Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have to prove the OS theorem. The OS theorem states that for some group $G$, acting on some set $X$, we get

$$ |G| = |\mathrm{Orb}(x)| \cdot |G_x| $$

To prove this, I said that this can be written as

$$ |\mathrm{Orb}(x)| = \frac{|G|}{|G_x|}$$

In order to prove the RHS, I can say that we can use Lagrange theorem, assuming that the stabiliser is a subgroup of the group G, which I'm pretty sure it is. I don't really know how I'd go about proving this though. Also, I was thinking, would this prove the whole theorem or just the RHS?

I didn't just want to Google a proof, I wanted to try and come up with one myself because it's easier to remember. Unless there is an easier proof?

share|cite|improve this question
up vote 5 down vote accepted

You’re on the right track. By definition $G_x=\{g\in G:g\cdot x=x\}$. Suppose that $g,h\in G_x$; then $$(gh)\cdot x=g\cdot(h\cdot x)=g\cdot x=x\;,$$ so $gh\in G_x$, and $G_x$ is closed under the group operation. Moreover, $$g^{-1}\cdot x=g^{-1}\cdot(g\cdot x)=(g^{-1}g)\cdot x=1_G\cdot x=x\;,$$ so $g^{-1}\in G_x$, and $G_x$ is closed under taking inverses. Thus, $G_x$ is indeed a subgroup of $G$. To finish the proof, you need only verify that there is a bijection between left cosets of $G_x$ in $G$ and the orbit of $x$.

Added: The idea is to show that just as all elements of $G_x$ act identically on $x$ (by not moving it at all), so all elements of a left coset of $G_x$ act identically on $x$. If we can also show that each coset acts differently on $x$, we’ll have established a bijection between left cosets of $G_x$ and members of the orbit of $x$.

Let $h\in G$ be arbitrary, and suppose that $g\in hG_x$. Then $g=hk$ for some $k\in G_x$, and $$g\cdot x=(hk)\cdot x=h\cdot(k\cdot x)=h\cdot x\;.$$ In other words, every $g\in hG_x$ acts on $x$ the same way $h$ does. Let $\mathscr{G}_x=\{hG_x:h\in G\}$, the set of left cosets of $G_x$, and let

$$\varphi:\mathscr{G}_x\to\operatorname{Orb}(x):hG_x\mapsto h\cdot x\;.$$

The function $\varphi$ is well-defined: if $gG_x=hG_x$, then $g\in hG_x$, and we just showed that in that case $g\cdot x=h\cdot x$.

It’s clear that $\varphi$ is a surjection: if $y\in\operatorname{Orb}x$, then $y=h\cdot x=\varphi(hG_x)$ for some $h\in G$. To complete the argument you need only show that $\varphi$ is injective: if $h_1G_x\ne h_2G_x$, then $\varphi(h_1G_x)\ne\varphi(h_2G_x)$. This is perhaps most easily done by proving the contrapositive: suppose that $\varphi(h_1G_x)=\varphi(h_2G_x)$, and show that $h_1G_x=h_2G_x$.

share|cite|improve this answer
Saying $G/G_x$ is a little dangerous here - $G_x$ won't in general be normal. Of course if you just mean the set of (left or right) cosets, then everything's fine, but I don't remember if $G/G_x$ is standard notation for that when it isn't a group. – Matthew Pressland Dec 27 '12 at 14:46
@Matt: You’re right: that was sloppy at best. Thanks for catching it. – Brian M. Scott Dec 27 '12 at 14:49
How do I go about proving the bijection? I asked a question on proving bijection yesterday as well, I can't seem to get the hang of it. – Kaish Dec 27 '12 at 14:50
@Kaish: Consider a specific left coset $hG_x$ of $G_x$; can you guess which element of the orbit of $x$ should correspond to $hG_x$? – Brian M. Scott Dec 27 '12 at 14:51
Give me a couple of minutes. I need to read up on cosets again... – Kaish Dec 27 '12 at 14:54

We can have the claim for $(G\mid\Omega)$ which is transitive and so we get $$|G|=|\Omega||G_{\omega}|$$ To see this, put $\Omega^*=\{G_{\omega}x\mid x\in G\}$and define the following map: $$f:\Omega^*\to\Omega,\;\;f(G_{\omega}x)=\omega^x$$ You can easily verify that $f$ is well-defined and is a bijective. Both $\Omega^*$ and $\Omega$ are finite and then $$|\Omega|=|\Omega^*|=[G:G_{\omega}]$$

No put the condition of being transitive aside, you will have your own claim as others confirmed.

share|cite|improve this answer
+1!${}{}{}{}{}{}{}\quad \ddot\smile\quad$ – amWhy Mar 1 '13 at 1:06

You should show that the identity of $G$ fixes $x$, and that if $g,h\in G$ both fix $x$, then $gh$ fixes $x$. This will establish that $G_x$ is a subgroup of $G$.

This won't finish the proof, although it will let you say that $\frac{\lvert G\rvert}{\lvert G_x\rvert}$ is the number of cosets of $G_x$ in $G$ (left or right, there's the same number either way). You still need to find a bijection between these cosets and the orbit of $x$ under $G$.

share|cite|improve this answer

Let $x\in X$. The set $H=\{g\in G:gx=x\}$ is a subgroup of $G$. Now consider the function $f:G/H\rightarrow Orb(x)$ that sends $g+H$ to $gx$. It is easy to verify that $f$ is a bijection. Hence, $Orb(x)=G/H$ . Finally use lagrange's theorem to get $|G|=|H||G/H|$

share|cite|improve this answer

Your Answer


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.