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Denote by $S_X$ the group of permutations of $X$, i.e. the group of bijections $f:X\to X$ with composition.
Do we want to show $[S_X : H] = S_X?$
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What is $H$? $\dots$ – Thomas Andrews Jan 21 '15 at 22:27
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Maybe $H$ is an assumed proper subgroup of $S_X$ and $[S_X,H]$ is its index. (of course that index, if it exists, is not the same kind of object as $S_X$ itself, so the "want to show" line isn't technically right.) – coffeemath Jan 21 '15 at 22:30
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The question doesn't make sense to me, somehow it's upvoted. – Zero Jan 21 '15 at 23:15
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@Zero Either $S_X$ has a subgroup of finite index, or it doesn't. So IMO the question makes sense... – coffeemath Jan 21 '15 at 23:27
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It suffices to prove there is no normal subgroup of finite index, and it follows from the Schreier-Ulam-Baer theorem that the unique maximal normal subgroup of $S_X$ is the subgroup $S_{|X|}(X)$ of permutations which move strictly fewer than $|X|$ elements. To see that $[S_X:S_{|X|}(X)]=\infty$ you could for instance check that there is an element in $S_X$ all of whose powers move $|X|$ elements.
I wouldn't be surprised to see a simpler proof though, basically using something that looks like a small fraction of the proof of Schreier-Ulam-Baer.
Sean Eberhard
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No there are other approaches. Here we just need the easier fact that $S_X$ is generated by its elements of order $p$ for any prime $p$ (for an arbitrary group this rules out proper subgroups of finite index). This is maybe a little easier than the theorem you quote (conversely the Schreier Ulam Baer theorem certainly implies it). – YCor Jan 23 '15 at 20:30