ B\in GL(n-1,\mathbb{R}) \right\}. \end{align*} This subgroup fixes the vector $e_1$ itself; it does not merely preserve the line spanned by $e_1$. [/example] A different problem arises when the object of interest is a subspace rather than a vector. Then the stabiliser should allow motion inside the subspace but forbid moving the subspace elsewhere; this is the right notion when the geometry depends on directions, planes, or flags rather than individual vectors. [definition: Stabiliser of a Subspace] Let $G$ be a group, let $V$ be a vector space over a field $k$, and let \begin{align*} \rho: G &\to GL(V) \end{align*} be a representation. For a subspace $W \subset V$, the stabiliser of $W$ is \begin{align*} G_W = \{g \in G : \rho(g)(W) = W\}. \end{align*} [/definition] The difference between fixing vectors and preserving subspaces is substantial. A group element may scale every vector in a line and hence preserve the line, while not fixing any nonzero vector on that line. [example: Vector Stabiliser Versus Line Stabiliser] Let $G=GL(2,\mathbb{R})$ act on $\mathbb{R}^2$ by matrix multiplication. Let \begin{align*} v=e_1= \begin{pmatrix} 1\\ 0 \end{pmatrix} \quad\text{and}\quad L=\operatorname{span}(e_1). \end{align*} We first compute the stabiliser of the vector $v$. Write \begin{align*} A= \begin{pmatrix} a & b\\ c & d \end{pmatrix} \in GL(2,\mathbb{R}). \end{align*} Then \begin{align*} Av &= \begin{pmatrix} a & b\\ c & d \end{pmatrix} \begin{pmatrix} 1\\ 0 \end{pmatrix}\\ &= \begin{pmatrix} a\\ c \end{pmatrix}. \end{align*} Thus $A$ fixes $v$ exactly when \begin{align*} Av&=v,\\ \begin{pmatrix} a\\ c \end{pmatrix} &= \begin{pmatrix} 1\\ 0 \end{pmatrix}, \end{align*} which is equivalent to \begin{align*} a=1 \quad\text{and}\quad c=0. \end{align*} So every element of $G_v$ has the form \begin{align*} \begin{pmatrix} 1 & b\\ 0 & d \end{pmatrix}. \end{align*} Since this matrix must lie in $GL(2,\mathbb{R})$, \begin{align*} \det \begin{pmatrix} 1 & b\\ 0 & d \end{pmatrix} &=1\cdot d-b\cdot 0\\ &=d \end{align*} must be nonzero. Conversely, if $d\ne 0$, then the matrix is invertible and \begin{align*} \begin{pmatrix} 1 & b\\ 0 & d \end{pmatrix} \begin{pmatrix} 1\\ 0 \end{pmatrix} &= \begin{pmatrix} 1\\ 0 \end{pmatrix} =v. \end{align*} Therefore \begin{align*} G_v = \left\{ \begin{pmatrix} 1 & b\\ 0 & d \end{pmatrix} : b\in\mathbb{R},\ d\in\mathbb{R}\setminus\{0\} \right\}. \end{align*} Now compute the stabiliser of the line $L$. Since \begin{align*} L=\left\{ \lambda \begin{pmatrix} 1\\ 0 \end{pmatrix} :\lambda\in\mathbb{R} \right\}, \end{align*} we have \begin{align*} A(L) &= \left\{ A\left(\lambda \begin{pmatrix} 1\\ 0 \end{pmatrix}\right) :\lambda\in\mathbb{R} \right\}\\ &= \left\{ \lambda A \begin{pmatrix} 1\\ 0 \end{pmatrix} :\lambda\in\mathbb{R} \right\}\\ &= \left\{ \lambda \begin{pmatrix} a\\ c \end{pmatrix} :\lambda\in\mathbb{R} \right\}\\ &= \operatorname{span} \begin{pmatrix} a\\ c \end{pmatrix}. \end{align*} Thus $A(L)=L$ exactly when \begin{align*} \operatorname{span} \begin{pmatrix} a\\ c \end{pmatrix} = \operatorname{span} \begin{pmatrix} 1\\ 0 \end{pmatrix}. \end{align*} This happens precisely when $c=0$ and $a\ne 0$: if the spans are equal, then $\begin{pmatrix}a\\c\end{pmatrix}\in L$, so $c=0$, and $a\ne 0$ because $A$ is invertible and therefore its first column is nonzero; conversely, if $c=0$ and $a\ne 0$, then \begin{align*} \operatorname{span} \begin{pmatrix} a\\ 0 \end{pmatrix} = \operatorname{span} \begin{pmatrix} 1\\ 0 \end{pmatrix}. \end{align*} With $c=0$, \begin{align*} \det \begin{pmatrix} a & b\\ 0 & d \end{pmatrix} &=a d-b\cdot 0\\ &=ad, \end{align*} so invertibility is equivalent to $a\ne 0$ and $d\ne 0$. Hence \begin{align*} G_L = \left\{ \begin{pmatrix} a & b\\ 0 & d \end{pmatrix} : a,d\in\mathbb{R}\setminus\{0\},\ b\in\mathbb{R} \right\}. \end{align*} Finally, consider \begin{align*} D= \begin{pmatrix} 2&0\\ 0&1 \end{pmatrix}. \end{align*} Then \begin{align*} Dv &= \begin{pmatrix} 2&0\\ 0&1 \end{pmatrix} \begin{pmatrix} 1\\ 0 \end{pmatrix}\\ &= \begin{pmatrix} 2\\ 0 \end{pmatrix} \ne \begin{pmatrix} 1\\ 0 \end{pmatrix}, \end{align*} so $D\notin G_v$. But \begin{align*} D(L) &= \operatorname{span} \begin{pmatrix} 2\\ 0 \end{pmatrix}\\ &= \operatorname{span} \begin{pmatrix} 1\\ 0 \end{pmatrix} =L, \end{align*} so $D\in G_L$. The same matrix can preserve the line through $e_1$ without fixing the vector $e_1$, so the acted-on set must be specified before a stabiliser can be interpreted. [/example] Representation theory also uses stabilisers through induced representations, little groups, and orbit methods. In each case the stabiliser is not just a bookkeeping device: it is the subgroup from which local data are extended to the larger group. For example, an induced representation starts with a representation of a subgroup $H \le G$ and builds a representation of $G$; geometrically, $H$ is often the stabiliser of a point in a transitive $G$-set. ## Beyond and Connected Topics Stabilisers are part of the general theory of [group actions](/page/Group%20Action), where they pair with orbits, kernels, and fixed-point sets. The [orbit-stabiliser theorem](/theorems/796) is the first major bridge from action language to subgroup counting, and it underlies many applications in finite group theory. In [Cambridge IA Groups](/page/Cambridge%20IA%20Groups), stabilisers appear in the standard development of group actions, conjugacy, centralisers, and counting arguments. That course is the natural place to see stabilisers used in Sylow theory, class equations, and permutation actions. In [Cambridge II Representation Theory](/page/Cambridge%20II%20Representation%20Theory), stabilisers reappear when groups act linearly on vector spaces or on bases, characters, and subspaces. The stabiliser of a vector or subspace often controls how a representation decomposes or how a representation is induced from a subgroup. In [Cambridge II Algebraic Topology](/page/Cambridge%20II%20Algebraic%20Topology), stabilisers enter through group actions on topological spaces, covering spaces, deck transformations, and group actions on simplicial or CW complexes. Free actions are especially important because quotient spaces behave better when point stabilisers are identity-only. In [Cambridge III Differential Geometry](/page/Cambridge%20III%20Differential%20Geometry), stabilisers are usually called isotropy subgroups. They describe homogeneous spaces $G/H$, symmetric spaces, and the local geometry of group orbits on manifolds. A final direction is geometric invariant theory, where stabilisers measure how far an orbit is from being stable under an algebraic group action. There the dimension and structure of stabilisers become part of the geometry of moduli spaces. ## References Androma, [Group Action](/page/Group%20Action). Androma, [Cambridge IA Groups](/page/Cambridge%20IA%20Groups). Androma, [Cambridge II Algebraic Topology](/page/Cambridge%20II%20Algebraic%20Topology). Androma, [Cambridge II Representation Theory](/page/Cambridge%20II%20Representation%20Theory). Androma, [Cambridge III Differential Geometry](/page/Cambridge%20III%20Differential%20Geometry). J. J. Rotman, *An Introduction to the Theory of Groups* (1995). J.-P. Serre, *Linear Representations of Finite Groups* (1977). M. Artin, *Algebra* (1991).