# How to show a certain group element must belong to the stabilizer of a set element

I'm studying for personal fun and culture group theory, specifically the orbit stabilizer theorem. I've then found this interesting problem:

Let a group $$G$$ acting on $$\Omega$$ and take $$\alpha,\beta\in\Omega$$, $$x\in G$$ and $$\alpha^x=\beta$$ (meaning $$x$$ acts on $$\alpha$$ to get $$\beta$$). Show that if $$y\in G$$ and $$\alpha^y=\beta$$ $$\implies \exists g\in \operatorname{Stab}(\alpha)$$ s.t. $$y=gx$$, i.e. $$y\in \operatorname{Stab}(\alpha)x$$.

In words, this is tantamount to claim all group elements moving $$\alpha$$ to $$\beta$$ belong to the group set given by the stabilizer elements of $$\alpha$$ followed by a guess element moving $$\alpha$$ to $$\beta$$.

To get a better visual mental grasp on it, I've used the rotational symmetries of a cube acting on the set of cube faces. In this case, given a rotation moving a face onto another one, we can get the full rotations which realise the same thing using the stabilizer subgroup of each face. This is the set of rotations around an axis perp to the face mid point, and is isomorphic to a cyclic group of order 4. So, taking each stabilizer rotation followed by the initial rotation from one face to the target one, gives you back the full list of moves.

That said, I've tried to show that such $$g$$ element exists (problem 3.3 of Groups and Characters book) but I'm not fully sure this demo is complete. It goes like this:

Let's take the element $$g\in G$$ s.t. $$y=gx$$, i.e. $$g=yx^{-1}$$. We can also claim $$\alpha^y=\alpha^x$$ so, replacing the y: $$\alpha^{gx}=\alpha^x \implies (\alpha^g)^x = \alpha^x$$ But this can be true only iff:$$\alpha^g=\alpha \implies g\in \operatorname{Stab}(\alpha) \implies y\in \operatorname{Stab}(\alpha)x$$

I was wondering if this is enough, or if I'm missing something. I mean: is this enough to proof that the elements $$\in \operatorname{Stab}(\alpha)x$$ are the only ones bringing $$\alpha$$ to $$\beta$$?

I was also wondering if there could be a reductio ad absurdum possible version of this, i.e. "Let's suppose such $$g \notin \operatorname{Stab}(\alpha)$$ . . . " and deriving from it a contradiction.

• I'm just assuming $g\in G$, and this without loss of generality, since given two distinct elements $x$ and $y$, belonging to $G$, there's always an element $g\in G$ s.t. $y=gx$. That element is exactly $g=yx^{-1}$. So that generic element always exists. The proof is exactly to show that element $\in Stab(\alpha)$. So, I'm not assuming it exists, since $g$ is just assumed to be generic, but I derived it $\in Stab(\alpha)$ as aimed. So I miss a bit your concern on that. About your second comment: $x$ and $y$ are assumed to be different, of course. – riccardoventrella Jan 12 at 16:37
• My bad. Couldn't realise it first time. – Thomas Shelby Jan 12 at 16:47
• no problem, I'm writing here exactly to get opinions like yours and it was highly appreciated. Let me know for any other concern on that, eventually, please. About $x$ and $y$, I abused above in writing "are assumed to be different". This is not the case, as you can see from above in which I merely quoted exactly the book exercise. So they could be the same element, or distinct ones, indeed. So in the former case you are right for sure in claiming to use the $e$ element, of course. But in the generic, the $g$ is derived to $\in Stab(\alpha)$ indeed (which $e$ belongs to of course, as subgroup) – riccardoventrella Jan 12 at 16:59

The proof is enough. Some further clarification:

Claim: $$\alpha^x = \alpha^y \implies y \in Stab(\alpha)x$$

Proof: As you noted, there exists $$g \in G$$ such that $$y = gx$$. Thus,

$$(\alpha^g)^x = \alpha^x$$

It suffices to show that $$\alpha ^ g = \alpha$$, as this would imply that $$y \in Stab(\alpha)x = \{ gx | \alpha^g = \alpha\}$$.

But this follows readily from the existence of an inverse $$x^{-1}$$ of $$x$$:

$$((\alpha^g)^x)^{x^{-1}} = (\alpha^x)^{x^{-1}}$$

so you would get

$$\alpha^g = \alpha$$

as desired.

• Thank @Netric. Your detail adds value and complete my proof, in which I skipped your $\alpha^g = \alpha$ rigorous derivation. In my case, I claimed instead "but this can be true only iff", using your inverse trick implicitly, while I should have written it clearly for a full reasoning. Thank for having pointed that out. – riccardoventrella Jan 12 at 17:04
• You're welcome! – Metric Jan 12 at 17:05