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I encountered this question but can't figure out the answer + the reasoning.

I'd also be interested in a more general answer to this question regarding homomorphisms from $S_n$ to $S_m$ for arbitrary $n,m$ if there is one. (Or at least: In which cases does such a homomorphism $f$ definitely not exist?)

2GR8
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  • There is a natural injection when $n\le m$. – lhf Feb 03 '24 at 13:25
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    The kernel would have to be trivial or $A_7$. Of course there are plenty of such maps that factor through the sign map. – lulu Feb 03 '24 at 13:26
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    I think you meant to add some restrictions to the map. Obviously, there is always the trival homomorphism, so I guess you meant "non-trivial". But do you consider maps that factor through sign to be trivial? They sure aren't very interesting. And that's really all you get in the case where you can't inject. – lulu Feb 03 '24 at 13:29
  • See also https://mathoverflow.net/questions/36452/explicit-description-of-all-morphisms-between-symmetric-groups – lhf Feb 04 '24 at 10:32

1 Answers1

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To summarize the discussion in the comments:

Assume $n≥5$ so the only normal subgroups of $S_n$ are the trivial ones and $A_n$.

It follows that maps $S_n\to S_m$ (or to any image group) are either injections, trivial, or factor through the quotient map to the group of order $2$. As $S_m$ will, generally, have many elements of order $2$, there are many maps of the last type.

For completeness, let's consider the consider the cases where $n≤4$. As before, this is just a matter of listing the non-trivial normal subgroups.

$S_2$ is abelian of order $2$, $S_3$ has only the cyclic group of order $3$ as a normal subgroup.

$S_4$ has $A_4$, of course, but it also has a normal subgroup isomorphic to $V_4$, the Klein group of order $4$. See this question for a discussion. The quotient is a copy of $S_3$ which gives us another class of homomorphisms from $S_4$ to $S_n$ (or to any image group). Of course, for $n≥3$, $S_n$ does have subgroups isomorphic to $S_3$.

lulu
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