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Let $G$ be an infinite group that acts transitively on an infinite set X. If $H$ is a subgroup of $G$ of finite index, does this imply there are only finitely many $H$-orbits when $H$ acts on $X$?

A professor of mine made a remark with the following specifications: We had $G=SL_2(\mathbb{Z})$, $X=\mathbb{Q}\cup\{\infty\}$, and $H$ a congruence subgroup of $SL_2(\mathbb{Z})$. I don't know why its true in this case either. Can someone shed some light?

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up vote 3 down vote accepted

Let $Y \subset X$ be an $H$-orbit and $g_1, ... g_n$ a set of representatives of the cosets $G/H$. Then I claim that $$\bigcup_{i=1}^n g_i Y = X.$$ Can you see how the conclusion follows from this?

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Yes, thank you. This was my initial guess, but I had somehow convinced myself something was wrong. – Sid Raval Jan 26 '12 at 22:56

Write a coset decomposition $G=\cup_{i=1}^rHg_i$. Fix an element $x\in X$ and let $x_i=g_i\cdot x$ for $i=1,\ldots,k$.

Since the action is transitive, for any $y\in X$ we can find $g\in G$ such that $y=g\cdot x$. Write $g=hg_i$ for some $h\in H$ following the above decomposition.

Then $y=h\cdot x_i$, i.e. $y$ is in the $H$-orbit defined by $x_i$. By the generality of $y$ we conclude that there at at most $r=[G:H]$ orbits.

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