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There is a well-known morphism $S_4\to S_3$, obtained by having $S_4$ act on the three partitions of $4$ objects into $2+2$. Similarly, given any $n$, one can devise a morphism $S_n\to S_k$ for some $k$ by having $S_n$ act on the partitions of $n$ objects into $n_1+n_2+\ldots+n_\ell$. One can further endow some or all of the element of the partition with an order, for example with $\ell=1$ one has $S_n$ acting on the set of ordered sequences of size $n$, and gets the left action of $S_n$ on itself, which is a morphism $S_n\to S_{n!}$ (the largest one that is irreducible).

Is that, are something close to that, the complete list of all morphisms $S_n\to S_k$ (up to conjugacy, of course)? I assume the answer is well-known.

Edit: the question was really naive, but I would like to know if some general information are nevertheless available on the Burnside ring of $S_n$ (which encodes the permutation representations of a finite group, but this is almost all I know).

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    $\begingroup$ A map $S_n \to S_k$ is equivalent to an index $k$ subgroup of $S_n$, which is equivalent to finding a group of order $n!/k$ which acts on $n$ objects. Classifying all pairs (finite group, finite set on which it acts) is basically hopeless, and thus so is your question. One might be able to give a good classification in the case that $k$ is not too much larger than $n$. $\endgroup$
    – David E Speyer
    Commented Aug 23, 2010 at 15:17
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    $\begingroup$ The statement of the question isn't entirely clear to me (for instance, how $k$ and $\ell$ are related). Aside from that, it's important to keep in mind that for $n \geq 5$, the alternating group is a simple nonabelian group of index 2 in $S_n$. In that case homomorphisms from $S_n$ have few possible kernels. $\endgroup$
    – Jim Humphreys
    Commented Aug 23, 2010 at 18:53
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    $\begingroup$ @David Speyer: it is even worse. A map Sn to Sk is actually a multi-set of conjugacy classes of subgroups whose index totals to k. Only the "irreducible" (transitive) maps correspond to a single subgroup. $\endgroup$
    – Jack Schmidt
    Commented Aug 24, 2010 at 1:19
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    $\begingroup$ The maximal subgroups of Sk that are isomorphic to Sn are classified, and are all natural actions of Sn on some construction on n points that ends up giving k points. Maximal subgroups is a long way from arbitrary group homomorphisms, so let me know if you want references for this much more modest classification. $\endgroup$
    – Jack Schmidt
    Commented Aug 24, 2010 at 1:20
  • $\begingroup$ @David Speyer: ok, I understand my question was really naive. @Jim: $k$ is the number of partitions corresponding to the given partition of unity (for example if $n=5,\ell=2$ and $n_1=2$, $n_2=3$ then $k=10$). In particular, except for a few case like $S_4\to S_3$, $k>n$ and the kernels are trivial. @Jack Schmidt: I would like a reference for this, thanks. $\endgroup$
    – Benoît Kloeckner
    Commented Aug 24, 2010 at 6:45

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Bret Benesh and Ben Newton determined all pairs $(m,n)$ such that $S_m$ contains a maximal subgroup isomorphic to $S_n$. They are either $(n+1,n)$ with the obvious inclusion (or mapping $S_5$ into the image of a point stabilizer under the outer automorphism of $S_6$); $(\binom{n}{k},n)$, coming from the action of $S_n$ on the subsets of $k$ elements of $\{1,2,\ldots,n\}$; and $((kr)!/(r!)^k k!, kr)$ with $1\lt k,r$, with $S_{kr}$ acting on the the right cosets of a maximal subgroups of the wreath product $S_k\wr S_r$. This appears in A classification of certain maximal subgroups of symmetric groups, J. Algebra 304 (no. 2) pp. 1108-1113, MR2265507.

Bret later also determined all pairs $(m,n)$ such that $S_m$ has a maximal subgroup isomorphic to $A_n$; such that $A_m$ has a maximal subgroup isomorphic to $S_n$; and such that $A_m$ has a maximal subgroup isomorphic to $A_n$. This appears in the book Computational Group Theory and the Theory of Groups, Contemporary Mathematics 470 (L-C Kappe, R. F. Morse, and me as editors), AMS 2008; the paper is A classification of certain maximal subgroups of alternating groups, pp. 21-26, MR2478411.

As pointed out by Jack, this does exhaust all possible embeddings of $S_n$ into $S_k$ (presumably you are okay with the maps that are not embeddings...)

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  • $\begingroup$ 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? $\endgroup$
    – Benoît Kloeckner
    Commented Aug 24, 2010 at 14:31
  • $\begingroup$ @Benoît: I agree with the first half of your comment, but in the second half you probably mean "whose image is not transitive". $\endgroup$
    – j.p.
    Commented Aug 24, 2010 at 15:41
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    $\begingroup$ @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. $\endgroup$
    – Arturo Magidin
    Commented Aug 24, 2010 at 17:26
  • $\begingroup$ @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. $\endgroup$
    – Benoît Kloeckner
    Commented Aug 25, 2010 at 7:58

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