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Aug 25, 2010 at 7:58 comment added Benoît Kloeckner @Arturo Magidin: got it. You're of course right about me mixing up quotient and subgroups, sorry. @jp: I was referring to the second comment by Jack Schmidt; if the image is not transitive, then you can simply decompose the action into orbits, so that seems no big deal to me. The point seems to be that there are a wild bunch of subgroups of $S_m$ that are isomorphic to $S_n$, but are contained into another proper subgroups. Those are not susceptible to be classified, if I am not mistaken.
Aug 24, 2010 at 17:26 comment added Arturo Magidin @Benoit: the attempt at a joke is merely that you asked about all morphisms, but it is really only the embeddings that are interesting (for $n\neq 4$ at least). Any morphism that is not an embedding will factor through $S_n/A_n$, so the image is either trivial or cyclic of order $2$, and of course those are no big deal. I am a bit puzzled by your phrasing of the comment, though: any homomorphism between symmetric groups that restricts to an embedding on $A_n$ is an embedding of $S_n$, so I don't understand your disjunction.
Aug 24, 2010 at 15:41 comment added j.p. @Benoît: I agree with the first half of your comment, but in the second half you probably mean "whose image is not transitive".
Aug 24, 2010 at 14:31 comment added Benoît Kloeckner I don't quite understand your last sentence: most homomorphism between symmetric groups are embeddings of $S_n$ or $A_n$, due to the simplicity of the latter when $n\geqslant 5$. Moreover, Jack's point is precisely that there are many embeddings whose image is not maximal, no?
Aug 24, 2010 at 14:17 history answered Arturo Magidin CC BY-SA 2.5