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I have seen a statement: let $S_n$ act transitively on a set with $m$ elements. Then, $m\leq 2$ or $n \leq m$. I was able to prove it, so I believe it. However, I find this strikingly unintuitive. For instance, $S_5$ contains $S_3$, which acts transitively on $\{1,2,3\}$, so shouldn't $S_5$ also act transitively? Where is the gap in my logic here?

$\textbf{Edit:}$ Can someone give me an example of a non-trivial action of $S_5$ on $\{1,2,3\}$ that is not transitive?

user302540
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  • Can't really tell where your error is if you don't show your argument. – Tim Raczkowski Jan 04 '16 at 03:34
  • If $S_5$ acts transitively on ${1,2,3}$ it has a subgroup of index $3$. The transitive action on the cosets induces a non-trivial homomorphism $S_5 \to S_3$. This is a contradiction as the kernel can only be $A_5$ or $S_5$. – user302540 Jan 04 '16 at 03:39
  • But by the argument in my question $S_5$ acts transitively on ${1,2,3}$, so I guess I've shown mathematics is inconsistent. But I don't want mathematics to be inconsistent so someone please prove me wrong. – user302540 Jan 04 '16 at 03:41
  • (G acts on X) and (a subgroup H acts transitively on X) $\Longrightarrow $ G acts transitively on X – p Groups Jan 04 '16 at 03:49
  • @user302540: The problem is that your subgroup of index $3$ does not exist. (If such a subgroup would exist your reasoning would be right.) – Jendrik Stelzner Jan 04 '16 at 03:51
  • (new edited question:) If $\sigma$ is even permutation, define $\sigma.i=i$ for $i=1,2,3$, and if $\sigma$ is odd, define $\sigma.1=2$, $\sigma.2=1$ and $\sigma.3=3$ – p Groups Jan 04 '16 at 03:55
  • The problem is that $S_5$ will not act in the natural way on ${1,2,3}$, as the image of $3$ under $(3,4)$ is not in the set, – ahulpke Jan 04 '16 at 15:51

2 Answers2

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You would need to extend the acton of $S_3$ on $\{1, 2, 3\}$ to an action of $S_5$ on $\{1, 2, 3\}$, i.e. you would need to extend the identity $\mathrm{id} \colon S_3 \to S_3$ to a group epimorphism $p \colon S_5 \to S_3$. This is not possible.

More generally, there exists no group epimorphism $p \colon S_n \to S_m$ if $n \geq 5$ and $n \neq m > 2$ because then $A_n$ is the only proper nontrivial normal subgroup of $S_n$. We would then have $\ker p = \{1\}$, and contradicting $n \neq m$, or $\ker p = A_n$, contradicting $m \neq 2$, or $\ker p = S_n$, contradicting $m \neq 1$.

PS: As we can see from this argument there are precisely four actions of $S_5$ on $\{1,2,3\}$, classified by the group homomorphisms $p \colon S_5 \to S_3$. The case $\ker p = S_5$ corresponds to the trivial action. If $\ker p = A_5$ then $\mathrm{im} \ p \cong S_2$, so we have three such actions, corresponding to the three transpositions of $S_5$: So we fix any of the three elements and let $S_5$ swap the other two where $A_5$ does nothing.

Jendrik Stelzner
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  • I already understood one can't extend the action but I'm still confused. Can you can give me an action of $S_5$ on ${1,2,3}$ that is not transitive? – user302540 Jan 04 '16 at 03:53
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Standard argument (above) is through group action and homomorphism. Here is a little different:

suppose $S_5$ acts on $\{1,2,3\}$ transitively. Then $|S_5|=|\mathrm{Orbit}(i)|.|\mathrm{Stab}(i)|=3.|\mathrm{Stab}(i)|$, hence $\mathrm{Stab}(i)$ is subgroup of index $3$ in $S_5$, so its order is $40$. In group of order $40$, Sylow-$5$ subgroup is normal, i.e. normalizer of Sylow-$5$ subgroup in $S_5$ contains $\mathrm{Stab}(i)$. So index of normalizer in $S_5$ of Sylow-$5$ subgroup is $\leq 3$, i.e. number of Sylow-$5$ subgroups in $S_5$ is $\leq 3$, contradiction (it is $6$).

Jendrik Stelzner
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p Groups
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