Let $G$ be a finite abelian group acting on a finite set $X$. Let $\mathcal{O}(x)$ be the orbit of $x$: $$\mathcal{O}(x)=\{g.x\colon g\in G\}.$$ Then obviously $G$ acts transitively on $\mathcal{O}(x)$.

Question: Does $G$ contain a subgroup $H$ which acts transitively on $\mathcal{O}(x)$ with $|H|=|\mathcal{O}(x)$?

  • $\begingroup$ I proceeded as follows: consider action of $G$ on $\mathcal{O}(x)$, which is transitive. Let $G_x=Stab(x)$. Then $G/G_x$ acts transitively on $\mathcal{O}(x)$, and both have same orders. But, since a subgroup of an abelian group may not have a complement, $G_x$ may not have a complement. $\endgroup$ – Groups Oct 4 '15 at 8:41
  • $\begingroup$ Let $H:=\{g \in G \ | \ gx = x\}$ i.e. the subgroup of G which fixes x. Then you can show $|G|/|H|=|\mathcal{O}(x)|$. But this is pretty much just the orbit stabiliser theorem. I would be interested to see an answer to this too. $\endgroup$ – User0112358 Oct 4 '15 at 9:00

Let $G = \mathbb Z/4\mathbb Z$ and let $X = \{0,1\}$. Let $G$ act on $X$ via the action $$g\cdot x = g+x\pmod 2$$

Then $G$ acts transitively on $X$. However, the only subgroup of $G$ of order $2$ is $\{0,2 \pmod 4\}$, which does not act transitively on $X$.

  • $\begingroup$ nice! I didn't think of such a simple example. Thanks for it. $\endgroup$ – Groups Oct 4 '15 at 9:26

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