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I am trying to construct an example of the stabilizer orbital theorem and trying to make sense as to why group actions are important. As I understand it, the stabilizer is also the centralizer. Could someone explain to me what the $|\text{Stabilzer}(\mathbb{Z_6})|$ and what the $|\text{Orbital}(\mathbb{Z_6})|$? I couldn't quite unravel the definitions to make them seem less abstract. I have an upcoming test and I just want to understand these materials to feel more confident.

Student
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    Group actions are important simply because most of the time groups appear in nature acting on stuff, not «abstractly». We had a related discussion here If you are trying to make sense of why group actions are important, that is a better direction to look, I think! – Mariano Suárez-Álvarez Oct 16 '12 at 07:16
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    In any case, what is the orbital of a group? if you mean the size of an orbit, then every group acting on any set is an example: pick any group, let is act in any possible way on a set, and you will find that the size of the group is the product of the size of tthe stabilizer of an element of the set times the size of its orbit. – Mariano Suárez-Álvarez Oct 16 '12 at 07:18
  • Your second comment cleared up some things so what if I wanted $\mathbb{Z_6}$ act on itself via conjugation? – Student Oct 16 '12 at 07:31
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    Try it! How else are you going to figure out your what if?! – Mariano Suárez-Álvarez Oct 16 '12 at 07:32
  • I took your advice and found out that the the conjugacy class of the element 1 is just 1 and the order of the centralizer of 1 is 6. Thanks. – Student Oct 16 '12 at 07:43

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Let a group $G$ act on a set $S$, and pick an element $s\in S$ to look at. There are two important sets which tell you about what $G$ does to $s$. $Stab_G(s)=\{x \in G : s\cdot x = s\}$ is the set of elements in $G$ that don't do anything to $s$ - whenever those elements act on $s$, $s$ just stays where it is. (Remember, $Stab_G(s)$ is comprised of elements of $G$, and in fact is a subgroup of $G$.) The other set is the orbit of $s$, $\mathcal{O}_s=\{s^x : x \in G\}$. This is the set of all the places in $S$ that $s$ can be sent to when being acted upon by some element of $G$. (So, $\mathcal{O}_s$ is comprised of elements of $S$.) As it turns out, $|Stab_G(s)|\cdot|\mathcal{O}_s|=|G|$. This post is about why you should care about this formula.

You're already familiar with centralizers, which are good motivation for group actions: say you've got some element $g$ in your group and you want to know everything that commutes with that element. You look at $C_G(g)=\{x \in G : gx = xg\}$. We can rewrite that condition as $x^{-1}gx=g$ or $g^x=g$ (that's how group theorists write conjugation).

Another way to look at $C_G(g)$ is to say, let $G$ act on $G$ by conjugation. The action is $g \cdot x = g^x$, and since this is such a commonly used group action, we give a special name to $Stab_G(g)=\{x \in G : g^x = g\}$ - it's $C_G(g)$.

But there are other nice group actions out there.

Say you let $G$ act on the set of all subgroups of $G$ by conjugation (i.e., $H\cdot g = H^g = \{h^g : h \in H\}$). Under this action, which I'll call "subgroup conjugation," $Stab_G(H)=\{x \in G : H^x = H\}$. This action is also pretty common so again we have a special name for the stabilizer - the "normalizer" of $H$, written $N_G(H)$. You can see that this works exactly the same as the centralizer; it's just a stabilizer for a different group action.

What is the orbit of $H$ under subgroup conjugation? It's every subgroup $K$ of $G$ so that $H^x=K$ for some $x\in G$ - the conjugate subgroups of $H$ in $G$. With the Orbit-Stabilizer theorem, you can immediately tell how many there are: $|\mathcal{O}_H|=|G|/|Stab_G(H)|=|G|/|N_g(H)|=[G:N_G(H)]$.

Now ask yourself, what is the size of the conjugacy class of an element $g\in G$? If you can answer this question by applying the same logic as I did talking about subgroup conjugation, you'll understand why group actions can make certain questions very intuitive.

Here's a couple of other things to think about.

  • What does it mean if $|\mathcal{O}_H|=1$ under the subgroup conjugation action?

  • How would you describe elements $g\in G$ with $|\mathcal{O}_g|=1$ under the conjugation action of $G$ on itself?

Things to keep in mind as you go forward in group theory are, how does a factor group $G/N$ act on a normal subgroup $N$? Can I use Orbit-Stabilizer with divisibility rules / congruences? Perhaps most importantly, can I find a way to show that a group acting on some set must give rise to something with orbit size 1 (a "fixed point")?

  • Are these answers to your question correct? 1) If the size of the orbit of $H$ is 1, then its normalizer must be $G$. 2) if $G$ acts on itself through conjugation, then the elements with orbit of size one are those that commute with every other element in $G$. – Sergio Sep 27 '22 at 07:37
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    @Sergio You are correct. To further follow up with 1, is there a term for subgroups whose normalizer is the whole group? – Alexander Gruber Sep 28 '22 at 06:01
  • If $G$ is Abelian the condition would be satisfied, but is it necessary and sufficient for $G$ to be Abelian so that the normalizer of a particular subgroup is all of $G$? – Sergio Sep 28 '22 at 09:24
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    @Sergio The name of a subgroup whose normalizer is the whole group is: a normal subgroup of G. You're right that abelian is sufficient, but not necessary for this. However you can still talk about abelian groups with this language if you refer to the Centralizer, instead of the Normalizer. The Centralizer fixes each element in the subgroup individually under the conjugation action, whereas the Normalizer fixes only the subgroup as a whole under the conjugation action. – Alexander Gruber Sep 28 '22 at 15:32
  • Oh yes, silly me. How could I not see it? Thank you for your explanations! – Sergio Sep 28 '22 at 15:44
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    This answer should be written in every group theory book. In at least three books, in Wikipedia and in many other places, I didn't find such clear explanation like this – Saul Berardo Oct 11 '22 at 04:40