I don't fully understand the definition of an Orbit. Mathematically, it is given by

$$ \operatorname{Orb}(x) = \{y = gx \mid g \in G\} $$

where $G$ is a group and $x \in X$, a set that is acted upon by the group $G$, but what does this actually mean? Is it the set of all elements $x$ after having been acted upon by some element $g$?

The stabilizer is given by $\{g \in G \mid gx = x\}$. So does this mean the set of all elements $g$, when after actings upon $x$ give you the element $x$? That is, the set of all elements $g$ whose action on $x$ doesn't change it?

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    $\begingroup$ Are you familiar with orbits of a permutation on a set S? That serves as a nice example what the orbits of a permutation mean. $\endgroup$
    – amWhy
    Dec 7, 2012 at 17:26
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    $\begingroup$ I think that the term "orbit" comes from the action of the rotation group on $\mathbb{R}^2$. $\endgroup$ Dec 7, 2012 at 17:42
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    $\begingroup$ @amWhy: What do you think about the notation of action teh OP used? I think $y=x^g$ could be better than $y=gx$. $\endgroup$
    – Mikasa
    Dec 7, 2012 at 17:52
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    $\begingroup$ @rschwieb: Well, if you identify $\mathbb{R}^2$ with $\mathbb{C}$ and consider the group $\mathbb{S}^1=\{z\in \mathbb{C}| \lvert z \rvert=1\}$ then you have an obvious group action of $\mathbb{S}^1$ on $\mathbb{C}$ such that the orbit of any point $z\ne 0$ is the circle with center at the origin and passing through $z$. That is, if $0$ is the Sun and $z$ is the Earth, the group-theoretic "orbit" is the same as the astronomical "orbit". I guess that the origin of the term comes from this, but it is only my guess. $\endgroup$ Dec 7, 2012 at 17:53
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    $\begingroup$ @Mikasa, re, certainly one can reasonably prefer ${^gx}$ or $x^{g^{-1}}$ for $gx$, but not, I would argue, $x^g$ for $gx$. In fact, Serre specifically calls this out on p. 42 of Cohomologie galoisienne for giving rise to "l'horrible formule $x^{(s t)} = (x^t)^s$." $\endgroup$
    – LSpice
    Mar 13, 2022 at 0:09

8 Answers 8


The orbit of $x$ is "everything that can be reached from $x$ by an action of something in $G$."

The stabilizer of $x$ is "the set of all elements of $G$ which don't move $x$ when they act on $x$".

Those already seem pretty intuitive... what else can be said?

I guess you might want to look at orbits and stabilizers for particular actions. For example, if a group is acting on itself by conjugation, then the orbit of an element is that element's conjugacy class. One element stabilizes another in this action exactly when they commute.


Some examples may illuminate these definitions.

Let $G$ be the circle group, $$G = \{z\in\mathbb{C}: |z|=1\}.$$ This is the unit circle in $\mathbb{C}$, and it is a group under multiplication. By the same token, $G$ acts on $X = \mathbb{C}$ by multiplication. Geometrically, multiplication by $z = e^{i\theta}$ acts on $\mathbb{C}$ by rotation by the angle $\theta$. Now if we fix some $x\in\mathbb{C}$, the orbit through $x$ will be exactly the circle of radius $|x|$ centred at the origin -- unless $x=0$, in which case the orbit will just be the point $\{0\}$.

How about stabilizers? Let $x\in\mathbb{C}$, and let $G^x$ be its stabilizer. If $x\neq 0$, then $zx = x$ if and only if $z=1$, and so $G^x=\{1\}$. On the other hand, if $x=0$ then $zx = x$ for all $z\in G$, and so $G^x=G$.

Here's an example with more interesting stabilizers. Let $G$ now be the dihedral group of order 8,

$$G = \{\rho,\tau: \rho^4=e, \tau^2=e, \tau\rho\tau = \rho^{-1}\}.$$

This acts on the set $X = \{1,2,3,4\}$ of vertices of a square

enter image description here

with $\rho$ acting by rotation counterclockwise by $\frac \pi 4$ radians and $\tau$ acting by reflection through the red line. In this way, the elements of $G$ consist of four rotations (through $0, \frac\pi 4, \frac\pi 2$, and $\frac{3\pi} 4$ radians) and four reflections (horizontally, vertically and through each of the two diagonals). Note that each vertex can be sent to any other given vertex by an appropriate rotation, and so the orbit of each vertex is $\{1,2,3,4\}$. What is the stabilizer of, say, $3$? None of the nontrivial rotations fix $3$, and the only reflection that fixes $3$ is $\tau$, and so the stabilizer of $3$ is $\{e,\tau\}$.

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    $\begingroup$ I like your illuminating examples. $\endgroup$
    – Mikasa
    Dec 7, 2012 at 17:54
  • $\begingroup$ Thank you for your examples. It helped me $\endgroup$
    – user296169
    Aug 27, 2017 at 15:48

A bit about group actions:

In algebra and geometry, a group action is a description of symmetries of objects using groups. The essential elements of the object are described by a set, and the symmetries of the object are described by the symmetry group of this set, which consists of bijective transformations of the set. In this case, the group is also called a permutation group (especially if the set is finite or not a vector space) or transformation group (especially if the set is a vector space and the group acts like linear transformations of the set).

A group action is an extension to the definition of a symmetry group in which every element of the group "acts" like a bijective transformation (or "symmetry") of some set, without being identified with that transformation. This allows for a more comprehensive description of the symmetries of an object, such as a polyhedron, by allowing the same group to act on several different sets of features, such as the set of vertices, the set of edges and the set of faces of the polyhedron.

If $G$ is a group and $X$ is a set then a group action may be defined as a group homomorphism $h$ from $G$ to the symmetric group of $X$. The action assigns a permutation of $X$ to each element of the group in such a way that the permutation of X assigned to the identity element of $G$ is the identity (do-nothing) transformation of $X$; a product gh of two elements of $G$ is the composition of the permutations assigned to $g$ and $h$.

Since each element of $G$ is represented as a permutation, a group action is also known as a permutation representation.

See also Gowers's blog entry for a "down to earth" discussion on group actions


The defining properties of a group guarantee that the set of orbits of (points x in) $X$ under the action of $G$ form a partition of X. The associated equivalence relation is defined by saying $x \sim y$ if and only if there exists a $g \in G$ with $gx = y.$ The orbits are then the equivalence classes under this relation; two elements $x$ and $y$ are equivalent if and only if their orbits are the same; i.e., $Gx = Gy$.

If you have the spare time, you might find this You Tube video, Orbits of group action helpful.

Fixed points and stabilizer subgroups

Given $g \in G$ and $x \in X$ with $gx=x$, we say $x$ is a fixed point of $g$ and $g$ fixes $x$.

For every $x \in X$, we define the stabilizer subgroup of $x$ as the set of all elements in $G$ that fix $x$:

$$G_x = \{g\in G\mid gx = x\}.$$

$G_x$ is a subgroup of $G$, though typically not a normal one.

Also, Wikipedia discusses orbits and stabilizers and how they relate, in its "Group Action" entry.

See also Gowers's follow-up blog entry for a "down to earth" discussion on group actions, orbits, and stabilizers:



Consider a sphere $S \subset \mathbb{R^3}$ and a group $G$ of (all) rotations along the OZ axis (north-south pole, as Earth).

$\hspace{70pt}$enter image description here

For every angle $\alpha \in [0, 2\pi)$ there is an element of the group $g_\alpha$ that would map a point $x$ of the sphere to the point $y = g_\alpha\cdot x$ such that the directed angle between would be $\alpha$. Observe that $S$ is a 3d object, where the group is almost like $[0,2\pi)$ - those are two different things, yet are joined by the "$\cdot$" operator, i.e. the group $G$ acts on sphere $S$.

What is an orbit then? See for yourself: $Orb(x) = \{ gx \mid g \in G\}$. That means the orbit of a point $x$ of the sphere $S$ is a set of all the points that a $G$ (i.e. rotations) can produce from it, in our case a circle, like the dotted one on the picture. Can you guess why it is called an orbit?


Stabilizers are subgroups that won't disturb your points. In our sphere example, we have two kind of points: poles and non-poles. Does any rotation moves a pole? No, so $G_{\text{pole}} = G$. However, every other point would be disturbed by any non-zero rotation, hence the stabilizer of a non-pole point is a trivial group, $G_{\text{non-pole}} = \{1\}$.

In a more involved example, group of permutations of $\{A,B,C\}$, we can pick some element, let it be $B$, and still have a non-trivial stabilizer $G_{B} = \{(A)(B)(C), (AC)(B)\} \sim \mathbb{Z}_2$.

In conclusion: stabilizer of $x$ is the biggest subgroup that won't disturb your $x$.

I hope it explained something ;-)


If you think of an action of a group $G$ on a set $X$ as a assigning to each $g \in G$ a mapping of $X$ to itself, that is to say for each $g \in G$ we have a function $g: X \rightarrow X$ then the orbits and stabilisers take on an intuitive meaning.

The orbit of $x\in X$, $Orb(x)$ is the subset of $X$ obtained by taking a given $x$, and acting on it by each element of $G$. It is not the set of all elements $x$ after being acted on by some element $g$, that would be the image of $g$ when considered as a mapping, written $Im(g)$ or $g(X)$. It is different for each $x$, and can be thought of as the set of all points that you could possibly send $x$ to if you are allowed to act on $x$ by any element of $G$.

The stabiliser of $x$, $Stab(x)$ is what you say it is. For a given $x$, it is the set of elements of $G$ which map $x$ to $x$, i.e. the ones that don't change it.

In summary, $Orb(x) \subset X$ and is called the orbit because it is the elements of $X$ that can be reached by acting on $x$ by some element of $G$. $Stab(x) \subset G$ and in fact is the subgroup (Try to prove that for yourself!) of $G$ that acts trivially on $x$ by sending it to itself. Notice that the intersection over all $x \in X$ of $Stab(X)$ is the set of all elements of $G$ that map every element of $X$ to itself, i.e. the set of elements of $G$ that act on $X$ as the identity function.


enter image description here

The group action ultimately is a map which takes a group element of a group G and element of a set X and maps back into elements of the set X (*). So, I have made two copies of the set and drawn the group actions as the threads connecting elements in each of the two sets.

Considering an element $ p \in X$ , those group elements which preserve $p$ through the group action are the stabilizers of $p$. This is what is shown in the left picture. Note that stabilizer is a subset of the group set.

On the right, we have the orbit of the same point $p \in X$. These are all the points which $p$ can be shot into other than itself in $X$ by the action of the group elements. Note that orbit is a subset of the set X.

Note that the orbit is a subset of $X$ while the stabilizer is a subset of $G$.

A few points to 'develop' interest

  1. All elements in the stabilizer correspond to one element in the orbit, 'x' itself!

  2. It turns out $X$ can be partitioned into disjoint orbits of it's elements! (Crazy..right?)

  3. Stabilizers turn out to be a subgroup!

  4. The orbit stabilizer theorem states that the product of the number of threads which map an element into itself (size of stabilizer set) and number of threads which push that same element into different elements (orbit) equals the order of the original group!

*:(...with some reasonable conditions on how the group acts on X)

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    $\begingroup$ Nice illustration! I am fond of doing Maths with shapes, figures,... +) $\endgroup$
    – Mikasa
    Mar 14, 2022 at 10:12

Ok so a group action on a set is basically a way of assigning for each element of the group a way to "mix up" the elements of the set.

For example when you take the group to be symmetries of a shape and the set to be "corners" of the shape then after performing each symmetry you jumble up the corners (and this is done in a consistent way).

The stabilizer of a specific thing in the set are the elements of the group that fix that object under the corresponding "mixing up".

The orbit of something in the set are the possible things the group "mixes it up" to.


The simplest (and perhaps historically the first?) example of action is the natural "action" of the set (forget about abstract groups here) of all the bijections on a finite set $X$, on $X$ itself. Take $|X|=n$ and $S_X$ the set (set) of all the bijections on $X$; then, bacause of the definition/properties of a bijection on one same set and the definition of composition between maps, we have: i) $Id_X(x)=x$ for every $x\in X$, and ii) $f(g(x))=(fg)(x)$ for every $f,g\in S_X$ and $x\in X$. In this case, the orbit of $x\in X$ under the action of $S_X$ is given by the set of the images of $x$ under all the bijections on $X$: of course we can always define a bijection $f$ sending $x$ to every other $y\in X$, by $(f(x)=y) \wedge (f(y)=x) \wedge (f(z)=z$ for $z\ne x,y)$: so we expect that the orbit of every $x\in X$ is the whole $X$ itself. And this what the orbit-stabilizer theorem confirms: the stabilizer of every $x\in X$ has cardinality (forget about orders here) $(n-1)!$, because this is the cardinality of the subset of $S_X$ made of the bijections which fix any single point; thence, the orbit by every $x\in X$ has size $n!/(n-1)!=n$, i.e. it is the whole $X$. So, the "prototypical" action turns out to be transitive.


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