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In proving the Orbit-Stabilizer Theorem, I couldn't help but notice its resemblance to the First Isomorphism Theorem for groups. Let $G$ be a finite group and $X$ be a finite set, and let $G$ act on $X$. Let $S_x = \{g\in G\mid g\cdot x=x\}$ be the stabilizer of $x$ for any $x\in X$, and let $G\cdot x = \{g\cdot x\mid g\in G\}$ be the orbit of $x$.

We can give a group structure on $G\cdot x$ by defining $(g\cdot x)(h\cdot x) = (gh)\cdot x$. Let's show that this is a group:

i) $(gh)\cdot x\in G\cdot x$, so the operation is closed.

ii) Let $g,h,k\in G$. Then, $$[(g\cdot x)(h\cdot x)](k\cdot x) = ((gh)\cdot x)(k\cdot x) = ((gh)k)\cdot x = (g(hk))\cdot x = (g\cdot x)((hk)\cdot x) = (g\cdot x)[(h\cdot x)(k\cdot x)],$$ so the operation is associative.

iii) $1\cdot x$ is the identity.

iv) $g^{-1}\cdot x$ is the inverse of $g\cdot x$.

So $G\cdot x$ is a group. Importantly, the map $\varphi:G\to G\cdot x$ defined by $\varphi(g)=g\cdot x$ is a group homomorphism, with kernel $S_x$. By the first isomorphism theorem, $$G/S_x\cong G\cdot x.$$ This means, specifically, that $G/S_x$ is a group, which is true if and only if $S_x$ is a normal subgroup of $G$. But this is clearly not the case generally, since the conjugation action of $S_3$ (permutation group) on itself does not give a normal stabilizer; for instance, the stabilizer of $(1\ \ 2)$ is $\{(1), (1\ \ 2)\}$, which is not a normal subgroup of $S_3$.

My question: where does my "proof" go wrong? How does it not show that $S_x$ is a normal subgroup generally? When is the stabilizer a normal subgroup of $G$; i.e. when is $\varphi:g\mapsto g\cdot x$ a group homomorphism?

IAAW
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  • ARe you sure that the multiplicative law is define ? if is inthe stabilizer $(gs).x=g.x$ but there is no reason that $(gs).h x= (gh)x$ . – Thomas Dec 22 '24 at 06:52
  • Certainly, the quotient is well defined if we check before that $S_x$ is normal. If $S_x$ is not normal, all bets are off. $G/S_x$ is a group if $S_x$ is normal, but I'm not sure about the "only if" part. What does $G/S_x$ even mean, otherwise? – AlvinL Dec 22 '24 at 06:52
  • @AlvinL I'm not sure I understand what you mean. $G/S_x = {gS_x}$ is the set of cosets of $S_x$ in $G$, which is well-defined since $S_x\leq G$. The argument that I'm making in the post is that $S_x$ is the kernel of a homomorphism out of $G$, which is always a normal subgroup of $G$. – IAAW Dec 22 '24 at 06:57
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    @AlvinL it is also the case that $G/H$ is a group if and only if $H$ is a normal subgroup of $G$: https://math.stackexchange.com/questions/1155714/if-g-h-is-a-group-then-does-h-have-to-to-be-normal – IAAW Dec 22 '24 at 07:14
  • @AlvinL what is the equivalence relation? I know that orbits themselves partition $G$, but $G\cdot x$ is specifically one such equivalence class: the orbit of a single element $x$. The multiplication I'm defining is on elements of $X$. I do share your suspicion that the operation $(g\cdot x)(h\cdot x) = (gh)\cdot x$ may not be well-defined... all my attempts so far have returned fruitless, in either direction. Is there a condition which makes this operation well-defined in general? – IAAW Dec 22 '24 at 07:27
  • As @AlvinL says: to get that your definition of $(g\cdot x)(h\cdot x)$ is well-defined you need to have a normal stabiliser of $x$. – ancient mathematician Dec 22 '24 at 07:27
  • @ancientmathematician can you explain why that is? – IAAW Dec 22 '24 at 07:28
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    Let $g\cdot x=g_1\cdot x$, then $g=g_1 s$ with $s\in S_x$, similarly $h=h_1 t$ when $h\cdot x=h_1 \cdot x$. Then $gh\cdot x= g_1 s h_1 t \cdot x= g_1 s h_1 \cdot x$, but to get that $h_1$ to the left of the $s$ you'll need $s h_1 =h_1 s'$ with $s'\in S_x$, which is equivalent to normality. – ancient mathematician Dec 22 '24 at 07:34
  • The following question doesn't answer your question, but does explain the true way in which they are analogues: https://math.stackexchange.com/questions/4761714/orbit-stabiliser-theorem-as-an-analogue-to-first-isomorphism-theorem – ancient mathematician Dec 22 '24 at 07:36
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    @IAAW As for your comment regarding $G/S_x$. Sure, we can regard it as the set cosets, but a priori there is no structure on it. Usually, the $/$ sign refers to quotient (hence you already assume the relation is a congruence). The only known reasonable structure $G/S_x$ can have is $gS_x \cdot hS_x := (gh) S_x$, but that requires normality. If you try to skip ahead of the well-definedness checks, you will often get punished for it somewhere down the line - been there myself. – AlvinL Dec 22 '24 at 08:03
  • $G\cdot x$ is not a group hence you can't just impose a group homomorphism like you say towards the end of your post. $G\cdot x$ is simply the set generated by $G$ at $x\in X$. – August Bowden Dec 22 '24 at 21:52

2 Answers2

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Recall that $g\cdot x=(g k)\cdot x$ for whatever $k\in S_x$. So, we want the result of your tentative multiplication to be independent on the arbitrary $k,k'\in S_x$, namely: $$(g k\cdot x)(h k'\cdot x)=(gkh k')\cdot x=(gh)\cdot x$$ $\iff h^{-1} khk' \in S_x$ $\stackrel{ k'\in S_x}{\iff}h^{-1} kh \in S_x$. By the arbitrariness of $h\in G$ and $k\in S_x$, the latter is precisely the normality condition for $S_x$.

Kan't
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In general $G \cdot x$ will just be a $G$-set, not a group. It's best to keep in mind some simple examples, which clears this sort of thing up immediately. Pretty much any non-abelian group action will do here.

For instance, let $G=S_n$, the group of permutations on $n$ elements, act on $X=[n]=\{1, 2, \ldots, n\}$ naturally. The action is obviously transitive, so $S_n \cdot 3 = [n]$, say. You claim to produce a group structure on $[n]$, though this is pretty suspicious on its face. Certainly $\mathbb{Z}/n\mathbb{Z}$ is an (additive) group, but as far as $X$ is concerned the numbers are just arbitrary labels, and $S_n$ knows nothing about addition, so how could this group action possibility produce that group? And if $n$ is prime, that's the only possibility at all.

Stress test your definitions by running through them in this example. Your claimed identity element is $3$. Let's compute $2 \ast 4$. Your recipe says to write $2 = u \cdot 3$ and $4 = v \cdot 3$ and then declare $2 \ast 4 = (uv) \cdot 3$. That is, pick $u(3) = 2, v(3) = 4$ and then set $2 \ast 4 = u(v(3)) = u(4)$. We immediately see the problem: $u(4)$ can be chosen to be anything, so this purported product is not well-defined in general.

As the other answers say, this product is well-defined if and only if the stabilizer of $x$ is normal.

Joshua P. Swanson
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  • Thank you for the amazing answer. Can we conclude, then, that the orbit of $x$ is a group (under the operation I defined) if and only if the stabilizer of $x$ is normal? This seems like a cool result – IAAW Dec 22 '24 at 16:50
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    @IAAW Given a group action and any orbit (or union of orbits, really), there is an induced action on that orbit. The kernel of the action is the core of a point-stabiliser or, equivalently, the intersection of all the point-stabiliser. This kernel is equal to a point-stabiliser if and only if the point-stabiliser is normal, so your observation is a special instance of this more general well-known phenomena. – verret Dec 22 '24 at 19:19