Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The Orbit-Stabilizer says that, given a group $G$ which acts on a set $X$, then there exists a bijection between the orbit of an element $x\in X$ and the set of left cosets of the stabilizer group of $x$ in $G$. In other words, that the cardinality of the orbit of an element $x\in X$ is equal to the index of its stabilizer subgroup in $G$.

I've seen two different texts present this, both of which explicitly say that this captures a very intuitive idea. I'm sorry if it's obvious, but I don't see the intuition behind this.

I've asked a few questions looking for intuition now, and have received outstanding advice. As such, again I'm looking to the community to share some of their insights on this idea, and how they think of this theorem. As always, any help is greatly appreciated. Thanks!

share|cite|improve this question
Additional comment: I'm really trying to view this in relation to the cosets of $K$. So would I be right in saying, if one element of a coset of $K$ advances and element $x$ through its orbit, then every element of said coset does. Similarly, would I be right in saying that the stabilizer subgroup of $x$ in $G$ is a union of left cosets of $K$? Because then the theorem would follow from 'grouping' cosets of $K$ – user45793 Nov 23 '12 at 4:46
It's not exactly what you asked for, but I wrote something about intuition about Orbit-Stabilizer here that you may be interested in reading. – Alexander Gruber Nov 23 '12 at 5:40
@AlexanderGruber thanks for the link. Definitely is helping me piece things together – user45793 Nov 23 '12 at 7:29
up vote 3 down vote accepted

The statement of the the theorem that I know says that if we partition $G$ into cosets of some $G_x,$ then we get an isomorphism of $G$-sets $G/G_x\to O(x),aG_x\mapsto a\cdot x.$

Here is my naive way of thinking about it. Given a $G$-space $X$ and finite $G,$ the orbit-stabilizer theorem implies the "counting theorem" $|G_x||O(x)|=|G|.$ Since the size of $G$ is fixed, this is telling us that the larger the orbit is, the fewer elements of $G$ can possibly fix $x,$ while the smaller the orbit is, the more elements of $G$ must fix $x.$ This seems pretty intuitive, but it's not obvious. For example, we might think it possible that $|O(x)|$ is small, while many elements of $G$ act nontrivially but in the same way on $x.$ The orbit-stabilizer theorem disallows this, letting us determine the size of the stabilizer from the size of the corresponding orbit, and vice-versa.

share|cite|improve this answer

This is not my answer, but Gowers explains the orbit-stablizer theorem in an excellent way.

Think about the symmetric group of a cube. Call this group $G$ and it acts on the cube. If you want to know how many elements there are in $G$, then you can think about it in two steps.

Step I: If you fix one face, there are 4 ways to move the cube because you can only rotate the cube now. (These are the stabilizers )

Step II: There are six possible choice where this face can go. (Orbit of the face).

So you figure out $|G|=4\cdot 6$. That is the intuition.

share|cite|improve this answer
Hmm yes I've read through his blog on it. Thanks for the link! – user45793 Nov 23 '12 at 7:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.