Let $H$ be a normal subgroup of $G$ and assume that $G$ is acting over a set $X$. Let $c$ be some element of $X$, is there any relationship among the size of the orbit of $c$ under the action of $H$ and the size of the orbit of $c$ under the action of $G$?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
Assuming $G$ is a finite group: If $G_c$ is the stabilizer of $c$, then $$ |Gc| = |G|/|G_c| $$ and $$ |Hc| = |H|/|H\cap G_c| = |HG_c|/|G_c| $$ so $$ \frac{|Gc|}{|Hc|} = \frac{|G|}{|HG_c|}. $$
EDIT: For the infinite case asked for in the comment, we can do this: By choosing representatives $gc$ for each element of the orbit of $c$, the map $(gc,s) \mapsto gs$ establishes a one-to-one correspondence showing $$ |Gc|\cdot |G_c| = |G|. $$ Similarly (using choice), we have $$ |Hc| \cdot |G_c| = |HG_c|. $$ Then we can say $$ |Gc|\cdot |HG_c| = |Gc|\cdot |Hc|\cdot |G_c| = |Hc|\cdot |G|. $$ I don't immediately see a way around making arbitrary choices, though.
answered Aug 15, 2019 at 19:55
-
$\begingroup$ I am actually not assuming that $G$ is finite. But I was hoping the same could be concluded with cardinal operations, i.e. $|HG_{c}| | G_{c}|= |G||H_{c}|. $\endgroup$ Commented Aug 15, 2019 at 20:06