3
$\begingroup$

Suppose that $G$ is a finite group, and let $\# G$ denote the number of elements in $G$. Let $m$ be a point of $M$, and consider the orbit of $m$ under $G$, which contains $\#(G\cdot m)$ elements. If $n$ is an element of this orbit, then $n=am$ for some $a$ in $G$. If $n=bm$ as well, then $a^{-1}b$ must lie in $G_m$. This means that to each element $n$ there are exactly $\#G_m$ group elements which map $m$ into $n$. Therefore, since every element of $G$ carries $m$ into some orbit element, we conclude that

$$\#G=\#(G\cdot m)\#G_m$$

Above is an excerpt from Sternberg, Group Theory and Physics. To be clear, $G\cdot m \subset M$ is the orbit of $m\in M$, and $G_m \subset G$ the isotropic group of each $m$. I know that for any two elements $a,b \in G$ that map $m$ to $n$, there are two elements $a^{-1}b$ and $b^{-1}a$ in $G_m$, but I don't understand the emphasized bit, since $n$ is any element in $G\cdot m$, while $\#G_m$ is the fixed number. We can't be sure if there exists $b$ in $G$ besides $a$ such that $bm=n$, and even if with this particular $n$ there are $\#G_m$ ways to map $m$ to $n$, then nothing precludes another $n'\in G\cdot m$ from having a different number of ways to be mapped from $m$.

So why is that?

$\endgroup$
0

2 Answers 2

3
$\begingroup$

What the text says is that if $b$ maps $m$ to $n=a\cdot m$, then $b\in aG_m$. In fact the converse is true : if $b=ag$ for some $g\in G_m$, then $$b\cdot m=(ag)\cdot m=a\cdot (g\cdot m)=a\cdot m=n.$$ Thus the set of elements of $G$ that maps $m$ to $n$ is exactly $aG_m$, which has the same number of elements as $G_m$. This is true for any $n$ in the orbit of $G$ (although of course with a different element in the orbit you will get a different coset of $G_m$); in particular, the number of ways to map $m$ to any element of its orbit only depends on $m$.

$\endgroup$
3
$\begingroup$

For any $n$ in the orbit of $m$, there obviously exists $a \in G$, and $a \notin G_m$, such that $am = n$. Now, note that $aG_mm = am = n$. Thus, you have exactly #$G_m$ elements that map $m$ to $n$ (note that $G_m$ contains $1$, so when you say #$G_m$, it counts $a$ as well).

There cannot be another $n' \in G.m$ that has a different number of ways. The same argument as above applies. If there is $a' \in G$ such that $a'm = n'$, then $a'G_mm = a'm = n'$. Note that there cannot be $x \in G, x \notin a'G_m$ such that $xm = n'$ because in that case, $$xm = n' \implies m = x^{-1}n' \implies a'G_mm = a'G_mx^{-1}n' \implies n' = a'G_mx^{-1}n' \implies a'G_mx^{-1} = 1 \implies x = a'G_m,$$ which is a contradiction.

This observation is the departure point for any and all applications of group actions. What you have is actually a bijection between $G/G_m$ and $G.m$, $aG_m \mapsto am$. This to me is one of the most remarkable expedients of group actions. When someone says a group is acting on a set, it immediately tells us a whole bunch of things about the group and the set. (When I read about group actions for the first time, I thought that one could, without much effort, define an action between any group and any set. This is NOT TRUE.)

The orbits of a group action are equivalence classes for the relation $x \sim x'$ if $x' = gx$ for some $g \in G$. The orbits partition the set, i.e. the set can be expressed as a disjoint union of orbits.

This image is what you must keep in mind when you know a group is acting on a set.enter image description here

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .