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Suppose $X$ a infinite set and $S_X$ is the permutation group of $X$. Prove that any proper subgroup of $S_X$ has infinite index.

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    $\begingroup$ It helps a lot if you say what you've already tried in proving this and where it didn't work. $\endgroup$ – qaphla Jun 24 '14 at 18:59
  • $\begingroup$ Presumably $s_x$ is the full symmetric group on $x$? $\endgroup$ – Derek Holt Jun 24 '14 at 20:57
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    $\begingroup$ @Derek Yes, it must be, since the finitary symmetric group on an infinite set admits the alternating subgroup of index two. $\endgroup$ – blue Jun 24 '14 at 21:07
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The nontrivial normal subgroups of the symmetric group on an infinite set $X$ consist of the finitary alternating and symmetric groups, together with the groups whose support is bounded above by a fixed infinite cardinality. (See, for example, Dixon & Mortimer, Permutation Groups, Thereom 8.1A.) So all proper normal subgroups have infinite index and hence the same applies to all proper subgroups.

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