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A friend recently asked me if a finite simple group acts transitively on a set, then is the set finite?

I want to say yes, since if the action is transitive, then the cardinality of the orbit of any element is the cardinality of the whole set, and the cardinality of such an orbit must divide the index of the stabilizer in the group, which is finite since the group is. (I don't think the hypothesis that the group is simple is used at all.)

My main concern is that maybe the orbit stabilizer theorem only applies when the set is already known to be finite. What's the answer here? Thanks.

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"My main concern is that maybe the orbit stabilizer theorem only applies when the set is already known to be finite." You needn't be concerned: the Orbit-Stabilizer Theorem applies to any group acting on any non-empty set. – Pete L. Clark Sep 28 '11 at 0:09
up vote 7 down vote accepted

You don’t need any fancy machinery, and you’re right that the group need not be simple.

If $G$ is finite, then every orbit $Gx$ (for $x\in X$) is finite. $G$ acts transitively on $X$ if and only if $Gx=X$ for each $x\in X$, so if $G$ acts transitively on $X$, $X$ is necessarily finite. In fact $\vert X\vert \le \vert G\vert$, since $\vert Gx\vert \le \vert G\vert$.

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Thanks, I might have been over thinking it a bit. – bdub Sep 24 '11 at 23:20

Given an element $x\in X$ and its stabilizer $G_x$, you always have a bijection (in fact, even am isomorphism of $G$-spaces) $ x^G \ni xg \leftrightarrow G_xg \in G/G_x$ between the orbit of $x$ and the the set of cosets of $G_x$. The proof only uses the equivalence $xg = xh \Leftrightarrow hg^{-1}\in G_x \Leftrightarrow G_xg=G_xh$, and hence does not depend on any assumptions about the cardinalities involved.

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