32

The notes I'm using to study group theory make a remark that another appropriate name for the "orbit stabiliser theorem" is the "first isomorphism theorem for group actions". For reference, here are the two theorems:

First isomorphism theorem: let $\phi: G \to H$ be a group homomorphism with kernel $K$. Then $G / K \cong \text{Image}(\phi)$. The isomorphism is given by $\psi: G / K \to \text{Image}(\phi)$ with $\psi(gK) = \phi (g)$.

Orbit stabiliser theorem: let $G$ be a group acting on a set $X$. Let $x \in X$. Then the map $\phi: G / \text{Stab} (x) \to \text{Orb} (x)$ where $g \: \text{Stab} (x) \mapsto gx$ is a bijection. Also, if $G$ is finite, then $|G| = | \text{Stab} (x) | | \text{Orb} (x) |$.

I can see why the orbit stabiliser theorem is like the first isomorphism theorem, especially when considering that the stabiliser fixes points (i.e. for the stabiliser, the group action is the same as the identity action), so the stabiliser is a kind of analogue to the kernel. Similarly, the orbit is a bit like the image of a map.

What I want is to understand how they're analogous in a more rigorous way, and maybe if any results can be derived for group actions in much the same way as the second and third isomorphism theorems and the correspondence theorem can be derived from the first isomorphism theorem.

Shaun
  • 47,747
omnicube
  • 423

2 Answers2

40

Welcome to MSE, this is a great question!

Say that $\phi : G \to H$ is a group homomorphism.

Then there is a group action $G \curvearrowright H$ defined by $g \cdot h = \phi(g) h$, using the existing multiplication in $H$. If it's not obvious, checking that this really is a group action whenever $\phi$ is a group homomorphism is a worthwhile exercise.

Now let's look at the orbit and stabilizer of the identity element $e_H$. We can compute

  • $\text{orb}_G(e_H) = \text{im}(\phi)$
  • $\text{stab}_G(e_H) = \text{ker}(\phi)$

Again, if these equalities aren't obvious, they make for great exercises!

But what does the orbit stabilizer theorem tell us in this case? It tells us that the map $\overline{\phi} : G \big / \text{stab}_G(e_H) \to \text{orb}_G(e_H)$ (that is, $\overline{\phi} : G \big / \text{ker}(\phi) \to \text{im}(\phi)$) given by $g \ \text{ker}(\phi) \mapsto g \cdot e_H$ is a bijection. But it's not hard to see that this is the same function as $g \ \text{ker}(\phi) \mapsto \phi(g)$, which is the bijection from the first isomorphism theorem!

I hope this helps ^_^

Chris Grossack
  • 42,257
  • Oh that's rather nice! Thank you for your explanation :) – omnicube Sep 01 '23 at 01:09
  • 1
    I might leave the question open for a bit longer to see if anyone has ideas on deriving more theorems – omnicube Sep 01 '23 at 01:10
  • 1
    @omnicube -- definitely do! It's a good idea to leave a question open for a few days so that multiple people have a chance to chime in. I will mention that I know of a couple things that look like the other isomorphism theorem that work for group actions, but I don't know of a way to derive them from orbit-stabilizer – Chris Grossack Sep 01 '23 at 01:45
6

One way to think about the analogy might be the following: the first isomorphism theorem says that, given a group $G$ you can construct all possible images of $G$ under group homomorphism up to isomorphism just from $G$ and its normal subgroups: if $\phi\colon G\to H$ is a group homomorphism, then $\phi(G)$ is isomorphic to $G/K$ where $K = \ker(\phi)$, and $\phi$ induces the isomorphism

If $G$ acts on a set $X$ and $x \in X$, then the orbit of $X$ is "isomorphic" as a $G$-set to the orbit of its stabilizer $S= \text{Stab}(x)$, the latter being being the set of all $S$-cosets of $S$ in $G$, and the "isomorphism" is induced by the action map: $gS \mapsto g.x\in X$.

Both theorems thus, in a sense, tell you how to construct the objects you are studying (images of $G$, transitive $G$-sets respectively) from data already contained in $G$.

krm2233
  • 7,230