# Abstract Nonsense

## 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.

Motivation

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.

Orbits

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

$\text{ }$

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

$\text{ }$

$\text{ }$

References:

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 -

1. […] By the Orbit Decomposition Theorem we know […]

Pingback by Review of Group Theory: Group Actions (Pt. IV Conjugation and the Class Equation) « Abstract Nonsense | January 6, 2011 | Reply

2. […] verified to be a group action with the orbit of equal to and isotropy subgroups . Thus, by earlier theorem we know that for every . If we say that is conjugate to . So, with one small lemma we are ready […]

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3. […] by the Orbit Decomposition Theorem […]

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4. […] but this clearly implies that , which is a contradiction. It follows then that for every with we have that . But, by the Orbit-Stabilizer Theorem […]

Pingback by Review of Group Theory: Alternate Proof of the Sylow Theorems (Pt. II) « Abstract Nonsense | January 12, 2011 | Reply

5. […] both sides and recalling by the orbit stabilizer theorem that we see […]

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6. […]  By first principles we know that is non-trivial. So, we may choose a non-trivial . From the orbit stabilizer theorem we know that where is the conjugacy class containing and ‘s centralizer. It follows from […]

Pingback by Representation Theory: Burnside’s Theorem « Abstract Nonsense | March 11, 2011 | Reply

7. […] to the conjugacy class of since it is the only trivial conjugacy class). Thus, . But, by the orbit-decomposition theorem that and so doing the simple computation we find that . To find we recall (considering the […]

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8. […] Let be a finite group which acts on the finite set , suppose that breaks up into oribts as where is transveral for the set of orbits . Then, if is permutation representation of on […]

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9. […] by the orbit-stabilizer theorem we have that . What we claim though is that . Indeed, if then for some and so and thus […]

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10. […] We use the orbit-stabilizer theorem we write […]

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11. […] commutes with every element of and since we have by b) that . Recall though that by the orbit-stabilizer theorem that and since we have that for some , and so if we must have that […]

Pingback by University of Maryland College Park Qualifying Exams (Group Theory and Representation Theory) (January-2004) (Pt. I) « Abstract Nonsense | May 6, 2011 | Reply

12. […] and the associated orbit under the action of on induced by . Then, is […]

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13. […] We thus see that must map a -cycle to a -cycle for which there are ways to do. Thus, if we fix a particular permutation of the -cycles then choosing a particular -cycle we see that must only cyclically permute the elements of this cycle, for which there are obviously ways to do, and thus within each fixed choice of permutations of the -cycles there are choices for ‘s action. Thus, for each there are choices and thus the total number of choices for is . Thus, we may conclude that from where the conclusion follows by the orbit-stabilizer theorem. […]

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14. […] for the action of on . We can obviously partition into the two sets of all with and . From the Orbit Decomposition Theorem we know […]

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