Suppose that $|G|$ and $|X|$ are finite. Show that the following are equivalent.
(a) $G$ acts transitively on $X$.
(b) For some $x \in X$, $|G_x| = |G| / |X|$.
(c) For every $x \in X$, $|G_x| = |G| / |X|$.
(Here, $G_x$ is the isotropy subgroup for an $x$ in $X$)
Been working on this for a while.
I know that if a group $G$ acts transitively on group $X$, then for every $x,y$ in $X$, there exists an $a$ in $G$ such that $ax=y$.
I also recognize that $|G_x| = |G| / |X|$ looks a lot like Lagrange's Theorem (If $K$ is a subgroup of a finite group $G$, then the order of $K$ divides the order of $G$. In particular, $|G| = |K| [G:K]$, or in other words, $|G|/|K| = [G:K]$ or $|G|/[G:K] = |K|$).
(Here, $[G:K]$ is the index of $K$ and $G$ and defined by the number of distinct right cosets of $K$ in $G$)
What I have done so far:
- I know that $G_x \lt G$ by definition.
- Next, I think I need to show that $[G:G_x]=|X|$ somehow.
- Finally use Lagrange's Theorem to say that $|G_x|=|G|/|G:G_x|$ and since $[G:G_x]=|X|$, $|G_x|=|G|/|X|$ and thus proven.
What I need help on is how can I show that $[G:G_x]=|X|$?