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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.

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    I took your $a$dvice 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.2012-10-16

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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")?