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The orbit-stabilizer theorem is completely encoded by the equation

$$|G| = |\operatorname{Orb}(x)||\operatorname{Stab_G}(x)| $$

Most books/online presentations I am reading jump straight into this equation after the definitions are introduced.

Note that Lagrange Theorem tells us

$$|G| = [G: \operatorname{Stab}_G(x) ]|\operatorname{Stab}_G(x)|$$

So what prompts us to suggest $[G:\operatorname{Stab}_G(x)]$ is bijective with $|\operatorname{Orb}(x)|$?

Is it observed via a few examples and conjectured later?

Note I am not asking for the proof

Lemon
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4 Answers4

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I'm going to answer what I think is the literal question; if I've misunderstood the question, perhaps you can correct me.

I believe what you are asking is: How can a mathematician discover this theorem, knowing nothing about it beforehand?

Well, the answer is that they can't. That's not what happens. Just learning the abstract definition of a group, and maybe the definition of a group action, and maybe the definition of orbits, and then expecting that anyone can just say "Hey, here's a theorem!"... well... that's not how any mathematical theorems ever get discovered.

Instead, someone learns about actual groups, and actual group actions. They learn examples. They observe patterns. They stumble upon this particular pattern, noticing that it holds in a few different examples: the order of the group is the size of an orbit times the size if a stabilizer of a point in that orbit. They think "Huh... is this just a coincidence?" They might look for more examples to bolster the point, they might look (unsuccessfully) for counterexamples, which bolsters the point even more.

They become more and more convinced that the pattern is true.

And when a mathematician becomes convinced that something is true, then they are very motivated to prove that it is true.

And fortunately, the proof is easy.

I really don't think there's much more than that to say.

Lee Mosher
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  • Along these lines, this is basically a theorem where if you run across/prove a specific example you basically have the proof of the orbit-stabilizer theorem except you need to phrase things in a more general way. Another lemma/theorem/observation which I think is similar in this way is the ping-pong lemma--once you prove one example you have all the ideas to do the general lemma. –  Sep 20 '19 at 04:10
  • So basically my guess is true. – Lemon Sep 20 '19 at 18:00
  • Indeed. In fact, what I am asserting is that this is universally true for all theorems proved by humans. – Lee Mosher Sep 20 '19 at 18:31
  • Well some theorems I feel can be anticipated, but this one doesn't seem to be one of those. – Lemon Oct 04 '19 at 00:59
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Here is one motivation. Call a set on which $G$ acts a "$G$-set". One particular example is that if $H$ is a subgroup, then the set $G / H$ of cosets of $H$ is a $G$-set in the obvious way: $g \cdot (xH) := gx H$. The "orbit-stabilizer" theorem is saying that this example is "canonical" in the sense that EVERY example is essentially this example.

What I mean is that another way of phrasing the theorem is that, viewing $\mathrm{Orb}(x)$ as a $G$-set, it is isomorphic to the coset space $G / \mathrm{Stab}(x)$. So every (transitive) $G$-set is equivalent to some coset space $G/H$. The version of the theorem that you mention is a corollary of this fact by taking the cardinality of both sets.

Nick
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Hint: the mapping $p$ from $G$ to $X$ defined by $p(g) = gx$, partitions $G$ by the equivalence relation $\simeq$ where $g \simeq h$ iff $p(g) = p(h)$. This equivalence relation has $|\mathrm{Orb}(x)|$ equivalence classes and each equivalence class has $|\mathrm{Stab}(x)|$ elements.

Rob Arthan
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  • Well I am actually trying to figure out the general case... – Lemon Sep 20 '19 at 00:49
  • The Lagrange theorem doesn't apply in the general case. Please clarify your question. – Rob Arthan Sep 20 '19 at 00:51
  • No that isn't my point. I am just giving motivation for writing down the equation. I am aware of how to prove this. My whole point is that in every literature that introduces group actions, they always introduce the definitions of what orbits and stabilizes are and then this theorem is immediately the first theorem to be proven. I want to know the gap between, what would make us think to even come up with this? – Lemon Sep 20 '19 at 00:57
  • When I wrote down Lagrange's theorem, I am just suggesting that might be a starting point. – Lemon Sep 20 '19 at 00:59
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    A starting point to what? What gap? Your comments are making your question seem less and less clear. – Rob Arthan Sep 20 '19 at 01:07
  • A starting point to reach the orbit-stabilizer theorem. There should be some motivation to reaching the equation in the theorem. I am just saying the Lagrange's theorem would be a starting pointing. When basic definitions are introduced, the followup theorems are usually motivated a bit, but this came straight out of the hat. From just the definitions alone, no one should be able to conjecture the equation (let alone the theorem) without something first. I am asking what that something is (and that is the gap i was taking about). – Lemon Sep 20 '19 at 01:15
  • If that is still confusing. Let me break it down like this. Forget every single thing you know about orbits and stabilizers. Now lets say you are (re)introduce to this entire subject again, I start with the definitions. I tell you that given a group $G$ and action $\theta$, i give the definition of the stabilizer and the orbits of the group. Immediately, I jumped straight to the theorem - the orbit-stabizer equation. So upon reading, one should want to know how does one come up with the equation in the first place. No I am not asking for the proof or any technical points. – Lemon Sep 20 '19 at 01:21
  • I've completely revised my answer. Does it help now? – Rob Arthan Sep 20 '19 at 01:37
  • Nothing motivates a theorem more than its truth. – Lee Mosher Sep 20 '19 at 03:34
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Let's try to motivate the theorem this way.

If $G$ acts on $X$ then clearly the orbits are important because if two elements are in the same orbit, then they "look the same with respect to $G$", and if they are different orbits, then they are different with respect to $G$.

So when studying the action of $G$ on $X$, it makes sense to study one orbit at a time because two different orbits are unrelated as far as $G$ is concerned.

So we pick a point $x \in X$ and consider its orbit in $G$. Now we ask ourselves, what can we say about this orbit? To say that $y$ is in the same orbit as $x$ means that there exists $g \in G$ such that $gx = y$. So it is a reasonable question to ask, how many such elements $g$ are there? From there, a little thought leads one to the conclusion that this number does not depend on $y$ (as long as it is in the orbit of $x$), because elements in the same orbit "look the same with respect to $G$". Thus we may take $y=x$, so we are studying elements such that $gx=x$, i.e., the stabilizer.

From here, it is not far to the orbit-stabilizer theorem.

I hope this was helpful.

Ted
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