2
$\begingroup$

Let $G$ be a symmetric group on a finite set acting on another finite set $X$ through a natural action $\alpha:G \times X \to X$, $\alpha(g,x)=gx$. Let $x \in X$ and consider the orbit $G \cdot x := \{gx: g \in G\}$. Assuming that $|G\cdot x| <|G|$, can we always find a subgroup $H$ of $G$ such that (i) $|H|=|G\cdot x|$ and (ii) $H \cdot x = G \cdot x$?

By always, in particular I mean even if the stabilizer group $G_x$ of $x$ is not normal. Thanks!

$\endgroup$
0

1 Answer 1

10
$\begingroup$

No. Take $G=S_5$. It has a subgroup $H$ of order $4$ and index $30$. Then the group $G$ acts transitively on $G/H$ but $G$ does not have a subgroup of order $30$.

$\endgroup$

You must log in to answer this question.

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