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Let $E$ be a finite set and $G$ a transitive group of permutations of $E$. For $x\in E$, let $S_x$ denote the stabilizer of $x$. Then, for any $x,y\in E$, the number of orbits of $E$ under the action of $S_x$ is equal to the number of orbits of $E$ under the action of $S_y$.

I have seen this mentioned in some books without proof like a triviality, but I can't find a proof. I know Burnside's lemma and I know that $G/S_x$ and $G/S_y$ (sets of left cosets) are isomorphic to $E$, and hence to each other, as $G$-sets. None of this seems to help. I'm probably missing something very obvious. Can you help me?

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  • $\begingroup$ $G$ acts component-wise on $E×E$. As $G$ is transitive, the diagonal $\Delta$ is an orbit. The number of orbits of $G$ on $E×E$ equals the number of orbits of $S_x$ on $E$. I think that as the same is obviously true for $y$, the authors of the books considered the equality you are looking for to be trivial. $\endgroup$
    – j.p.
    Aug 13, 2011 at 7:44

2 Answers 2

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You mean $x, y \in E$, right? Let $g \in G$ be such that $gx = y$. For the purposes of counting orbits, relabeling the elements of $E$ shouldn't change the number of orbits, so let's relabel every $a \in E$ with the new label $ga$.

In this new labeling, the group action looks like the following: where we used to have $ha = b$ (with $h \in G$), we now have

$$(ghg^{-1}) ga = gb.$$

In other words, in this new "coordinate system," $h \in G$ acts as $ghg^{-1}$.

Now restricting to $S_x$ in the old coordinate system is the same as restricting to $g S_x g^{-1} = S_y$ in the new coordinate system. In particular the two group actions are isomorphic, so have the same number of orbits.

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Since $G$ acts transitively on $E$, then for any $x,y\in E$ there exists $g\in G$ such that $gx=y$. Then $S_y = gS_xg^{-1}$:

Indeed, since $gx=y$, then $g^{-1}y=x$. Thus, if $ghg^{-1}\in gS_xg^{-1}$, we have $ghg^{-1}y = ghx = gx = y$, so $ghg^{-1}\in S_y$; hence $S_y\subseteq gS_xg^{-1}$. Using the same argument, we get that $S_x\subseteq g^{-1}S_yg$, which gives the other inclusion.

So $S_x$ and $S_y$ are conjugate, and in particular there is an automorphism of $G$ that maps $S_x$ to $S_y$; this automorphism induces an isomorphism of the cosets of $S_x$ to the cosets of $S_y$ as $G$-sets.

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  • $\begingroup$ About your last sentence: maybe you are using the fact that the automorphism is inner? I'm not sure whether a general automorphism sending a subgroup $K$ to a subgroup $H$ will really give rise to a $G$-set isomorphism $G/K \to G/H$. $\endgroup$ Dec 2, 2016 at 0:26

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