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I found this in example for one specific group but I think it works in general, but I can't prove why:

Let $G$ be group that acts on set $S$ (both have infintely many elements) and $G$ has only one orbit on $S$, ie. for every $s, s'\in S$ there is $g \in G$ such that $g\cdot s=s'$, ie. acts transitively on $S$. If $H<G$ is a subgroup of finite index then $H$ can have only finitely many orbits when it acts on $S$.

Short proof from the book was: "if $H$ has infinitely many orbits then $G$ couldn't be a union of finitely many $g_jH$, left cosets of H".

I tried to understand it this way:

Let suppose that $H$ has infinitely many orbits. If $s$ and $s'$ are two elements such that $Hs$ and $Hs'$ orbits are different then $gHs$ and $gHs'$ orbits are different. Which means that every left coset $gH$ has infinitely many orbits? Can it be used to prove this?

Also, I found a statement but I also don't know if it is true for groups $G$ in general or just somme specific groups:

Let G act transitively on $S$ and $H<G$ if finite index. Then the nuber of orbits of $H$ acting on $S$ is at most equal to index $[G:H]$

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  • $\begingroup$ Just a remark, if $H$ is normal, then the number of its orbits even divide $|G : H|$ as these form a system of blocks on which $G$ acts. $\endgroup$ – StefanH Jan 13 at 12:58
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Or simply as follows: Let $Hg_1,\ldots,Hg_n$ be the $n=[G:H]$ left cosets, let $s\in S$. Then $$S=Gs=\bigcup_{k=1}^nHg_ks$$ shows that $S$ is the union of (at most) $n$ $H$-orbits.

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