Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

After reading some notes about permutation groups, I have tackled with this really simple question I hope it is not a ridiculous question. :)

From the first chapters of any book about these kinds of groups and knowing how a group $G$ acts on a set $\Omega$, we are faced to an important structure, named $G_\alpha$ where is $\alpha \in \Omega$.

Why do we frequently regard $G_\alpha$ acting on set $\Omega-\{\alpha\}$. Why this set,$\Omega-\{\alpha\}$? Of course with this assumption a majority of theorems will be carried out properly and desirably. Thanks.

share|cite|improve this question
What is $G_{\alpha}$? – Qiaochu Yuan Jun 27 '12 at 21:07
Presumably, it is the stabilizer of the element $\alpha$ wrt to the action of $G$ on $\Omega$ – mixedmath Jun 27 '12 at 21:07
Well, doesn't it act on $\Omega\setminus\{\alpha\}$? It acts on $\Omega$, because all of $G$ does, and throwing away the fixed point doesn't change that fact? When dealing with $G_\alpha$ this action becomes interesting basically because that group does nothing to elements outside $\Omega\setminus\{\alpha\}$. – Jyrki Lahtonen Jun 27 '12 at 21:13
Not sure I understand the reason for the question, but $G_{\alpha}$ fixes $\alpha,$ so permutes the remaining elements of $\Omega$ among themelves. In general, there is no a priori reason to suppose that $G_{\alpha}$ acts on any proper subset of $\Omega \backslash \{ \alpha \},$ and if $G$ acts doubly transitive on $\Omega$ it won't. – Geoff Robinson Jun 27 '12 at 21:15
Babak, you often consider $S_5$ acting on the set $\{1,2,3,4,5\}$, right? You are not interested about the numbers $6,7,8,\ldots$ then, because the group "does nothing" there. It's the same thing here - you leave out the element the group $G_\alpha$ does nothing to. Sorry, I don't quite understand, what the problem is either. – Jyrki Lahtonen Jun 27 '12 at 21:28
up vote 1 down vote accepted

Consider two transitive groups on the same set: $G = \langle (1,2,3) \rangle$ and $H=\langle (1,2), (1,2,3) \rangle$ both acting on $\Omega = \{1,2,3\}$.

How well do $G$ and $H$ swirl the points of $\Omega$? Well they both do a very thorough job! They are both transitive groups. If the boss wants 1 moved to 3, no problem, $(1,2,3)^2$ will do the job in both $G$ and $H$. Now move 3 back to 2? No problem! $(1,2,3)^2$ again will do the trick.

What if the boss goes mad with the power? What if he wants you to move 1 to 2 and at the same time leave 3 alone? In $G$ this cannot be done. Every element other than the identity moves all the points ($G$ is regular). In $H$ this is easy, $(1,2)$ does the job.

What is the difference between these groups? They are both transitive, but once you start saying where one point goes, they react very differently about what they can do to the rest.

In other words, $G_3=\{()\}$, but $H_3=\langle (1,2) \rangle$. The first group cannot do anything, but the second is transitive on $\Omega \setminus \{3\}$.

To understand a group action, it is not enough to just ask if it is transitive. We often want to know how the group can move pairs of points.

Consider the actions of $G$ and $H$ on $\Omega \times \Omega$, the set of ordered pairs $\{ (i,j) : 1 \leq i,j \leq 3 \}$. Clearly these actions are not transitive, because $(i,i)^g = (i^g, i^g)$. One orbit is always the "diagonal orbit" $\{ (i,i) : i \in \Omega \}$. What about the rest of $\Omega \times \Omega$?

Well for $G$ there are two orbits: $\{ (1,2), (2,3), (3,1) \}$ and $\{ (2,1), (3,2), (1,3) \}$. $G$ moves them in a circle, and does not change clockwise versus counter-clockwise. We say $G$ is a rank three permutation group.

For $H$ there is only one orbit: any $i,j$ with $i \neq j$ can be sent to any other such pair. We say $H$ is a 2-transitive permutation group.

For groups acting on graphs, we restrict what part of $\Omega \times \Omega$ is allowed and of concern, and "transitive" becomes "vertex transitive" while "2-transitive" becomes "edge-transitive".

share|cite|improve this answer

Since $G_{\alpha}$ acts on $\Omega$ as a group of bijections, but every single one of those bijections maps $\alpha$ to $\alpha$, then the restriction of elements of $G_{\alpha}$ to $\Omega-\{\alpha\}$ will yield bijections from $\Omega-\{\alpha\}$ to itself. So $G_{\alpha}$ also acts on $\Omega-\{\alpha\}$. And given that we know exactly how it acts on $\{\alpha\}$, then understanding the action of $G_{\alpha}$ is equivalent to understanding the action on $\Omega-\{\alpha\}$. Moreover, for doubly transitive actions (which are of importance), having decided one of the actions (mapping $\alpha$ to $\alpha$), you now have a "leftover" transitive action on $\Omega-\{\alpha\}$.

share|cite|improve this answer
Honestly, I just wanted to know that why $G_\alpha$ is acting on the set as above? Why it does not act on any certain subsets of $\Omega$. We always are facing theorems in this area with this assumption that “Let $G$ acts on a set $\Omega$ , so $(G_\alpha|\Omega-\{\alpha\})$ for $\alpha\in\Omega$ and…". Jyrki did the first spark through comments and here you two sketched a picture describing the answer in details. Thanks for your time. – Babak S. Jun 28 '12 at 5:57
@BabakSorouh: It may act on other subsets as well, but this is one subset in which we are guaranteed it acts. – Arturo Magidin Jun 28 '12 at 5:58

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.