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Let $G$ be a group and $S$ a $G$-set with action $(g,s) \mapsto gs$. For some $s \in S$, let the stabilizer of $s$, $G_s=\{g \in G\,|\,gs=s\}$ be normal in $G$. What does this let us say about the action of $G$ on $S$?

I thought it might be interesting to look at an action of $G/G_s$ on $S$. However, something like $(gG_s,s) \mapsto gs$ isn't even well-defined in general.

Are there situations in which we can recover anything interesting?

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A stupid remark: because of the formula $G_{gs} = gG_sg^{-1}$, if the action is transitive, it factors through the quotient $G/G_s$. –  PseudoNeo Feb 14 '13 at 3:54

3 Answers 3

up vote 6 down vote accepted

What normality of the stabiliser says is exactly that every group element $g\in G$ that fixes $s$ also fixes the entire orbit $Gs$ pointwise. Conversely any $g\in G$ that fixes any element of the orbit $Gs$ will also fix $s$.

These two parts are equivalent, although the first sentence says that every conjugate of $G_s$ contains $G_s$, while the second sentence says that any conjugate of $G_s$ is contained in $G_s$. Even though a subgroup $H$ may strictly contain a conjugate ${}^gH$ of itself, if it contains all its conjugates then it must be equal to them all, in other words normal (should $H\supsetneq {}^gH$ then ${}^{g^{-1}}H\supsetneq H$, and the hypothesis excludes this).

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The stabilizer of a point $s \in S$ is normal in $G$ if and only if every element that stabilizes $s$ stabilizes the orbit $G\cdot s$ of $s$ pointwise.

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Assume the group action is transitive. Then all stabilizers are conjugate to one another. Therefore, if the stabilizer is normal, then all elements of $S$ have the same stabilizer. This is equivalent to saying that if $g$ fixes one point of $s$, then it fixes them all. So the action is regular: For every $s,t \in S$, there is a unique $g \in G$ such that $gs=t$. In particular, $|G| = |S|$.

(Even if the action is not transitive, then you can still say all of the above about the orbit of $s$, a point for which $G_s$ is normal.)

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