Abstract Nonsense

Crushing one theorem at a time

Review of Group Theory: Group Actions (Pt. II Orbits and the Orbit Decomposition Theorem)

Point of post: In this post we discuss more in-depth the concept of the orbits of a G-action and show how it carves up a group into a partition, which for finite groups gives us a lot of information.


In prior posts we’ve discussed Lagrange’s theorem and some of the profound consequences it can have on the study of finite groups. Look closely though and one will see that Lagrange’s theorem was really just a consequence of group actions. Namely, we had G was a finite group and H\leqslant G then H acted naturally on G by h\cdot g=hg. One can quick check that the orbits of these actions are the right cosets. The reason why this action was powerful was the way in which it carved up G into disjoint subsets, which we could exploit to tell us interesting things combinatorially/number theoretically about the order of G. It turns out that this wasn’t a coincidence. In particular, every G-action on a set S carves S up in a particularly nice way. This post will be devoted to studying in which way precisely this is.


Recall that if S is a G-space with G-action \cdot then for s\in S we define the orbit of S to be

\text{ }

\mathcal{O}_s=\left\{g\cdot s:g\in G\right\}

 \text{ }

and the isotropy subgroup of s to be

\text{ }

G_s=\left\{g\in G:g\cdot s=s\right\}

 \text{ }

(also known as the stabilizer subgroup and denoted \text{stab}(s)). We begin our discussion by showing that every G-space can be decomposed by its orbits in a way made precise by the theorem. Namely:

Theorem(Orbit Decomposition Theorem): Let S be a G-space with G-action \cdot. Then, the relation

\text{ }

x\sim y\Leftrightarrow \text{there exists }g\in G\text{ such that }y=gx

\text{ }

is an equivalence relation on S and [s]=\mathcal{O}_s where [s] is the equivalence class of s under \sim. In particular if \text{Orb}\left(S\right) denotes the set of (distinct) orbits on S under \cdot then

\text{ }

\displaystyle S=\biguplus_{\mathcal{O}_s\in\text{Orb}\left(S\right)}\mathcal{O}_s

\text{ }

Proof: To see that \sim is an equivalence relation it suffices, by definition, to show that it’s reflexive, symmetric, and transitive. The first of these is clear since x=e\cdot x and so x\sim x. Suppose next that x\sim y then there exists g\in G such that y=g\cdot x, then g^{-1}\cdot y=x and so y\sim x. Finally, if x\sim y and y\sim z then there exists g,g'\in G such that y=g\cdot x and z=g'\cdot y and so z=g'\cdot y=g'\cdot (g\cdot x)=(g'g)\cdot x and so x\sim z. It  follows that \sim is an equivalence relation.

\text{ }

To see that [s]=\mathcal{O}_s we merely note that

\text{ }

\displaystyle \begin{aligned}[s] &= \left\{x\in S:s\sim x\right\}\\ &= \left\{x\in S:\text{ there exists }g\in G\text{ such that }x=g\cdot s\right\}\\ &= \left\{g\cdot s:g\in G\right\}\\ &= \mathcal{O}_s\end{aligned}

 \text{ }

from where the conclusion follows. This last part follows since \text{Orb}\left(S\right) being the set of equivalence classes of an equivalence relation forms a partition of S. \blacksquare

Corollary: Let S be a finite G-space then

\text{ }

\displaystyle \#\left(S\right)=\sum_{\mathcal{O}_s\in\text{Orb}\left(S\right)}\#\left(\mathcal{O}_s\right)

 \text{ }

\text{ }

The problem with the above theorem (ostensibly) is that it may, in most cases, be nigh impossible to calculate \#\left(\mathcal{O}_s\right). Our next theorem says that this is not the case, in fact calculating \#\left(\mathcal{O}_S\right) is as easy as calculating the index of a series of subgroups of G. More precisely:

\text{ }

Theorem(Orbit-Stabilizer Theorem): Let S be G-space with G-action \cdot. Then, for each fixed s\in S the mapping

\text{ }

f:\mathcal{O}_s\to G/G_s:g\cdot s\mapsto gG_s

\text{ }

is a well-defined bijection (where here G/G_s just denotes the set of left cosets, it isn’t endowed with a group structure since it’s feasible that G_s is not normal in G).

Proof: The issue for why f might not be well-defined is that it’s conceivable that g\ne g' yet g\cdot s=g'\cdot s and so it’s theoretically possible that g\mathcal{O}_s\ne g'\mathcal{O}_s. To see that this can’t happen we merely note that if g\cdot s=g'\cdot s then \left(g^{-1}g'\right)\cdot s=s and thus g^{-1}g'\in G_s and therefore

\text{ }

g\left(g^{-1}g'\right)=g'\in gG_s

 \text{ }

and thus gG_s=g'G_s. Moreover, to see that f is injective note that if gG_s=g'G_s then g=g'y for some y\in G_s and thus

\text{ }

g\cdot s=(g'y)\cdot s=g'\cdot\left(y\cdot s\right)=g'\cdot s

 \text{ }

Noting that f is clearly surjective finishes the argument. \blacksquare

Corollary: Let S be a finite G-space with G-action \cdot. Then, if  \Gamma is a set of one element from each distinct member of \text{Orb}\left(S\right) then

\text{ }

\displaystyle \#\left(S\right)=\sum_{s\in\Gamma}\left(G:G_s\right)

\text{ }

and if G is finite

\text{ }

\displaystyle \#(S)=|G|\sum_{s\in\Gamma}\frac{1}{|G_s|}

\text{ }

\text{ }


1.  Lang, Serge. Undergraduate Algebra. 3rd. ed. Springer, 2010. Print.

2. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 200


January 5, 2011 - Posted by | Algebra, Group Theory, Uncategorized | , , , , ,


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