I was trying to give an answer to this very same question by means of group actions (so I think this is not a duplicate, at least as long as the answers therein do not use group actions).

My attempt: If such a $H$ exists, then there is a homomorphism from $S_5$ onto $S_4$ with kernel $H$. But then, there is an action of $S_5$ on $X:=\{1,2,3,4\}$ with kernel $H=\bigcap_{i=1}^4\operatorname{Stab(i)}$. $X$'s partitioning into orbits can be any of the following:

a) $4=1+1+1+1$,

b) $4=1+1+2$,

c) $4=2+2$, and

d) $4=4$.

By the Orbit-Stabilizer Theorem, case a) corresponds to $\operatorname{Stab(i)}=S_5, \forall i\in X$, whence $H=S_5$: contradiction; cases b) and c) correspond to $|H|=5!/2=60$: contradiction. Finally, case d) corresponds to a transitive action and hence the $4$ stabilizers are conjugate in $S_5$.

How can I conclude from here?

Edit. I've just realized that also the case $4=1+3$ must be addressed.

Edit#2. Also the cases b) and c) are not so plain as I thougth, since the subgroups of index $2$ might not be unique, in principle.

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    $\begingroup$ You’re using the action induced by $S_4$, right? That action is transitive so only case (d) arises. $\endgroup$
    – user208649
    Sep 14 '20 at 18:59
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    $\begingroup$ One thing you should note is that you need to use properties of $S_5$, other than its order (as $S_4\times\mathbb{Z}_5\twoheadrightarrow S_4$). You don't seem to be using such properties (unless I am missing something). $\endgroup$
    – user1729
    Sep 14 '20 at 20:33
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    $\begingroup$ @user1729, I see, indeed I've used so far $S_5$ as any abstract group of order $5!$, rather than a group of permutations. $\endgroup$
    – user810157
    Sep 14 '20 at 20:42
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    $\begingroup$ I'd like to know the answer to this one (couldn't find an argument scribbling on the back of an envelope). I think it's interesting because if the answer works generally, it will tell us something about when (or not) S_n can have a transitive action on n-1 items (and I think it can't, with one or two possible exceptions). $\endgroup$
    – user214962
    Sep 15 '20 at 12:45
  • $\begingroup$ @user214962 If $n\geq5$ then the only proper normal subgroup of $S_n$ is $A_n$ (see here). So yes, there are no transitive actions. $\endgroup$
    – user1729
    Sep 15 '20 at 16:41

Let me have a punt at this. Sounds like you are looking for a proof with a different flavour than the other similar post. Perhaps this will do it.

You've set it up nicely with an onto homomorphism $\phi:S_5\rightarrow S_4$ with kernel $H$. This subgroup $H$ has order 5!/4! = 5, hence it is a cyclic subgroup. Now we play hunt of the contradiction. The fact that every non-trivial member of $H$ has to have order 5 is strong and will give us something to try and contradict with.

We know $S_4$ acts transitively on the set of four elements, so if this homomorphism exists, then $S_5$ acts transitively on the set of four elements too using the induced action of $\phi$. By this I mean the action of $g\in S_5$ on {1,2,3,4} is taken to be the action of $\phi(g)$. Keep in mind this action might be non-standard, eg the cycle (123)$\in S_5$, say, might map to some totally different cycle in $S_4$.

Now $S_4$ has transpositions, so take any $g\in S_5$ mapping to one of these transpositions. So $\phi(g)$ has order 2, so $g^2$ lies in the kernel $H$. So, because $H$ was cyclic of order 5, $g^2$ has order 5 or $g^2$ is the identity. If $g^2$ has order 5 then $g$ has order 10, which is a contradiction as no element of $S_5$ has order 10. Hence $g^2$ is the identity, ie $g$ is also a transposition.

Next bit, I'd like to show that $\phi$ is injective on the transpositions. Let $g$ and $h$ be transpositions in $S_5$ with $\phi(g) = \phi(h)$. Then $gh^{-1}\in H$. Again either $gh^{-1}$ has order 5 (which it doesn't) or it is the identity, which implies $g=h$. Hence injective on the transpositions.

Now the contradiction: there are fewer transpositions in S_4 than S_5, and we've just shown $\phi$ is an injective map from the larger to the smaller set of transpositions.

Let me know if I've made a mistake or anything not clear.

  • $\begingroup$ Your intro is correct, and in fact I've posted an answer to the other similar post, in the spirit of yours here. I'll go through your answer asap. It looks promising, though. $\endgroup$
    – user810157
    Sep 15 '20 at 21:22

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