Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $G$ act transitively on a (non-empty) set $S$, and fix $s ∈ S$. Then $S$ is in bijection with the set $G/G_s$ of cosets $gG_s$ of the isotropy group $G_s$ of $s$ in $G$, by $gs \longleftrightarrow gG_s$ Thus, $\text{card} S = [G : G_s]$ Proof: If $hG_s = gG_s$, then there is $x ∈ G_s$ such that $h = gx$, and $hs = gxs = gs$. On the other hand, if $hs = gs$, then $g^{−1}hs = s$, so $g^{−1}h ∈ Gs$, and then $h ∈ gG_s$.

I am not sure how this leads to the conclusion that $S$ is in bijection with the set $G/G_s$. How does $h$ being in $gG_s$ and possibility that there is $x$ in $G_s$ that $hs=gs$ lead to bijection conclusion?

Is the proof saying that all of $gG_s$ is in bijection with $h$?

share|improve this question
    
The bijection is between $S$ and the set of cosets. So $g G_s$ is being regarded as an element of a set here, not as a set in its own right. –  Zhen Lin Aug 23 '12 at 3:38
add comment

1 Answer 1

Actually the proof is incomplete.

Here's a complete proof:

If $hG_s = gG_s$, then $hs = gs$. Hence there exists a map $\psi\colon G/G_s \rightarrow S$ such that $\psi(gG_s) = gs$. Suppose $\psi(gG_s) = \psi(hG_s)$. Then $gs = hs$. Hence $gG_s = hG_s$ Hence $\psi$ is injective.

Let $t \in S$. Since $G$ acts transitively on $S$, there exists $g \in G$ such that $gs = t$. Hence $\psi(gG_s) = t$. Hence $\psi$ is surjective.

Therefore $\psi$ is bijective.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.